Jekyll2020-10-26T17:50:11-04:00https://mufan-li.github.io/feed.xmlMufan (Bill) Li{"name"=>"", "avatar"=>"Profile.jpg", "bio"=>nil, "location"=>nil, "employer"=>nil, "pubmed"=>nil, "googlescholar"=>"https://scholar.google.com/citations?user=9dSlc_cAAAAJ", "email"=>"mufan.li@mail.utoronto.ca", "researchgate"=>nil, "uri"=>nil, "bitbucket"=>nil, "codepen"=>nil, "dribbble"=>nil, "flickr"=>nil, "facebook"=>nil, "foursquare"=>nil, "github"=>"mufan-li", "google_plus"=>nil, "keybase"=>nil, "instagram"=>"mufan.li", "impactstory"=>nil, "lastfm"=>nil, "linkedin"=>"mufan-bill-li-35749833", "orcid"=>nil, "pinterest"=>nil, "soundcloud"=>nil, "stackoverflow"=>nil, "steam"=>nil, "tumblr"=>nil, "twitter"=>"mufan_li", "vine"=>nil, "weibo"=>nil, "xing"=>nil, "youtube"=>nil, "wikipedia"=>nil}mufan.li@mail.utoronto.caThe Auffinger-Chen Representation2019-02-13T00:00:00-05:002019-02-13T00:00:00-05:00https://mufan-li.github.io/auffinger_chen<p>Equivalent representation results contribute not only a connection between different concepts, but also a new set of proof techniques. Indeed, <a href="https://en.wikipedia.org/wiki/Stochastic_calculus">stochastic analysis</a> has offered a number of alternative proofs to many problems. Occasionally the proof can simplify drastically. In this post, we will discuss a particularly elegant application by Auffinger and Chen (2015), for an otherwise very difficult problem in spin glass.</p> <h2 id="the-problem-statement">The Problem Statement</h2> <p>For the sake of writing a self-contained blog post, we will not attempt to provide a description of spin glass models. Instead, we will state the problem in the most mathematically interesting form, without explaining where the quantities came from.</p> <p>Let $$\xi:[0,1]\to \mathbb{R}$$ be twice differentiable and strictly increasing and strictly convex (i.e. $$\xi', \xi'' &gt; 0$$), also let $$\zeta:[0,1] \to [0,1]$$ be a cumulative distribution function (CDF). We will consider the Parisi partial differential equation (PDE) defined as follows</p> $\begin{cases} \partial_t \Phi = \frac{ - \xi''(t) }{2} \left[ \partial_{xx} \Phi + \zeta(t) \left( \partial_x \Phi \right)^2 \right], \\ \Phi(1,x) = \log \cosh(x) \,, \end{cases}$ <p>where the time derivative is defined by the right limit for when $$\zeta(t)$$ is discontinuous.</p> <p>It is well known that we can solve this PDE backwards in time using a <a href="https://en.wikipedia.org/wiki/Burgers%27_equation#Solution_of_viscous_Burgers'_equation">Hopf-Cole transformation</a>; in fact, we will provide a sketch in a <a href="#sketch-of-strict-convexity">a later section</a>. This allows us to state an optimization objective as follows:</p> $\inf_{\zeta} \Phi_\zeta(0,x),$ <p>where we are minimizing over the set of all CDFs on $$[0,1]$$ for each $$x \in \mathbb{R}$$. Finally we can state the question as follows:</p> <p><strong>Question</strong> Does there exist a unique minimizer to the optimization problem $$\inf_{\zeta} \Phi_\zeta(0,x)$$ for each $$x\in\mathbb{R}$$?</p> <p>The main difficulty comes from the unclear dependence on $$\zeta$$, even if we can write down a closed form solution to the Parisi PDE. At the very least, it would be extremely unpleasant and tedious to work with. Additionally, we remark that the problem is already stated in a simplified form, as opposed to the original framing in spin-glass.</p> <p>Before we jump into the main results, we observe that existence of a minimizer is straight forward to prove. Since we are restricted to the domain $$[0,1]$$, any sequence of probability measures is <a href="https://en.wikipedia.org/wiki/Tightness_of_measures">tight</a>. It is then sufficient to consider any sequence of probability measures $$\{\zeta_n\}$$ that minimizes $$\Phi_\zeta(0,x)$$, and tightness implies there exist a converging subsequence such that $$\zeta_{n_k} \to \zeta^*$$ weakly, which is a minimizer of $$\Phi(0,x)$$.</p> <h2 id="the-auffinger-chen-representation">The Auffinger-Chen Representation</h2> <p>To complete the proof, it is sufficient to show $$\Phi(0,x)$$ is strictly convex in $$\zeta$$. In this section, we will use a stochastic representation to show convexity, which is the main difficulty of the problem. Readers unfamiliar with stochastic analysis can find a brief introduction in a <a href="https://mufan-li.github.io/stone_ito/">previous blog post</a>, in particular we will use <a href="https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma">Itô’s Lemma</a> in the upcoming proofs.</p> <p>We start by defining $$B_t := W_{\xi'(t)}$$, where $$\{W_t\}$$ is a standard Brownian motion. Let $$\{\mathcal{F}_t\}_{t\geq 0}$$ be $$\{W_t\}$$’s canonical filtration, and then we define a collection of processes</p> $\mathcal{D} := \left\{ (u_t)_{0 \leq t \leq 1} : u_t \text{ is adapted to } \mathcal{F}_t, |u_t| \leq 1 \right\}.$ <p>For simplicity of notation, we will write $$\sigma(t) = \sqrt{\xi''(t)}$$ for this section. At this point we will state the main result.</p> <hr /> <p><strong>Theorem (Auffinger-Chen Representation)</strong> For all $$\zeta$$ a probability distribution on $$[0,1]$$, we have the following</p> <p>\begin{equation} \begin{aligned} \Phi(0,x) = \max_{u \in \mathcal{D}} \bigg[ \mathbb{E} &amp; \Phi\left(1, x + \int_0^1 \sigma^2(s) \, \zeta(s) \, u_s \, ds + \int_0^1 \sigma(s) \, dW_s \right) \\\<br /> &amp;- \frac{1}{2} \int_0^1 \sigma^2(s) \, \zeta(s) \, \mathbb{E} u_s^2 \, ds \bigg]. \end{aligned} \end{equation}</p> <p>In particular, we have the maximizer is unique, and is given by $$u_s = \partial_x \Phi(s, x + X_s)$$, where $$X_s$$ is the strong solution of the following stochastic differential equation (SDE)</p> $dX_s = \sigma^2(s) \, \zeta(s) \, \partial_x \Phi(s, x + X_s) \, ds + \sigma(s) \, dW_s, \quad X_0 = 0.$ <hr /> <p><strong>Remark</strong> Before we begin the proof, we will observe that $$\Phi(0,x)$$’s convexity <strong>follows directly from this representation</strong>. Firstly both integral terms containing $$\zeta$$ are linear in $$\zeta$$. Since $$\Phi(1,x) = \log \cosh (x)$$ is convex in $$x$$, we have the $$\Phi$$ term is convex in $$\zeta$$. Next the expectation over the sum of two convex functions remain convex. Finally, a maximum (or supremum) over convex functions remain convex, proving the desired convexity result!</p> <p>Before we start, we will state several technical (but not difficult to prove) Lemmas. To guarantee a <a href="https://en.wikipedia.org/wiki/Stochastic_differential_equation#Existence_and_uniqueness_of_solutions">strong solution of the SDE</a>, it is sufficient to have $$\partial_x \Phi(s, x)$$ be Lipschitz in $$x$$. We will omit the proof of these results as they are not important to the main goal of this blog post. Instead we will state the following Lemma containing the desired estimates.</p> <hr /> <p><strong>Lemma (Derivative Estimates)</strong> For all $$\zeta$$ probability distributions on $$[0,1]$$, we have that</p> $|\partial_x \Phi(t, x)| \leq 1, |\partial_{xx} \Phi(t,x)| \leq 1.$ <hr /> <p>Another important result we will omit is the continuity of $$\Phi$$ in $$\zeta$$.</p> <hr /> <p><strong>Lemma (Lipschitz in $$L^1$$)</strong> For any discrete distributions $$\zeta_1, \zeta_2$$, and for all $$k \in \mathbb{N}$$, we have that</p> <p>\begin{align} \left| \Phi_{\zeta_1} - \Phi_{\zeta_2} \right| &amp;\leq \xi’‘(1) \int_0^1 |\zeta_1(t) - \zeta_2(t)| dt, <br /> \left| \partial_x^k \Phi_{\zeta_1}(t,x) - \partial_x^k \Phi_{\zeta_2}(t,x) \right| &amp;\leq c_k \, \xi’‘(1) \int_0^1 |\zeta_1(t) - \zeta_2(t)| dt. \end{align}</p> <hr /> <p>Since we can approximate any distributions in $$L^1$$ by discrete distributions, then we can extend the definition of $$\mathcal{P}(\cdot)$$ and $$\Phi(t,x)$$ to all distributions by continuity. Therefore it is sufficient to prove the result for only finitely supported distributions.</p> <p><em>proof (of the Auffinger-Chen representation):</em> The proof will be a straight forward application of Itô’s Lemma, and the results follow almost directly from invoking the Parisi PDE.