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## Homework Statement

Let [tex]X_1, X_2,...[/tex] be independent identically distributed continous random variables. We call [tex]n[/tex] the first time of increase if

[tex]X_1>X_2>X_3>...>X_{n-1}<X_n[/tex]

Let [tex]N[/tex] be the time until the first increase. Show that [tex]E[N]=e[/tex].

## Homework Equations

Given for a discrete random variable with values in the positive integers:

[tex]E(X)=\sum_{k=1}^\infty P(X>=k)[/tex]

## The Attempt at a Solution

Well I know that [tex]X_i[/tex] are exchangeable so the desired outcome has only 1 in [tex]n![/tex] chance of happening.

[tex]P(N=n)=P(X_1>\cdots>X_{n-1}\mbox{ and }X_{n-1}<X_n)[/tex]

[tex]=P(X_1>X_2)P(X_2>X_3)P(X_3>X_4)...P(X_{n-2}>X_{n-1})P(X_{N-1}<Xn)[/tex]

[tex]=\frac{1}{2} \times \frac{1}{2^2} \times \frac{1}{2^3}...\frac{1}{2^{n-1}}[/tex]

Ok clearly I don't know what I'm doing. Maybe if someone could tell me what this evaluates to I could find my way to it? I just don't see how an [tex]e[/tex] can pop out whereas the only two ways to calculate [tex]e[/tex] I know are

[tex]\lim_{k\to\infty}(1+\frac{1}{k})^k[/tex] and [tex]e^z=\sum_{i=1}^\infty \frac{z^i}{i!}[/tex]

Also, I have no idea how the tail sum formula is suppose to help. How can I make use of it when the random variables in question are continuous?

Thanks guys!