What is the distribution of the position of the maximum of a Brownian bridge?

Question

What is the distribution of the position of the maximum of a Brownian bridge?

Answer 1

The position of the maximum is uniformly distributed.

Let $(X_t)_{t \in [0, 1)}$ be a Brownian bridge. Fix $k \in [0, 1)$ and define the process $(Y_t)_{t\in[0,1)}$ by $$Y_t\equiv X_{(t+k) \mod 1} - X_k.$$

Claim: $Y$ is a Brownian bridge.

Proof: $Y$ is a continuous Gaussian process. $Y$ is zero-mean, like $X$. After some computation, we see that $$\begin{align}\text{cov}(Y_s,Y_t)&=s\wedge t - st\\ &=\text{cov}(X_s,X_t),\end{align}$$ so $Y$, being Gaussian with the same mean and covariance structure as $X$, has the same law.

Clearly $$\text{argmax}(Y)=\text{argmax}(X)+k\mod 1,$$ so, since $X$ and $Y$ are both Brownian bridges, the law of the position of the maximum is invariant under cyclic translations. Hence it is the Haar measure on the circle, which is the uniform distribution.