</p> <p>We start with discrete $$\zeta$$, i.e $$\zeta$$ is a piecewise constant function. Let $$u \in \mathcal{D}$$, and define</p> $dX_s := \sigma^2(s) \, \zeta(s) \, u_s \, ds + \sigma(s) \, dW_s, \quad X_0 = 0,$ <p>and let $$Y_s := \Phi(s, x + X_s)$$. Then we observe that</p> $X_1 = \int_0^1 \sigma^2(s) \, \zeta(s) \, u_s \, ds + \int_0^1 \sigma(s) \, dW_s$ <p>appears exactly inside the first $$\Phi$$ term of the Auffinger-Chen representation.</p> <p>At this point we adopt concise notation and write $$\Phi := \Phi(s, x + X_s)$$, and apply Itô’s Lemma to $$Y_s$$ to get</p> $dY_s = \left[ \partial_s \Phi + \sigma^2(s) \, \zeta(s) \, u_s \, \partial_x \Phi + \frac{1}{2} \sigma^2(s) \, \partial_{xx} \Phi \right] ds + \sigma(s) \partial_x \Phi \, dW_s.$ <p>Here we note while the time derivative $$\partial_s \Phi$$ does not exist at finitely many points, we will eventually only use it in integral form. Using the Parisi PDE at points of continuity, we can make the following substitution</p> $\partial_s \Phi + \frac{1}{2} \sigma^2(s) \partial_{xx} \Phi = - \frac{1}{2} \sigma^2(s) \, \zeta(s) \, (\partial_x \Phi)^2.$ <p>We will make the substitution and complete the square to get</p> <p>\begin{equation} \begin{aligned} dY_s &amp;= \left[ \sigma^2(s) \, \zeta(s) \, u_s \, \partial_x \Phi - \frac{1}{2} \sigma^2(s) \, \zeta(s) \, (\partial_x \Phi)^2 \right] ds + \sigma(s) \, \partial_x \Phi \, dW_s <br /> &amp;= \left[ \frac{1}{2} \sigma^2(s) \, \zeta(s) \, u_s^2 - \frac{1}{2} \sigma^2(s) \, \zeta(s) \, (u_s - \partial_x \Phi )^2 \right] ds + \sigma(s) \, \partial_x \Phi \, dW_s. \end{aligned} \end{equation}</p> <p>Next we write this equation as an integral over $$[0,1]$$, and taking expectation to remove the martingale term we get</p> <p>\begin{equation} \begin{aligned} \mathbb{E} \Phi(1, x + X_1) - \Phi(0,x) =&amp; \int_0^1 \frac{1}{2} \sigma^2(s) \, \zeta(s) \, \mathbb{E} u_s^2 \, ds <br /> &amp;- \int_0^1 \frac{1}{2} \sigma^2(s) \, \zeta(s) \, \mathbb{E} (u_s - \partial_x \Phi)^2 ds. \end{aligned} \end{equation}</p> <p>Since $$\Phi, \partial_x \Phi$$ are continuous in $$\zeta$$, we can extend this equation to all $$\zeta$$. Furthermore, since the second integral is always positive, we must have the inequality</p> $\Phi(0,x) \geq \mathbb{E} \Phi(1, x + X_1) - \int_0^1 \frac{1}{2} \sigma^2(s) \, \zeta(s) \, \mathbb{E} u_s^2 \, ds,$ <p>and the inequality must be strict unless $$u_s = \partial_x \Phi$$ almost surely.</p> <p>Observe this proves the inequality of the representation. Since $$|\partial_x \Phi| \leq 1$$, we have $$u_s = \partial_x \Phi \in \mathcal{D}$$, hence achieving the equality in the representation.</p> <p>$$\tag*{\Box}$$.</p> <h2 id="sketch-of-strict-convexity">Sketch of Strict Convexity</h2> <p>At this point, the author believes the goal of the blog post is already achieved: we have demonstrated the key technique with only very basic manipulations. That being said, to complete the story, we will provide a short sketch on how to prove strict convexity - hence proving there is a unique minimizer of $$\Phi_\zeta(0,x)$$.</p> <p>We once again start with a key technical lemma.</p> <hr /> <p><strong>Lemma (Strict Convexity in $$x$$)</strong> For all $$\zeta$$ a probability distribution on $$[0,1]$$, and for all $$s \in [0,1]$$, we have</p> $\partial_{xx} \Phi(s,x) &gt; 0.$ <hr /> <p>Here we remind the reader that strict convexity in $$x$$ does not directly imply strict convexity in $$\zeta$$. We could just take this result for granted, but there is a nice proof using the Hopf-Cole transform and another stochastic representation, so why not?</p> <p><em>sketch (of Lemma):</em> Since $$\Phi(t,x)$$ is continuous in $$\zeta$$, we will only consider a discrete $$\zeta$$. Then using an appropriate time change and time reversal, we can get a new PDE</p> $\partial_t \Phi = \frac{1}{2 \widehat \zeta(t)} \partial_{xx} \Phi + \frac{1}{2} (\partial_x \Phi)^2,$ <p>with <strong>initial conditions</strong> (as opposed to terminal conditions) $$\Phi(0,x)=\log \cosh(x)$$, and $$\widehat \zeta(t) = \zeta(1 - t)$$ changed due to time reversal. To simplify the PDE, we use the Hopf-Cole transformation to substitute $$\phi = \exp\left( \frac{\Phi}{\widehat\zeta(t)} \right)$$, which leads to the simplified linear PDE</p> $\partial_t \phi = \frac{1}{2 \widehat \zeta(t)} \partial_{xx} \phi,$ <p>with initial conditions $$\phi(0,x) = \frac{1}{\widehat\zeta(t)} \log \exp\left( \frac{\log \cosh(x)}{\widehat \zeta(t)} \right) = \cosh(x)^{1/\widehat \zeta(t)}$$. Using another time change, we can also remove the $$\widehat \zeta(t)$$ above.</p> <p>Here we can use any of the reader’s favourite method: the <a href="https://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula">Feynman-Kac Representation</a>, the <a href="https://en.wikipedia.org/wiki/Kolmogorov_backward_equations_(diffusion)">Kolmogorov backward equation</a>, or the <a href="https://en.wikipedia.org/wiki/Heat_kernel">heat kernel</a> to write</p> $\phi(t,x) = \mathbb{E} \cosh( x + W_t )^{1/\widehat \zeta(t)},$ <p>where $$W_t$$ is a standard Brownian motion, and $$\widehat \zeta$$ is constant in $$[0,t)$$. At this point it is sufficient to show strict convexity for this $$t$$, since we can piece together the constant intervals later. To this end, we will write</p> $\Phi(t,x) = \frac{1}{\widehat \zeta(t)} \log \mathbb{E} \cosh(x + W_t)^{1/\widehat \zeta(t)},$ <p>and define</p> $\langle f(W_t) \rangle := \frac{ \mathbb{E} f(W_t) \cosh(x + W_t)^{1/\widehat \zeta(t)} }{ \mathbb{E} \cosh(x + W_t)^{1/\widehat \zeta(t)} } \, ,$ <p>where we observe since $$\cosh(x) &gt; 0$$ and $$\mathbb{E}\cosh(x + W_t) &lt; \infty$$, we have that $$\langle \cdot \rangle$$ defines a new probability measure. In particular, we have Jensen’s inequality.</p> <p>With this we can take the second derivative of $$\Phi$$ to get</p> <p>\begin{equation} \begin{aligned} \partial_{xx} \Phi(t,x) &amp;= \frac{-1}{\widehat \zeta(t)} \left\langle \frac{1}{\widehat \zeta(t)} \tanh(x + W_t) \right\rangle^2 <br /> &amp; \quad+ \frac{1}{\widehat \zeta(t)} \left\langle \left( \frac{1}{\widehat \zeta(t)} \tanh(x + W_t) \right)^2 + \frac{1}{\widehat \zeta(t)} \left( 1 - \tanh(x + W_t)^2 \right) \right\rangle <br /> &amp;\geq \frac{1}{\widehat \zeta(t)} \left\langle \frac{1}{\widehat \zeta(t)} \left( 1 - \tanh(x + W_t)^2 \right) \right\rangle <br /> &amp;&gt; 0 \, , \end{aligned} \end{equation}</p> <p>where we used Jensen’s inequality and the fact that $$\tanh(x)^2 &lt; 1$$.</p> <!-- Observe that since $$\widehat \zeta(t) \in [0,1]$$, the second term in the square brackets $$[\cdots]$$ are always greater than 1. Then we can use Jensen's inequality to write $$\partial_{xx} \Phi(t,x) \geq \frac{1}{\widehat \zeta(t)} \mathbb{E} \frac{1}{\widehat \zeta(t)} \log \cosh (x + W_t).$$ Since $$\log \cosh(x) > 0$$ unless $$x = 0$$, we have the right hand side must be strictly positive, hence implying $$\partial_{xx}\Phi(t,x) > 0$$. --> <hr /> <p>Finally we return to strict convexity in $$\zeta$$.</p> <p><em>sketch (of Strict Convexity in $$\zeta$$):</em> We will start by introducing quantities related to convexity. Let $$\zeta_1 \neq \zeta_2$$, and let $$\zeta = \lambda \zeta_1 + (1-\lambda) \zeta_2$$ for some $$\lambda \in (0,1)$$. Recalling $$\Phi(1,x) = \log \cosh (x)$$, and using the optimal $$u_s = \partial_x \Phi_\zeta(s, x + X_s)$$, where $$X_s$$ defined with respect to $$\zeta$$. Note this $$u_s$$ is not necessarily optimal for $$\zeta_1, \zeta_2$$.</p> <p>Since $$\log\cosh(x)$$ is convex, we can write</p> $\Phi_\zeta(0,x) \leq \lambda_1 A_1 + \lambda_2 A_2,$ <p>where each $$A_i$$ is defined as</p> <p>\begin{equation} \begin{aligned} A_i := \mathbb{E}&amp; \, \log \cosh \left( x + \int_0^1 \sigma^2(s) \, \zeta_i(s) \, u_s \, ds + \int_0^1 \sigma(s) dW_s \right) <br /> &amp; - \frac{1}{2} \int_0^1 \sigma^2(s) \, \zeta_i(s) \, \mathbb{E} u_s^2 \, ds. \end{aligned} \end{equation}</p> <p>Since $$\log \cosh(x)$$ is strictly convex, the inequality is strict unless</p> $\int_0^1 \sigma^2(s) \, \zeta_1(s) \, u_s \, ds = \int_0^1 \sigma^2(s) \, \zeta_2(s) \, u_s \, ds,$ <p>almost surely. Using the Auffinger-Chen representation, we have that $$A_i \leq \Phi_{\zeta_i}(0,x)$$. Therefore to prove the convexity is strict, it is sufficient to prove a gap in the first inequality, which is equivalent to saying that</p> $Z := \int_0^1 \sigma^2(s) \, (\zeta_1(s) - \zeta_2(s)) \, u_s \, ds$ <p>has positive variance. The variance can be computed as</p> $\text{Var}(Z) = \int_0^1 \int_0^1 \varphi(s) \, \varphi(t) \, \text{Cov}(u_s, u_t) \, ds dt,$ <p>where $$\varphi(s) = \sigma^2(s) \, (\zeta_1(s) - \zeta_2(s))$$.