Is there a shortcut? I already have E[X] where X=Y+1, thus I can separate it into E[Y]+E[1], and for Y, it's number of tries is n, and the probability p.
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1There is no shortcut along the same lines – Ben Grossmann Apr 17 '17 at 12:25
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:3_\ have mercy on me, I'm only helping a friend – Althea C Apr 17 '17 at 12:29
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How about first doing something simpler? Find $\int_1^2 \frac{1}{x},dx = \ln 2$ from $\int_1^2 x,dx = 1$. – GEdgar Apr 17 '17 at 12:48
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As long as $X$ is nonzero, you should be able to apply the Law of the Unconscious Statistician to compute
$E[1/X] = \sum_j \frac{1}{x_j} p(x_j)$
or in the continuous case
$E[1/X] = \int \frac{1}{x} p(x) dx$.
Edit: it seems OP is looking for the case when Y~bino(n,p). So from the definition and the law we have $E[1/X] = E[1/(Y+1)] = \sum_{j=0}^n \frac{1}{j+1} p(n,j)$ where $p(n,j) = \binom{n}{j}p^j(1-p)^{n-j}$. I'll leave the fun part of computing the sum up to OP!
Matt
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@AltheaC the pdf, probability density function or pmf, the probability mass function: https://en.wikipedia.org/wiki/Probability_density_function https://en.wikipedia.org/wiki/Probability_mass_function – Matt Apr 17 '17 at 12:03
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Sorry if I sound rude, but because I'm just helping a friend out, if my E[Y] is already np, and E[Y+1]=np+1, is there any way I can calculate E[1/Y+1] just like that? – Althea C Apr 17 '17 at 12:08
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Sure, so $Y$ is a binomial random variable, no? I'll update my post in a moment. – Matt Apr 17 '17 at 12:11
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