We initially have M drunk guys located on the x-axis at positions $\mu_1,...,\,u_M$. As they are all completely "wasted", they will just randomly walk around in this 1D space for a while. After a few minutes, we want to find out which guy is located the furthest to the right on the x-axis.
Let us assume that their final locations are given by a gaussian distribution $\mathcal{N}(\mu_j,\sigma_j)$, centered around their initial position $\mu_j$ with a standard deviation $\sigma_j$, where $j \in \{1,...,M\}$. What is the probability of the $i^{th}$ being furthest to the right at the end?
My solution: Do a very ugly multidimensional integral, which is not numerically solvable, but reducible to a single integral:
$p_i = \int_{-\infty}^{\infty} \frac{1}{\sigma_i\sqrt{2\pi}} e^{-\frac{(x_i-\mu_i)}{2\sigma_i^2}} dx_i \Big[ \prod_{k \in \{1,...,i-1,i+1,...,M\}} \int_{-\infty}^{x_i} \frac{1}{\sigma_k\sqrt{2\pi}} e^{-\frac{(x_k-\mu_k)}{2\sigma_k^2}} \Big]$
As this is really ugly, I am wondering if there is a more elegant way of solving this problem?
PS: for clarification. this integral simply calculates the probability of each value of $x_k$ being smaller than $x_i$. This is why all integrals (except of the $i^{th}$) stop at $x_i$.