</p> <p>While we omit the technical details, it’s not hard to believe $$u_s = \partial_x \Phi(s, x + X_s)$$ satisfy the following SDE (from Itô’s Lemma and differentiating the Parisi PDE)</p> $du_s = \sigma(s) \partial_{xx} \Phi(s, x + X_s) dW_s.$ <p>Observing that $$u_s$$ is a martingale with independent increments, we can compute $$\text{Cov}(u_s, u_t)$$ as</p> $\text{Cov}(u_s, u_t) = \text{Var}(u_{s \wedge t}) = \int_0^{s \wedge t} \sigma^2(v) \mathbb{E} (\partial_{xx} \Phi(v, x + X_v))^2 dv,$ <p>where the last step followed from <a href="https://en.wikipedia.org/wiki/It%C3%B4_isometry">Itô’s Isometry</a>. Defining $$\tau(s) := \text{Var}(u_s)$$, we can also write $$\text{Cov}(u_s, u_t) = \tau(s) \wedge \tau(t)$$. With a bit of algebra we can derive</p> $\text{Var}(Z) = \int_0^1 \left( \int_v^1 \varphi(s) ds \right)^2 \tau'(v) dv.$ <p>Since $$\tau'(v) = \sigma^2(v) \mathbb{E} (\partial_{xx} \Phi(v, x + X_v))^2$$, the desired result follows from the fact $$\partial_{xx} \Phi &gt; 0$$.</p> <h2 id="final-comments">Final Comments</h2> <p>Recall the original problem of $$\inf_\zeta \Phi_\zeta(0,x)$$. We have shown that while the dependence structure is unclear, we are able to prove its convexity with it easily using a stochastic representation. The author would like to point out that most techniques used here are quite basic, which is surprising for an originally very difficult problem.</p> <p>The author would also like to point to a more general variational stochastic representation by <a href="https://projecteuclid.org/euclid.aop/1022855876">Boué and Dupuis (1998)</a>, perhaps more useful for other applications.</p> <p>Finally the post would not be possible without attending an excellent graduate course on spin glass taught by Dmitry Panchenko, where he has done a much better job explaining this topic. In particular, Dmitry has written an excellent book (Panchenko, 2013) with a bonus chapter covering this topic that can be found <a href="https://drive.google.com/file/d/0B6JeBUquZ5BwRFpLVjdVd3IwV1E/view?usp=drive_open">online</a>. I would also highly recommends Dmitry’s <a href="https://sites.google.com/site/panchenkomath/lecture-notes">notes on probability theory</a>, which has been in general very helpful to the author’s studies and research.</p> <h2 id="references">References</h2> <ul> <li>Auffinger, A., &amp; Chen, W. K. (2015). The Parisi formula has a unique minimizer. Communications in Mathematical Physics, 335(3), 1429-1444.</li> <li>Boué, M., &amp; Dupuis, P. (1998). A variational representation for certain functionals of Brownian motion. The Annals of Probability, 26(4), 1641-1659.</li> <li>Panchenko, D. (2013). The Sherrington-Kirkpatrick model. Springer Science &amp; Business Media.</li> </ul>{"name"=>"", "avatar"=>"Profile.jpg", "bio"=>nil, "location"=>nil, "employer"=>nil, "pubmed"=>nil, "googlescholar"=>"https://scholar.google.com/citations?user=9dSlc_cAAAAJ", "email"=>"mufan.li@mail.utoronto.ca", "researchgate"=>nil, "uri"=>nil, "bitbucket"=>nil, "codepen"=>nil, "dribbble"=>nil, "flickr"=>nil, "facebook"=>nil, "foursquare"=>nil, "github"=>"mufan-li", "google_plus"=>nil, "keybase"=>nil, "instagram"=>"mufan.li", "impactstory"=>nil, "lastfm"=>nil, "linkedin"=>"mufan-bill-li-35749833", "orcid"=>nil, "pinterest"=>nil, "soundcloud"=>nil, "stackoverflow"=>nil, "steam"=>nil, "tumblr"=>nil, "twitter"=>"mufan_li", "vine"=>nil, "weibo"=>nil, "xing"=>nil, "youtube"=>nil, "wikipedia"=>nil}mufan.li@mail.utoronto.caEquivalent representation results contribute not only a connection between different concepts, but also a new set of proof techniques. Indeed, stochastic analysis has offered a number of alternative proofs to many problems. Occasionally the proof can simplify drastically. In this post, we will discuss a particularly elegant application by Auffinger and Chen (2015), for an otherwise very difficult problem in spin glass.Stone-Weierstrass and an Alternative Proof of Itô’s Lemma2018-07-15T00:00:00-04:002018-07-15T00:00:00-04:00https://mufan-li.github.io/stone_ito<p>In a similar sense to <a href="https://en.wikipedia.org/wiki/Line_integral">line integrals</a>, <a href="https://en.wikipedia.org/wiki/Stochastic_calculus">stochastic calculus</a> extends the classical tools to working with <a href="https://en.wikipedia.org/wiki/Stochastic_process">stochastic processes</a>. One of the most elegant and useful result is the change of variable formula for stochastic integrals, commonly known as <a href="https://en.wikipedia.org/wiki/It%C3%B4%27s_lemma">Itô’s Lemma</a> (see <a href="#an-interesting-story-to-wrap-up">end of this post</a> for a discussion on Doeblin’s contribution). While this lemma is quite easy to use, the proof usually relies heavily on technical lemmas, hence difficult to develop intuition, especially for the first time reader.</p> <p>With this motivation in mind, it was quite pleasant to discover a set of excellent <a href="http://statslab.cam.ac.uk/~jpm205/teaching/lent2016/lecture_notes.pdf">lecture notes</a> by Jason Miller (2016), which contained an alternative proof built on the idea of <a href="https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem">Stone-Weierstrass Theorem</a>. We shall see that not only do we have a more interpretable proof, the technique is also generalizable beyond stochastic calculus. In particular, this blog post intends to illustrate the technique in detail through Itô’s Lemma.</p> <h2 id="a-brief-background-on-stochastic-calculus">A Brief Background on Stochastic Calculus</h2> <p>We will introduce (without too much rigour) some basic definitions and results to support the proofs in later sections. The reader need not to carefully analyze the technical details here to understand the proofs to come. Readers familiar with stochastic calculus may <a href="#overview-of-the-alternative-approach"> skip to the next section</a>.</p> <p>First we let $$(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t\geq 0}, \mathbb{P})$$ be a <a href="https://en.wikipedia.org/wiki/Probability_space">probability space</a> equipped with a <a href="https://en.wikipedia.org/wiki/Filtration_(probability_theory)">filtration</a> (also satisfying the <a href="https://en.wikipedia.org/wiki/Usual_hypotheses">usual conditions</a> to be rigorous). With this we can define several useful objects.</p> <p><strong>Definition</strong> A stochastic process $$X := \{X_t\}_{t\geq 0}$$ is said to be a <strong>martingale</strong> if</p> <p>(i) $$\forall t \geq 0$$, we have $$X_t$$ is measurable with respect to $$\mathcal{F}_t$$, denoted $$X_t \in \mathcal{F}_t$$;</p> <p>(ii) $$\forall 0 \leq s \leq t$$, we have $$\mathbb{E}[ X_t | \mathcal{F}_s ] = X_s$$ a.s.</p> <p><strong>Definition</strong> We say a random variable $$\tau:\Omega \to [0,\infty]$$ is a <strong>stopping time</strong> if $$\forall t \geq 0, \{\tau \leq t \} \in \mathcal{F}_t$$.</p> <p>An important property of stopping time is that if $$X_t$$ is a martingale and $$\tau$$ a stopping time, then $$X_{t \wedge \tau}$$ is also a martingale.</p> <p><strong>Definition</strong> Let the interval $$[0,T]$$ be partitioned using increments of $$2^{-n}$$, i.e. $$\{t_k^n\}_{k=0}^{\lceil T 2^n \rceil}$$, where $$t_k^n = k 2^{-n} \wedge T$$. Let $$X_t$$ be a continuous martingale, and $$f_t$$ be a continuous (possibly stochastic) process. We define the <strong>Itô integral</strong> as</p> <p>$\int_0^T f_t \, dX_t := \lim_{n\to\infty} \sum_{k=0}^{\lfloor T 2^n \rfloor} f_{t_k^n} (X_{t_{k+1}^n} - X_{t_k^n}),$</p> <p>if the limit converges u.c.p. (<a href="https://almostsure.wordpress.com/2009/12/22/u-c-p-convergence/">uniformly on compact intervals in probability</a> to be precise).