# Drunk guys in a race, who will win? (random walk)

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 $$x_i$$.

• You can make it look slightly more elegant by writing it as $$p_i = \int_{-\infty}^\infty f_i(x) \prod_{j \ne i} F_j(x)\; dx$$ where $f_i$ and $F_j$ are the pdf's and cdf's of these normal distributions, but I don't think you can get a better formula to compute. Mar 27, 2019 at 15:07
• @RobertIsrael Yes, thank you. This is a simpler version of the integral, but still doesn't solve the problem :( Mar 27, 2019 at 15:11