A sweet probabilistic proof that zeta(2)=…

June 24, 2012 at 5:19 pm (Brownian motion, probability, Stochastic analysis)
Tags: "brownian conformal invariance", zeta series probabilistic proof martingale euler

Here is a delightful probabilistic proof that ${\zeta(2)=\frac{\pi^2}{6}}$ extracted from a recent article by Greg Markowsky. This starts with pretty lemma (which can originally be found in this article) on the exit time of a ${2}$ dimensional Brownian motion ${B_t}$ . Suppose that ${f:\mathbb{C} \rightarrow \mathbb{C}}$ is an analytic function on the neighbourhood of the unit disk. This function maps the unit disk ${\mathbb{D}}$ to ${f(\mathbb{D})}$ with boundary ${\partial f(\mathbb{D}) = f(\partial \mathbb{D})}$ where ${\partial \mathbb{D}= \big\{e^{i\theta} : 0 \leq \theta \leq 2 \pi \big\}}$ . A two dimensional brownian motion started at ${f(0)}$ takes on average

$\displaystyle \mathbb{E}[ \, \tau \, ] \; = \; \sum_{k \geq 1} |a_k|^2$

to exit the domain ${f(\mathbb{D})}$ where ${f(z) = \sum_{k \geq 0} a_k z^k}$ and ${\tau = \inf \big\{ \, t>0: B_t \in \partial f(\mathbb{D}) \, \big\}}$ is the hitting time of the boundary ${\partial f(\mathbb{D})}$ . Indeed, since the situation is invariant by translation one can always suppose that ${f(0)=a_0=0}$ . The proof of this result is a simple martingale argument. Indeed, since ${M_t = \|B_t\|^2 - 2t}$ is a martingale, the optional stopping theorem shows that

$\displaystyle \mathbb{E}[ \, \|B_{\tau}\|^2 \, ] \; = \; 2 \mathbb{E}[\, \tau \, ].$

Then, since ${f(\cdot)}$ is an analytic function, Ito’s formula shows that the trajectories of ${B_t}$ are the same as the trajectories of ${f(W_t)}$ , up to a time change, where ${W_t}$ is another ${2}$ dimensional Brownian motion started at ${z=0}$ . Consequently, ${B_{\tau} = f(Z_{\rho})}$ where ${\rho}$ is the hitting time ${\rho = \inf \big\{ \, t>0: W_t \in \partial \mathbb{D} \, \big\}}$ . This shows that

$\displaystyle \mathbb{E}[ \, \|B_{\tau}\|^2 \, ] = \mathbb{E}[ \, \|f(W_{\rho})\|^2 \, ] = \frac{1}{2\pi} \int_{\theta=0}^{2\pi} |f(e^{i\theta})|^2 \, d\theta.$

Since Parseval‘s theorem (or a direct calculation) shows that the last quantity also equals ${\sum_{k \geq 0} |a_k|^2}$ , the conclusion follows.

To find a nice identity, it thus suffices to find an easy domain where one can compute explicitly the brownian exit time ${\mathbb{E}[ \, \tau \, ]}$ . The first thing that comes to mind is a strip ${S_a = \big\{ x+iy \, : \, -a < y < a\big\}}$ since it is classical that a Brownian motion started at ${z=0}$ takes on average ${T=a^2}$ to exit the strip ${S_a}$ . Since ${f(z) = \log\big( \frac{1-z}{1+z} \big) = -2(z+z^3/3+z^5/5 + z^7/7 + \ldots)}$ maps the unit disk ${\mathbb{D}}$ to the strip ${S_{\frac{\pi}{2}}}$ it follows that

$\displaystyle \frac{\pi^2}{4} = 2(1+3^{-2} + 5^{-2} + 7^{-2} + \ldots),$

which is indeed equivalent to the celebrated identity ${\zeta(2) = \frac{\pi^2}{6}}$ .

2 Comments

TheBridge said,

December 28, 2012 at 12:51 pm

Hi, nice and original proof of zeta(2).
By the way, when you say that $E [||f(W_\rho)||^2]=\frac{1}{2\pi}\int_0^{2\pi}|f(e^{i.\theta}|^2d\theta$ , you are using the argument that rotation of a 2d BM is invariant in distribtuion so it has to be uniform on the circle, right ?

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Alekk said,

December 28, 2012 at 5:59 pm

thanks!
And yes, that’s one way of proving that.

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