</p> <p><strong>Remark</strong> Observe the above definition uses a <strong>left Riemann sum</strong> to define the integral, where as <em>other choices</em> will lead to <em>different</em> integrals. This is opposed to deterministic integrals, where the all choices are equivalent.</p> <p><strong>Definition</strong> Consider the same partition $$\{t_k^n\}$$ as above. Let $$M,N$$ be two continuous martingales, we define the <strong>quadratic covariation</strong> as</p> <p>$[M,N]_T := \lim_{n\to\infty} [M,N]^n_T := \lim_{n\to\infty} \sum_{k=0}^{\lfloor T 2^n \rfloor} (M_{t_{k+1}^n} - M_{t_k^n}) (N_{t_{k+1}^n} - N_{t_k^n}),$</p> <p>where the limit is also u.c.p. We also define the <strong>quadratic variation</strong> as $$[M]_T := [M,M]_T$$.</p> <!-- **Definition** A stochastic process $$X_t$$ is said to **locally bounded** if there exists a sequence $$\{S_n\}$$ of stopping times such that $$S_n \to \infty$$ a.s., and $$X_{t\wedge S_n}$$ is bounded for all $$n$$. --> <p>Several useful results are stated next.</p> <hr /> <p><strong>Proposition (Finite Variation)</strong> Let $$X,Y$$ be continuous stochastic processes such that $$X$$ has finite variation, i.e.</p> $\lim_{n\to\infty} \sum_{k=0}^{\lfloor T 2^n \rfloor} | X_{t_{k+1}^n} - X_{t_k^n} | &lt; \infty,$ <p>and $$[Y]_t &gt; 0$$ a.s. Then we have</p> $[X,Y]_t = 0 \;\text{a.s.}$ <p><strong>Proposition (Itô’s Product Rule)</strong> Let $$X,Y$$ be continuous martingales, then we have</p> $X_t Y_t - X_0 Y_0 = \int_0^t X_s dY_s + \int_0^t Y_s dX_s + [X,Y]_t \,.$ <p><strong>Proposition (Fundamental Theorem)</strong> Let $$X,Y,Z$$ be continuous martingales, then we have</p> $\int_0^t X_s d\left( \int_0^s Y_u dZ_u \right) = \int_0^t X_s Y_s dZ_s.$ <p><strong>Proposition (Kunita-Watanabe Identity)</strong> Let $$X,Y,Z$$ be continuous martingales, then we have</p> $\left[ \int_0 X_s dY_s, Z \right]_t = \int_0^t X_s d[Y,Z]_s,$ <p>where both uses of $$[\;,\;]_t$$ denotes the covariation.</p> <p><strong>Proposition (Itô’s Isometry)</strong></p> <p>Let $$M$$ be a continuous martingale, and $$H$$ be a continuous stochastic process. Then we have</p> $\mathbb{E} \left[ \left( \int_0^t H_s dM_s \right)^2 \right] = \mathbb{E} \int_0^t H_s^2 d[M]_s.$ <hr /> <h2 id="the-lemma-and-the-classical-approach">The Lemma and the Classical Approach</h2> <p>For the purpose of the blog post, we will only state and prove a much simpler version of the lemma, but it is not difficult to adapt to more general conditions.</p> <hr /> <p><strong>Theorem (Itô’s Lemma)</strong> Let $$X_t$$ be a continuous martingale, and $$f \in C^2(\mathbb{R})$$. Then we have</p> <p>$f(X_t) = f(X_0) + \int_0^t \frac{\partial f}{\partial x}(X_s) dX_s + \frac{1}{2} \int_0^t \frac{\partial^2 f}{\partial x^2} (X_s) d[X]_s.$</p> <hr /> <p>Here we will sketch the proof from Karatzas and Shreve (1991).</p> <p><em>proof sketch:</em> We start by defining a stopping time $$\tau_r := \inf \{t \geq 0 : |X_t| + [X]_t &gt; r\}$$, and replace $$X_t$$ with $$X_{t \wedge \tau_r}$$. This <em>localization</em> technique will allow us to only consider the function $$f$$ in the interval $$B_r := [-r, r]$$ (or a ball in higher dimensions), which has bounded derivatives.</p> <p>By observing the lemma’s statement, the reader may notice the formula appears like the second order Taylor expansion of $$f(X_t)$$. Indeed we can write</p> \begin{align*} f(X_t) - f(X_0) =&amp; \lim_{n\to\infty} \sum_{k=0}^{\lfloor t 2^n \rfloor} f(X_{t_{k+1}^n}) - f(X_{t_{k}^n}) \\ =&amp; \lim_{n\to\infty} \sum_{k=0}^{\lfloor t 2^n \rfloor} \Big\{ \frac{\partial f}{\partial x}(X_{t_{k}^n}) [X_{t_{k+1}^n} - X_{t_{k}^n}] \\ &amp;+ \frac{1}{2} \frac{\partial^2 f}{\partial x^2} (\eta_k^n) [X_{t_{k+1}^n} - X_{t_{k}^n}]^2 \Big\}, \end{align*} <p>where $$\eta_k^n \in [X_{t_{k}^n}, X_{t_{k+1}^n}]$$ is chosen as part of Taylor’s theorem to satisfy the above equality. It’s not difficult to see the first sum converges to the first stochastic integral, then it remains to show the second term converges.</p> <p>To this goal, we will define</p> \begin{align*} J_1^n &amp;:= \sum_{k=0}^{\lfloor t 2^n \rfloor} \frac{\partial^2 f}{\partial x^2} (\eta_k^n) [X_{t_{k+1}^n} - X_{t_{k}^n}]^2, \\ J_2^n &amp;:= \sum_{k=0}^{\lfloor t 2^n \rfloor} \frac{\partial^2 f}{\partial x^2} (X_{t_{k}^n}) [X_{t_{k+1}^n} - X_{t_{k}^n}]^2, \\ J_3^n &amp;:= \sum_{k=0}^{\lfloor t 2^n \rfloor} \frac{\partial^2 f}{\partial x^2} (X_{t_{k}^n}) \{ [X]_{t_{k+1}^n} - [X]_{t_{k}^n} \}, \end{align*} <p>where observe $$J_3^n$$ converges to the desired integral. Next we will use the following technical inequality. Let $$|X_s| \leq K &lt; \infty, \forall s \leq T$$ be a martingale, then we have</p> $\mathbb{E} ([X]^n_T)^2 \leq 6 K^4.$ <p>Without stating the details, using this and Cauchy-Schwarz inequality, we can show</p> $\lim_{n\to\infty} |J_1^n - J_2^n| = 0 \; \text{a.s.}$ <p>To complete the proof, we will need one more technical lemma. Let $$|X_s| \leq K &lt; \infty, \forall s \leq T$$, then we have</p> $\lim_{n\to\infty} \mathbb{E} \sum_{k=0}^{\lfloor t 2^n \rfloor} [ X_{t_{k+1}^n} - X_{t_k^n} ]^4 = 0.$ <p>Then once again omitting the details, we can get</p> $\mathbb{E} |J_2^n - J_3^n| \leq 2 \sup_{x \in B_r} \left| \frac{\partial^2 f}{\partial x^2}(x) \right|^2 \mathbb{E} \left[ \sum_{k=0}^{\lfloor t 2^n \rfloor} [ X_{t_{k+1}^n} - X_{t_k^n} ]^4 + [X]_t \max_{k} ( [X]_{t_{k+1}^n} - [X]_{t_{k}^n} ) \right],$ <p>which combined with the previous lemma and bounded convergence theorem, we get the desired result</p> $\lim_{n\to\infty} |J_2^n - J_3^n| = 0 \; \text{a.s.}$ <p>Putting everything together gives us the desired formula as stated.</p> <p><strong>Remark</strong> The use of the propositions listed in the previous section is implicit in the two technical lemmas we stated above, where we also hide most of the proof difficulty in.</p> <p><strong>Interpretation</strong> This proof naturally leads to an interpretation that Itô’s Lemma as a consequence of Taylor’s expansion. However this proof provides no clear intuition on why the second order approximation is the correct order, and pushes the justification to complicated technical details. Probably the most troubling consequence is that a different integration scheme (e.g. <a href="https://en.wikipedia.org/wiki/Stratonovich_integral">Stratonovich</a> which rises from a mid-point Riemann sum) leads to a different change of variable formula, therefore the Taylor expansion intuition can lead to further confusion.</p> <h2 id="overview-of-the-alternative-approach">Overview of the Alternative Approach</h2> <p>At this point, we will first take a step back from Itô’s Lemma and look at a rough sketch of the proof technique.</p> <p>Suppose we want to prove a collection of functions (e.g. $$C^2([a,b])$$) satisfy a certain property $$(P)$$, we will start by defining $$\mathcal{A}$$ as the subset of $$C^2([a,b])$$ that satisfies the desired property $$(P)$$.</p> <p>(Step 1) We will identify a certain algebraic structure such that $$\mathcal{A}$$ is closed under, e.g. for an <a href="https://en.wikipedia.org/wiki/Algebra_over_a_field">algebra (over a field)</a> we have if $$f,g \in \mathcal{A}$$, then $$cf + g, fg \in \mathcal{A}$$. In other words, an algebra is a vector space with an associative vector multiplication.</p> <p>(Step 2) Then we can say that the collection $$\mathcal{A}$$ (or a dense subset) is generated by some very simple functions, e.g. under an algebra, the functions $$\{1, x\}$$ generate the entire collection of polynomials.</p> <p>(Step 3) At this point, we use a density argument such as Weierstrass approximation to show $$\mathcal{A}$$ is dense in $$C^2([a,b])$$. Specifically, $$\forall f \in C^2([a,b])$$, $$\exists \{f_n\}_{n \geq 1} \subset \mathcal{A}$$ such that $$f_n \to f$$ with respect to some metric $$\rho$$.</p> <p>(Step 4) Finally, it is sufficient to show $$\mathcal{A}$$ is closed under this metric $$\rho$$. I.e. if $$\{f_n\}_{n \geq 1}$$ all satisfy $$(P)$$ are such that $$f_n \to f$$ in $$\rho$$, then we have $$f$$ also satisfies $$(P)$$, hence $$f \in \mathcal{A}$$.