# Random walk with gaussian increments - Probability that it falls below 0

Suppose $\{Z_{i}\}_{i=1,2,\ldots}$ are normally distributed (identically and independent) random variables with mean $\mu>0$ and positive variance $\sigma^{2}$. Suppose we want to calculate the probability, that $$U(n)=u+\sum_1^nZ_i\le0 \text{ for some }n\in\mathbb{N}\text{, where }u>0\text{.}$$

According to risk of ruin wikipedia page, the probability for such setting can be approximated by

$$\left( \frac{2}{1+\frac{\mu}{r}}-1 \right)^{u/r}$$ where $$r=\sqrt{\mu^2+\sigma^2}\text{.}$$ However, I cannot find any proof/derivation for this result. I am grateful for any advice on how to approach this problem!

## 1 Answer

They simply take the exact formula for the ruin probability of a simple random walk and use it as an approximation. Ruin probability of a simple random walk that jumps to $+1$ with probability $p$ and to $-1$ with probability q is given by $(q/p)^z$ if $p>q$. Then, take the random walk that starts at $u=Cz$ and jumps to $+C$ or $-C$ with the same probabilities. Pick $C$, $p$ and $q$ in such a way that the mean and variance of simple random walk are equal to that of the Gaussian random walk. This gives the following equations $$C(p-q)=\mu$$ and $$C^2(p+q)= \mu^2+\sigma^2.$$ Of course, $p+q=1$. Then the probability of ruin of this random walk will be $(p/q)^{u/C}$ and after finding $p$ and $q$ will be exactly the formula that you gave.

Of course, this is not really a proof, just an explanation how one can obtain this approximation. I doubt that a rigorous proof exists. However, as $u\to \infty$, one can use Cramer asymptotics for probability of ruin.