A sweet probabilistic proof that zeta(2)=…

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}}.


  1. 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 ?

  2. Alekk said,

    December 28, 2012 at 5:59 pm

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

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