</p> <p><strong>Remark</strong> The reader may already recognize that the sketch above was intentionally phrased in a very general sense, so we can observe the flexibility of the technique. In fact we can even generalize beyond function spaces, as long as we have an equivalent approximation technique.</p> <h2 id="the-proof-in-detail">The Proof in Detail</h2> <p>We start by stating the key theorem.</p> <hr /> <p><strong>Theorem (Stone-Weierstrass, Real Numbers)</strong> Let $$S$$ be a compact <a href="https://en.wikipedia.org/wiki/Hausdorff_space">Hausdorff space</a>, and $$\mathcal{A} \subset C(S, \mathbb{R})$$ an algebra which contains a non-zero constant function. Then $$\mathcal{A}$$ is dense in $$C(S, \mathbb{R})$$ if and only if it separates points.</p> <hr /> <p>Clearly, if we let $$S = B_r$$, we have a compact Hausdorff space, and the collections of polynomials contains the functions $$\{1,x\}$$ and separates points. Therefore we have $$\mathcal{A}$$ is dense in $$C(B_r, \mathbb{R}), \forall r &gt; 0$$ with respect to the sup-norm.</p> <p>Applying the same theorem to the derivatives, we then have the same result for $$C^2(B_r, \mathbb{R})$$ with respect to a similar norm</p> $\| f \|_{B_r} := \sup_{x \in B_r, \, m = 0,1,2} \left| \frac{\partial^m f}{\partial x^m} (x) \right|.$ <p><em>proof (of Itô’s Lemma):</em> We will similarly use a localization argument, i.e. define $$\tau_r := \inf \{t \geq 0 : |X_t| + [X]_t &gt; r \}$$, and replace $$X_t$$ with $$X_{t \wedge \tau_r}$$.</p> <p>(Step 1, 2) Let $$\mathcal{A} \subset C^2(\mathbb{R})$$ be the collection of functions where Itô’s Lemma is satisfied. Trivially we have that $$\{1,x\}$$ are in $$\mathcal{A}$$, and $$\mathcal{A}$$ forms a vector space.</p> <p>Next we show that $$\mathcal{A}$$ forms an algebra. In particular, suppose $$f,g \in \mathcal{A}$$, and define $$F_t := f(X_t), G_t := g(X_t)$$. Using the product rule gives us</p> $F_t G_t - F_0 G_0 = \int_0^t F_s dG_s + \int_0^t G_s dF_s + [F,G]_t \,.$ <p>Using the Fundamental Theorem and Itô’s Lemma on $$g$$, we get</p> $\int_0^t F_s dG_s = \int_0^t f(X_s) \frac{\partial g}{\partial x}(X_s) dX_s + \frac{1}{2} \int_0^t f(X_s) \frac{\partial^2 g}{\partial x^2}(X_s) d[X]_s \,.$ <p>and observe the same is true switching the order of $$F,G$$. Next we use Itô’s Lemma and expand with the Kunita-Watanabe identity to get</p> $[F,G]_t = \int_0^t \frac{\partial f}{\partial x}(X_s) \frac{\partial g}{\partial x}(X_s) d[X]_s \, ,$ <p>where the extra terms are zero because the covariation with one finite variation process is zero, i.e. $$[ \,[X]\, ,Y ]_t = 0$$ as $$[X]_t$$ has finite variation. By grouping the integrals by the integrators (e.g. $$d[X]_t$$), we get that $$fg$$ satisfies Itô’s Lemma or simply $$fg \in \mathcal{A}$$.</p> <p>(Step 3) Here we can apply <em>the Stone-Weierstrass Theorem</em> to get that $$\mathcal{A}$$ is dense in $$C^2(B_r)$$ with respect to the norm $$\|\cdot\|_{B_r}$$.</p> <p>(Step 4) It remains to show that $$\mathcal{A}$$ is closed with respect to $$\|\cdot\|_{B_r}$$. In particular, let $$(f_n)_{n \geq 1}$$ be a sequence in $$\mathcal{A}$$ such that $$f_n \to f$$ in $$\|\cdot\|_{B_r}$$. Then we have</p> $\int_0^t \left| \frac{\partial^2 f_n}{\partial x^2}(X_s) - \frac{\partial^2 f}{\partial x^2}(X_s) \right| d[X]_s \leq \|f_n - f\|_{B_r} [X]_t \, .$ <p>At the same time, we also have by Itô’s Isometry</p> \begin{align*} \mathbb{E} \left( \int_0^t \frac{\partial f_n}{\partial x}(X_s) - \frac{\partial f}{\partial x}(X_s) dX_s \right)^2 &amp;= \mathbb{E} \int_0^t \left(\frac{\partial f_n}{\partial x}(X_s) - \frac{\partial f}{\partial x}(X_s) \right)^2 d[X]_s \\ &amp;\leq \|f_n - f\|_{B_r} [X]_t \, . \end{align*} <p>Since the process is localized we have that $$[M]_t \leq r$$, and therefore we can pass the limit in the Itô formula and get</p> \begin{align*} f(X_t) - f(X_0) &amp;= \lim_{n\to\infty} f_n(X_t) - f_n(X_0) \\ &amp;= \lim_{n\to\infty} \int_0^t \frac{\partial f_n}{\partial x}(X_s) dX_s + \frac{1}{2} \int_0^t \frac{\partial^2 f_n}{\partial x^2}(X_s) d[X]_s \\ &amp;= \int_0^t \frac{\partial f}{\partial x}(X_s) dX_s + \frac{1}{2} \int_0^t \frac{\partial^2 f}{\partial x^2}(X_s) d[X]_s \,. \end{align*} <p>Finally, since Itô’s Lemma hold for all $$r&gt;0$$, we can simply take $$r\to\infty$$ to complete the proof.</p> $\tag*{\Box}$ <p><strong>Remark</strong> Clearly the alternative proof is <em>not necessarily easier</em>, however let us observe a couple of advantages.</p> <p>Firstly, none of the steps above were very complicated, as most steps followed directly from useful (and well known) propositions. Notably, a first time reader of this subject will have a much easier time following the steps and seeing the bigger picture, rather than getting trapped by technical details.</p> <p>Secondly, we now have an additional interpretation of the second integral in the formula, which clearly arises as a consequence of Itô’s product rule and Kunita-Watanabe identity. For the readers that have not seen the proof, it follows almost directly from the definition, i.e. a direct consequence of choosing <em>the left Riemann sum</em>.</p> <!-- ## Another Application Since we have already seen a detailed proof, for this section we shall leave out the technical details, and only sketch the technique on a high level. In fact the complete proofs for this section are much more difficult, therefore we direct the rigorous reader to Dudley (2002). Suppose we are interested in proving the following well known theorem in probability theory. --- **Theorem (Prohorov)** Let a sequence of probability measures $$\{\mathbb{P}_n\}_{n\geq 1}$$ be [uniformly tight](https://en.wikipedia.org/wiki/Tightness_of_measures), i.e. $$\forall \epsilon > 0, \exists K \subset \Omega$$ compact such that $$\mathbb{P}_n(K) \geq 1 - \epsilon, \forall n \in \mathbb{N}.$$ Then there exists a subsequence such that $$\mathbb{P}_{n_k} \to \mathbb{P}$$ weakly, where $$\mathbb{P}$$ is a tight Borel probability measure. --- Here we can define $$\forall f \in C_b(S)$$ (i.e. continuous and bounded) $$I(f) := \lim_{n\to\infty} \int f d\mathbb{P}_n.$$ It is not too difficult to show that $$I$$ is a continuous linear functional on $$C_b(S)$$, and it follows from [Riesz Representation Theorem](https://en.wikipedia.org/wiki/Riesz%E2%80%93Markov%E2%80%93Kakutani_representation_theorem) that there exists a regular Borel measure $$\mathbb{P}$$ such that $$I(f) = \int f d\mathbb{P}.$$ Then it remains to show that $$\mathbb{P}$$ is a probability measure. A curious reader may already suspect that there exists a related version of Riesz Representation specific to probability measures, indeed we will investigate such a theorem - and of course using the same Stone-Weierstrass type technique. (Step 1) We start by defining a slightly different algebraic structure. **Definition** A collection of functions $$\mathcal{A} \subset \{f: S \to \mathbb{R}\}$$ is called a **Stone vector lattice** if $$\forall f,g \in \mathcal{A}$$ we have (i) $$cf + g \in \mathcal{A},$$ (ii) $$f\wedge g, f\vee g \in \mathcal{A},$$ (iii) $$f\wedge 1 \in \mathcal{A}.$$ Now we are ready to state the main theorem. --- **Theorem (Stone-Daniell)** If $$\mathcal{A}$$ is a Stone vector lattice and $$I:\mathcal{A} \to \mathbb{R}$$ is a continuous linear functional such that $$f \geq 0 \implies I(f) \geq 0.$$ Then exists a unique measure $$\mu$$ on the minimal $$\sigma$$-algebra such that all elements of $$\mathcal{A}$$ are measurable, so that $$\forall f \in \mathcal{A}$$ we have $$I(f) = \int f d\mu.$$ --- Before we dive into the technique, observe that Stone-Daniell Theorem applies to all functions, where as Prohorov's Theorem only needs $$C_b(S)$$. Therefore we are taking a "detour" trying to prove a much harder result. The author suspects there may be a simpler more direct approach, perhaps addressed in a future post. Next we provide a brief sketch that *skips most of the details* unrelated to the Stone-Weierstrass type technique. To start we define (not quite a measure) for $$f \leq g$$ in $$\mathcal{A}$$ $$\nu([f,g)) := I(g - f),$$ and eventually define the desired measure as $$\mu(A) := \nu([0, 1_A)).