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Suppose I have a series of independent, identically distributed random variables $X_1, X_2, ...$, each of which has the exponential distribution with parameter λ.

How can I find the expected value $n$ such that $X_n$ is the first variable with value greater than or equal to some constant Y?

I see that the probability that the first $m$ variables do not satisfy the criteria is $(1-e^{-\lambda Y})^m$, but I'm not really sure where to go from here.

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You are almost there.

Let $1 - (1 - e^{-\lambda Y}) = e^{-\lambda Y} = x$ be the probability of success in single trial (so that $X_i \ge Y$). What is the probabilty that the first success is at position $n$? Well, it is

$$(1-x)^{n-1}x$$

Now try to express it as a random variable and calculate the expected value.

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    $\begingroup$ I'll give you an A on the OP's homework assignment. $\endgroup$ Feb 18, 2017 at 23:42
  • $\begingroup$ @MarkL.Stone Yes you are right, but thanks for the A. I eddited my answer $\endgroup$ Feb 18, 2017 at 23:58
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    $\begingroup$ The OP can easily look at the edit history and see the original version. Perhaps you should delete your answer and then post a new answer. $\endgroup$ Feb 19, 2017 at 0:57
  • $\begingroup$ @Mark Well, the OP can now that you told them how to do it. I don't think it's necessary to delete the answer as long as the issue is kept in mind for next time. $\endgroup$
    – Glen_b
    Feb 19, 2017 at 1:31

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