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Math Help - comparison Gaussian-Exponential distribution

  1. #1
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    comparison Gaussian-Exponential distribution

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

    \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.

    Thanks for your help.
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  2. #2
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    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 \eta 's are less than x0. You can calculate this from integrating the pdf to find the probability that one \eta is less than x0 and taking the n'th power (everything is independent).

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