$$ At this point, we just need to check $$\mu$$ satisfies all the property we desire in the statement of the theorem. (Step 2,3) Intuitively, constructing a Lebesgue integral only requires approximation by simple functions; similarly for this proof, it turns out we only need to approximate indicators functions from sets of the type $$f^{-1}((1, \infty))$$. To this goal, we can define $$g_n := [ n( f - f \wedge 1) ] \wedge 1,$$ which is also contained in $$\mathcal{A}$$ and $$g_n \to 1_{f^{-1}((1, \infty))}$$. (Step 4) As for closure of $$\mathcal{A}$$, it is even easier as we have all the limit theorems (e.g. monotone convergence) to work with. By these means, (the author promises) we can construct the integral $$I(f) = \int f d\mu$$ with the desired properties. Finally, to complete the proof of Prohorov's Theorem, the reader may recognize that we can indeed recover a regular Borel probability measure by testing the case $$f = 1$$ and showing that $$I(f) = 1$$. --> <h2 id="summary">Summary</h2> <p>We have shown the Stone-Weierstrass Theorem is not only a strong result on its own, but leads to a powerful technique in general. In particular, we saw a nice alternative proof of Itô’s Lemma with much better interpretations. <!-- At this point, the author will have to apologize for the length of the blog post; this turned out to be much longer than intended. --> Ideally, the author would have liked to add another example, but the post is already quite long at this point. Hopefully the readers will still have enjoyed an interesting blog post, and added another proof technique in their arsenal.</p> <p>Please comment below (new feature!) for any questions or feedback!</p> <h2 id="an-interesting-story-to-wrap-up">An Interesting Story to Wrap Up</h2> <p>For the longest time, the lemma was credited to <a href="https://en.wikipedia.org/wiki/Kiyosi_It%C3%B4">Kiyosi Itô</a> alone in his <a href="https://projecteuclid.org/euclid.nmj/1118764702">1950 paper</a>. This was until the 1990s with a resurgence of interests in the late French-German mathematician <a href="https://en.wikipedia.org/wiki/Wolfgang_Doeblin">Wolfgang Doeblin</a>, who was well known to be quite gifted. The interests led to a demand to open the remaining “pli cacheté” (sealed envelope) held by the French Academy of Sciences, which he submitted just before he passed away in 1940 - he burned his notes and took his own life so the German soldiers cannot take advantage of his work. To everyone’s surprise, Doeblin’s letter contained significant research progress ahead of his time, including a statement of the same change of variables formula! To honour his contribution, the result is sometimes referred to as the Itô-Doeblin Lemma.</p> <p>For the interested readers, I would strongly recommend an excellent <a href="https://link.springer.com/article/10.1007/s780-002-8399-0">commentary</a> by Bernard Bru and Marc Yor (2002) for further details on this topic.</p> <h2 id="references">References</h2> <ul> <li>Bru, B. &amp; Yor, M. (2002). Comments on the life and mathematical legacy of Wolfgang Doeblin.. Finance and Stochastics, 6, 3-47. <!-- - Dudley, R.M. (2002). Real Analysis and Probability. Cambridge University Press --></li> <li>Karatzas, I. &amp; Shreve, S.E. (1991). Brownian Motion and Stochastic Calculus. Springer New York</li> <li>Miller, J. (2016). Stochastic Calculus, Lent 2016 Lecture Notes. Retrieved from http://statslab.cam.ac.uk/~jpm205/teaching/lent2016/lecture_notes.pdf</li> </ul>{"name"=>"", "avatar"=>"Profile.jpg", "bio"=>nil, "location"=>nil, "employer"=>nil, "pubmed"=>nil, "googlescholar"=>"https://scholar.google.com/citations?user=9dSlc_cAAAAJ", "email"=>"mufan.li@mail.utoronto.ca", "researchgate"=>nil, "uri"=>nil, "bitbucket"=>nil, "codepen"=>nil, "dribbble"=>nil, "flickr"=>nil, "facebook"=>nil, "foursquare"=>nil, "github"=>"mufan-li", "google_plus"=>nil, "keybase"=>nil, "instagram"=>"mufan.li", "impactstory"=>nil, "lastfm"=>nil, "linkedin"=>"mufan-bill-li-35749833", "orcid"=>nil, "pinterest"=>nil, "soundcloud"=>nil, "stackoverflow"=>nil, "steam"=>nil, "tumblr"=>nil, "twitter"=>"mufan_li", "vine"=>nil, "weibo"=>nil, "xing"=>nil, "youtube"=>nil, "wikipedia"=>nil}mufan.li@mail.utoronto.caIn a similar sense to line integrals, stochastic calculus extends the classical tools to working with stochastic processes. One of the most elegant and useful result is the change of variable formula for stochastic integrals, commonly known as Itô’s Lemma (see end of this post for a discussion on Doeblin’s contribution). While this lemma is quite easy to use, the proof usually relies heavily on technical lemmas, hence difficult to develop intuition, especially for the first time reader.Connected by Poincaré Inequality2017-12-30T00:00:00-05:002017-12-30T00:00:00-05:00https://mufan-li.github.io/poincare_inequality<p>While studying two seemingly irrelevant subjects, probability theory and partial differential equations (PDEs), I ran into a somewhat surprising overlap: the <a href="https://en.wikipedia.org/wiki/Poincar%C3%A9_inequality">Poincaré inequality</a>. On one hand, it is not out of the ordinary for analysis based subjects to share inequalities such as <a href="https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality">Cauchy-Schwarz</a> and <a href="https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality">Hölder</a>; on the other hand, the two forms of Poincaré inequality have quite different applications.</p> <p>In this blog post, I hope to put together some excellent content I studied recently, specifically from:</p> <ul> <li><em>Concentration Inequalities</em> by Boucheron, Lugosi, and Massart (2013)</li> <li><em>Partial Differential Equations</em> by Evans (2010) <!-- - *Functional Analysis, Sobolev Spaces, and Partial Differential Equations* by Haim Brezis (2011) --></li> </ul> <!-- Both books are great references, I would strongly recommend both for an intuitive yet rigorous read in their respective topics. --> <h2 id="a-simple-inequality">A Simple Inequality</h2> <p>We first state a very simple version of the inequality:</p> <hr /> <p><strong>Theorem (A Simple Poincaré Inequality)</strong> Let $$\Omega \subset \mathbb{R}^n$$ be open and bounded, and let $$f \in C^1_c(\Omega)$$ (differentiable with compact support). Then there exists a constant $$C$$ that depends only on $$\Omega$$ such that:</p> <p>$\left\lVert f \right\rVert_{L^2(\Omega)} \leq C \lVert \nabla f \rVert_{L^2(\Omega)}$</p> <!-- where $$\overline{f}$$ is the average of the function $$f$$ in domain $$\Omega$$. --> <hr /> <p>Quick aside: we say a function $$f$$ has <strong>compact support</strong> if the set $$S = \{ x \in \Omega : f(x) \neq 0 \}$$ has compact closure. This implies $$f(x) = 0$$ near the boundary.</p> <p>Observe that the inequality simply bounds the $$L^2$$-norm of a function in terms of the $$L^2$$-norm of its gradient instead. Note the compact support here is an important assumption when we are integrating with respect to the Lebesgue measure. Consider for example a constant function, then this inequality would fail as the gradient is zero. The reader may be comforted that a general form will require much fewer assumptions, and can be generalized to all $$L^p$$ norms.