1. ## comparison Gaussian-Exponential distribution

Let $\displaystyle \xi_1,\xi_2,...$ be independent identically distributed Gaussian variables with mean zero and variance one. Let $\displaystyle \eta_1,\eta_2,...$ be independent identically distributed exponential random variables with mean one. How do you prove that there is $\displaystyle n>0$ such that:

$\displaystyle \mathbb{P}(max(\eta_1,...,\eta_n)\geq max(\xi_1,...,\xi_n))>0.99$

I thought a proper application of the Law of Total Probability would do the trick, but things don't seem that simple. Otherwise, some fancy convolution... But I'm probably wrong.

2. ## An idea

I haven't worked this out, but this might lead to a proof -

Consider a cutoff value x0.

Calculate the probability that all $\displaystyle \eta 's$ are less than x0. You can calculate this from integrating the pdf to find the probability that one $\displaystyle \eta$ is less than x0 and taking the n'th power (everything is independent).

Now calculate the probability that all $\displaystyle \xi 's$ are less than x0. Subtract this probability from 1. This is the probability that some $\displaystyle \xi$ exceeds x0.

When you compare the 2 expressions, you should see that it's likely your relation holds.

What's left is figuring out how the value of n depends on x0 and your 0.99 value.