</p> <p>The reason we start with this inequality is because the proof is quite straightforward:</p> <p><em>proof (of the Simple Poincaré Inequality):</em></p> <p>Without loss of generality, we let $$\Omega \subset [0,M]^n$$ for some large $$M &gt; 0$$, and by the Cauchy-Schwarz inequality we have</p> <p>$\vert f(x) \vert^2 \leq \left\vert \int_0^{x_1} \frac{\partial }{\partial x_i} f(y_1, x_2, \ldots) dy_1 \right\vert^2 \leq \left[ \int_0^M 1^2 dy_1 \right] \left[ \int_0^M \left\vert \frac{\partial f}{\partial x_1} \right\vert^2 dy_1 \right]$</p> <p>Summing over all $$n$$ possible derivatives, and integrating over $$\Omega$$ we have</p> <p>$n \int_\Omega \vert f(x) \vert^2 \leq \int_\Omega \sum_{i=1}^n M \int_0^M \left\vert \frac{\partial f}{\partial x_i} \right\vert^2 dy_i = \sum_{i=1}^n M \left\lVert \frac{\partial f}{\partial x_i} \right\rVert^2_{L^2(\Omega)}$</p> <p>where in the last step we exchanged the order of integration, and used the fact that the $$L^2$$ norm is a constant. Rewriting the above we get the desired result</p> <p>$\lVert f \rVert_{L^2(\Omega)} \leq \frac{M}{\sqrt{n}} \lVert \nabla f \rVert_{L^2(\Omega)}$</p> $\tag*{\Box}$ <h2 id="as-a-concentration-inequality">As a Concentration Inequality</h2> <p>We now state the inequality in a form most useful for probability theory, see Theorem 3.20 from Boucheron, Lugosi, Massart (2013):</p> <hr /> <p><strong>Theorem (Gaussian-Poincaré Inequality)</strong> Let $$X = (X_1, \ldots, X_n)$$ be a vector of i.i.d. standard Gaussian random variables. Let $$\, f : \mathbb{R}^n \to \mathbb{R}$$ be any continuously differentiable function. Then</p> <p>$\text{Var}[f(X)] \leq \mathbb{E}\left[ | \nabla f(X)|^2 \right]$</p> <hr /> <p>Observe that the inequality is slightly different. Firstly this time the norm is centered, although centering in this case is not an issue since $$Var[X] \leq \mathbb{E}X^2$$. Secondly due to the measure being a probability measure, we have a much smaller constant on the inequality $$C=1$$. In combination, we were also able to drop the compact support assumption.</p> <p>An immediate consequence is to consider $$f$$ Lipschitz with coefficient $$1$$, i.e. $$| f(x) - f(y) | \leq \|x - y\|$$, then we have</p> <p>$\text{Var}[f(X)] \leq 1$</p> <p>In other words, we just found a constant bound on the variance for a huge class of random functions! In general, we can consider $$f$$ to be a smooth estimator based on a dataset with noise $$X$$. The Poincaré inequality will provide a very useful bound on estimation error. <!-- If this doesn't excite you, I don't know what does :) --></p> <p>To prove this inequality, we will use a famous result from 1981 (Theorem 3.1 in Boucheron, Lugosi, Massart (2013)):</p> <hr /> <p><strong>Theorem (Efron-Stein Inequality)</strong> Let $$X = (X_1, \ldots, X_n)$$ be a vector of i.i.d. random variables and let $$Z = f(X)$$ be a square-integrable function of $$X$$. Then</p> <p>$\text{Var}(Z) \leq \sum_{i=1}^n \mathbb{E} \left[ \left( Z - \mathbb{E}^{(i)}Z \right)^2 \right]$</p> <p>where $$\mathbb{E}^{(i)}Z = \int f(X_1, \ldots, X_{i-1}, x_i, X_{i+1},\ldots) d\mu_i(x_i)$$, i.e. the expectation over $$X_i$$ only.</p> <hr /> <p>The Efron-Stein inequality can be proved by decomposing the variance as a sum of telescoping differences of conditional expectations, and applying Jensen’s inequality to the individual terms. While we omit the proof here, we should remark that the simple Efron-Stein inequality has wide ranging applications; we will only look at one such use for the proof of the Poincaré inequality, taken from Theorem 3.20 in Boucheron, Lugosi, Massart (2013):</p> <p><em>proof (of Gaussian-Poincaré Inequality):</em></p> <p>First we observe that a direct application of the Efron-Stein inequality can reduce the problem down to $$n=1$$, i.e. it is sufficient to show</p> <p>$\mathbb{E}^{(i)} \left[ \left( Z - \mathbb{E}^{(i)}Z \right)^2 \right] \leq \mathbb{E}^{(i)} \frac{\partial f}{\partial x_i}(X)^2$</p> <p>From here we assume without loss of generality $$n=1$$. Then we notice that it is sufficient to prove this inequality for compactly supported, twice differentiable functions, i.e. $$f \in C_c^2(\mathbb{R})$$, since otherwise we can just take a limit to the original function.</p> <p>Here we let $$\epsilon_1,\ldots,\epsilon_n$$ be i.i.d. Rademacher random variables, i.e. $$\mathbb{P}[\epsilon_j = 1] = \mathbb{P}[\epsilon_j = -1] = \frac{1}{2} \,\forall j \in \{ 1,2,\ldots,n \}$$, and we define</p> <p>$S_n = n^{-1/2} \sum_{j=1}^n \epsilon_j$</p> <p>Observe that for every $$i$$ we have</p> <p>$\text{Var}^{(i)}[f(S_n)] = \frac{1}{4} \left[ f\left( S_n + \frac{1-\epsilon_i}{\sqrt{n}} \right) - f\left( S_n + \frac{1+\epsilon_i}{\sqrt{n}} \right) \right]^2$</p> <p>Applying the Efron-Stein inequality, we get</p> <p>$\text{Var}[f(S_n)] \leq \frac{1}{4} \sum_{i=1}^n \mathbb{E} \left[ \left( f\left( S_n + \frac{1-\epsilon_i}{\sqrt{n}} \right) - f\left( S_n + \frac{1+\epsilon_i}{\sqrt{n}} \right) \right)^2 \right]$</p> <p>Let $$K = \sup_x \vert f''(x) \vert$$, then we have that</p> <p>$\left|f\left( S_n + \frac{1-\epsilon_i}{\sqrt{n}} \right) - f\left( S_n + \frac{1+\epsilon_i}{\sqrt{n}} \right)\right| \leq \frac{2}{\sqrt{n}} |f’(S_n)| + \frac{2K}{n}$</p> <p>which implies</p> <p>$\frac{n}{4} \left( f\left( S_n + \frac{1-\epsilon_i}{\sqrt{n}} \right) - f\left( S_n + \frac{1+\epsilon_i}{\sqrt{n}} \right) \right)^2 \leq f’(S_n)^2 + \frac{2K}{\sqrt{n}} | f’(S_n) | + \frac{K^2}{n}$</p> <p>Finally the central limit theorem then imply the desired result</p> <p>$\limsup_{n\to\infty} \frac{1}{4} \sum_{i=1}^n \mathbb{E} \left[ \left( f\left( S_n + \frac{1-\epsilon_i}{\sqrt{n}} \right) - f\left( S_n + \frac{1+\epsilon_i}{\sqrt{n}} \right) \right)^2 \right] = \mathbb{E} \left[ f’(X)^2 \right]$</p> $\tag*{\Box}$ <p><strong>Remark</strong> There are also Poincaré type inequalities for non-Gaussian random variables, for example if $$X\sim$$Poisson$$(\mu)$$:</p> <p>$\text{Var}[f(X)] \leq \mu \mathbb{E}\left[ (f(X+1) - f(X))^2 \right]$</p> <p>Or if $$X$$ is double exponential i.e. with density $$\frac{1}{2}e^{-\vert x \vert}$$, then we have:</p> <p>$\text{Var}[f(X)] \leq 4 \mathbb{E}\left[ (f’(X))^2 \right]$</p> <h2 id="an-application-to-pdes">An Application to PDEs</h2> <p>To do proper justice for the theory of PDEs, we will need a significant background in functional analysis. In this section, we will try to side-step the technical details and focus on one single application, that is showing the existence and uniqueness of a weak solution for <strong>Poisson’s equation</strong>:</p> <p>$-\Delta u = f \; \text{ in } \Omega$ $u = g \; \text{ in } \partial \Omega$</p> <p>where $$\Omega \subset \mathbb{R}^n$$ is open bounded with smooth boundaries $$\partial \Omega$$. By <strong>weak solution</strong>, we meant there exists a $$u \in C^1(\Omega)$$ such that $$\forall v \in C^1_c(\Omega)$$ we have</p> <p>$B[u,v] := \int_\Omega \nabla u \cdot \nabla v = \int_\Omega f v$</p> <p>Note if $$u$$ is a solution to the (original) Poisson’s equation, then we have the above weak equation by <a href="https://en.wikipedia.org/wiki/Green%27s_identities">Green’s identity</a>. The main tool we will use to prove existence and uniqueness is the following result:</p> <hr /> <p><strong>Theorem (Lax-Milgram)</strong> Let $$H$$ be a Hilbert space, $$B: H^2 \to \mathbb{R}$$ be a continuous, coersive, bilinear form. Then $$\forall \varphi \in H^*$$, there exists a unique $$u\in H$$ such that</p> <p>$B(u,v) = &lt;\varphi, v&gt; \quad \forall v \in H$</p> <p>where $$&lt;\varphi,v&gt;$$ is the linear functional $$\varphi$$ applied to $$v$$.</p> <hr /> <p><strong>Remark</strong> Before going into the definitions and technical details, we observe that Lax-Milgram Theorem gives us exactly what we want - the existence and uniqueness! Now we just have to fill in the blanks:</p> <ul> <li>define a <a href="https://en.wikipedia.org/wiki/Hilbert_space">Hilbert space</a> $$H$$ where the solution lives in</li> <li>show that $$B(u,v)$$ is continuous and coersive (we will define later)</li> <li>let $$&lt;\varphi, v&gt; = \int_\Omega f v$$</li> </ul> <p><strong>Step 1</strong> To define our Hilbert space, we will consider the following <a href="https://en.wikipedia.org/wiki/Inner_product_space">inner product</a>:</p> <p>$(u,v) := \int_\Omega [uv + \nabla u \cdot \nabla v]$</p> <p>which corresponds to the following <strong>Sobolev norm</strong>:</p> <p>$\lVert u \rVert_{H^{1}(\Omega)} := (u,u)^{1/2} = \left[ \lVert u \rVert_{L^2(\Omega)}^2 + \lVert \nabla u \rVert_{L^2(\Omega)}^2 \right]^{1/2}$</p> <p>By equipping the space $$C^1_c(\Omega)$$ with the above inner product, we almost have a Hilbert space! Here we will simply take the completion of $$C^1_c(\Omega)$$ with respect to the Sobolev norm, i.e. add all the limit points to the space. We call this (completed) Hilbert space $$H_0^1(\Omega)$$.</p> <!-- At the same time, we should note the form on the right hand side is just a geometric mean of two quantities related by the Poincar&eacute; inequality! This observation will be key to our proof. --> <p><strong>Step 2</strong> We now turn our attention to $$B(u,v)$$, the bilinear form (fancy term for separately linear in both inputs). Then we say $$B$$ is <strong>continuous</strong> if</p> <p>$\exists C_1 &gt; 0 : \forall u,v \in H, \vert B(u,v) \vert \leq C_1 \lVert u \rVert_{H^1(\Omega)} \lVert v \rVert_{H^1(\Omega)}$</p> <p>Note this is an immediate consequence of Cauchy-Schwarz inequality</p> <p>$\vert B(u,v) \vert \leq \lVert \nabla u \rVert_{L^2(\Omega)} \lVert \nabla v \rVert_{L^2(\Omega)} \leq \lVert u \rVert_{H^1(\Omega)} \lVert v \rVert_{H^1(\Omega)}$</p> <p>We say $$B$$ is <strong>coersive</strong> if</p> <p>$\exists C_2 &gt; 0 : \forall u \in H, B(u,u) \geq C_2 \lVert u \rVert_{H^1(\Omega)}^2$</p> <p>We notice this is the only non-trivial condition left to check, and to prove this we will finally use <em>Poincaré inequality</em>! Start by rewriting</p> <p>$B(u,u) = \int_\Omega \nabla u \cdot \nabla u = \lVert \nabla u \rVert_{L^2(\Omega)}^2$</p> <p>Applying the Poincaré inequality on half of the norm we have</p> <p>$\frac{1}{2} \lVert \nabla u \rVert_{L^2(\Omega)}^2 \geq \frac{1}{2C} \lVert u \rVert_{L^2(\Omega)}^2$</p> <p>Therefore</p> <p>$B(u,u) \geq \frac{1}{2} \lVert \nabla u \rVert_{L^2(\Omega)}^2 + \frac{1}{2C} \lVert u \rVert_{L^2(\Omega)}^2 \geq \min\left(\frac{1}{2}, \frac{1}{2C}\right) \lVert u \rVert_{H^1(\Omega)}^2$</p> <p>And voilà, we have existence and uniqueness! A rigorous and careful reader may notice that $$u$$ does not necessarily have compact support - this is correct. However every $$u \in H_0^1(\Omega)$$ is a limit of compactly supported functions, therefore we just need to take a limit to get our result!</p> <p><strong>Remark</strong> In fact, we can use similar Lax-Milgram based methods to show existence and uniqueness for a large subset of <a href="https://en.wikipedia.org/wiki/Elliptic_partial_differential_equation">elliptical PDEs</a>. We should note that the fact we can “convert” between $$\|u\|$$ and $$\|\nabla u\|$$ is highly useful for studying Sobolev norms. We refer curious readers to Evans (2010) for an excellent chapter on <a href="https://en.wikipedia.org/wiki/Sobolev_space">Sobolev spaces</a> and related inequalities.</p> <h2 id="final-words">Final Words</h2> <p>I have a weak spot for connections between different fields, probably because it’s always surprising, and surprises are intriguing in math! I hope to have to presented a readable introduction to the inequality and its applications in both topics, without drowning readers in technical details. On this note, I should remark that to study Sobolev spaces rigorously, the reader will need to go through all the details carefully!</p> <p>As this is my first blog post, any constructive feedback or suggestions on future topics will be appreciated!</p> <h2 id="references">References</h2> <ul> <li>Boucheron, S., Lugosi, G., &amp; Massart, P. (2013). Concentration inequalities: A nonasymptotic theory of independence. Oxford university press.</li> <li>Evans, L. C. (2010). Partial differential equations. Providence, R.I.: American Mathematical Society.</li> </ul> <!-- This nice inequality is named after the brilliant [Henri Poincar&eacute;](https://en.wikipedia.org/wiki/Henri_Poincar%C3%A9), who contributed much more to mathematics than a blog post can contain. --> <!-- Observe that these inequalities now have *different constant coefficients*! As we will see soon, the more general PDE version of Poincar&eacute; inequality does in fact have a constant coefficient. --> <!-- ## Some Background in Sobolev Spaces I will introduce a few technical definitions so I can state the inequality rigorously. Note these technicalities are only necessary for applications in PDEs. Let $$\Omega \subset \mathbb{R}^n$$ be bounded with a smooth ($$C^1$$) boundary, and let $$u,v \in L^1_{\text{loc}}(\Omega)$$ (i.e. integrable on compact subsets). We say $$u$$ has a **weak derivative** $$v$$ with respect to $$x_i$$ if $$\forall \varphi \in C^\infty_c(\Omega)$$ (compact support) we have \$\int_\Omega u \frac{\partial \varphi}{\partial x_i} = \int_\Omega v \varphi \$ Then if $$u$$ has all weak derivatives of degree 1, we can define a $$L^p$$ like norm as follows \$\lVert u \rVert_{W^{1,p}(\Omega)} := \left[ \lVert u \rVert_{L^p(\Omega)}^p + \sum_{i=1}^n \left\lVert \frac{\partial u}{\partial x_i} \right\rVert_{L^p(\Omega)}^p \right]^{1/p} \$ The above **Sobolev norm** can be interpreted as a "sum" of all the weak derivatives' $$L^p$$ norms. We then define the normed-space of such functions a **Sobolev space**, denoted $$W^{1,p}(\Omega)$$. Note we can extend the definition up to the $$k^\text{th}$$ degree weak derivative, which is denoted $$W^{k,p}$$ instead. An important space we will look at is the $$W^{1,p}$$-closure of $$C^\infty_c(\Omega)$$, denoted $$W^{1,p}_0(\Omega)$$. In other words, all the limit points of $$C^\infty_c(\Omega)$$ in the norm we defined above. ## In the Context of Sobolev Spaces Finally we can state the second form of Poincar&eacute; inequality, (quoting Corollary 9.19 from Brezis (2011)): --- **Theorem (Poincar&eacute; Inequality)** Suppose $$1 \leq p < \infty$$, $$\Omega$$ bounded open. Then there exists a constant $$C$$ depending on $$\Omega, p$$ such that: \$\lVert u \rVert_{L^p(\Omega)} \leq C \lVert \nabla u \rVert_{L^p(\Omega)} \quad \forall u \in W^{1,p}_0(\Omega) \$ --- **Remark** To connect with the previous inequality, we need to consider squaring both sides, taking $$p=2$$, and change the Lebesgue measure to a probability measure (note the Radon-Nikodym derivative is just the density, which is bounded for Gaussian). Lastly, there is in fact a version of this inequality with the left hand side centered - i.e. $$\|u - \overline{u}\|_{L^p(\Omega)}$$ where $$\overline{u}$$ is the average. We will discuss this at the end. The following proof is based on Proposition 9.18 from Brezis (2011): *proof (of Poincar&eacute;'s Inequality)* Firstly since $$u \in W^{1,p}_0(\Omega)$$, there exists a sequence $$u_k \in C^\infty_c(\Omega) : u_k \to u$$ in $$W^{1,p}(\Omega)$$. \\ Using Green's identity from calculus, and the fact that $$u_k$$ have compact support we have \$\left\vert \int_\Omega u_k \frac{\partial \varphi}{\partial x_i} \right\vert = \left\vert \int_\Omega \frac{\partial u_k}{\partial x_i} \varphi \right\vert \leq \left\lVert \frac{\partial u_k}{\partial x_i} \right\rVert_{L^p(\Omega)} \lVert \varphi \rVert_{L^q{\Omega}} \quad \forall \varphi \in C^\infty_c(\Omega) \$ where we used H&ouml;lder's inequality in the last step. -->{"name"=>"", "avatar"=>"Profile.jpg", "bio"=>nil, "location"=>nil, "employer"=>nil, "pubmed"=>nil, "googlescholar"=>"https://scholar.google.com/citations?user=9dSlc_cAAAAJ", "email"=>"mufan.li@mail.utoronto.ca", "researchgate"=>nil, "uri"=>nil, "bitbucket"=>nil, "codepen"=>nil, "dribbble"=>nil, "flickr"=>nil, "facebook"=>nil, "foursquare"=>nil, "github"=>"mufan-li", "google_plus"=>nil, "keybase"=>nil, "instagram"=>"mufan.li", "impactstory"=>nil, "lastfm"=>nil, "linkedin"=>"mufan-bill-li-35749833", "orcid"=>nil, "pinterest"=>nil, "soundcloud"=>nil, "stackoverflow"=>nil, "steam"=>nil, "tumblr"=>nil, "twitter"=>"mufan_li", "vine"=>nil, "weibo"=>nil, "xing"=>nil, "youtube"=>nil, "wikipedia"=>nil}mufan.li@mail.utoronto.caWhile studying two seemingly irrelevant subjects, probability theory and partial differential equations (PDEs), I ran into a somewhat surprising overlap: the Poincaré inequality. On one hand, it is not out of the ordinary for analysis based subjects to share inequalities such as Cauchy-Schwarz and Hölder; on the other hand, the two forms of Poincaré inequality have quite different applications.