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Let's $x$ be a random variable with a binomial distribution ($x \sim B(n,p)$). I know that the expected value of a binomial is $E(x) = n \cdot p $ but the inverse of a binomial?

  • $E\big(\frac{1}{x}\big)$ = ???

EDIT

$Y \sim unknown$ and $X \sim B(n,p)$

$E\big(\frac{Y}{X}\big) = E(Y) \cdot E\big(\frac{1}{X}\big)$

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    $\begingroup$ I wonder whether such general result exists! $\endgroup$
    – hola
    Jun 27, 2014 at 14:44
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    $\begingroup$ What is $1/x$ when $x=0$? $\endgroup$
    – Stefan Hansen
    Jun 27, 2014 at 14:50
  • $\begingroup$ @pushpen.paul I don't know if such a general result exist but since i need to compute the expected value of a ratio between two random variables I found this problem. $\endgroup$
    – gmeroni
    Jun 27, 2014 at 15:17
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    $\begingroup$ In that case, the expectation is obviously infinite, since there is positive probability that $x=0$. $\endgroup$
    – Stefan Hansen
    Jun 27, 2014 at 15:36
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    $\begingroup$ It takes only a single infinite case to make the expectation infinite! $\endgroup$
    – whuber ♦
    Jun 27, 2014 at 15:36

1 Answer 1

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Since $\Pr[X = 0] = p^n > 0$, what is $${\rm E}\left[\frac{1}{X}\right] = \sum_{i=0}^n \frac{1}{i} \Pr\left[X = i\right] = ?$$

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    $\begingroup$ Check your expression carefully. Should $i$ start from $0$? Doesn't it make the whole expression meaningless? $\endgroup$
    – hola
    Jun 29, 2014 at 11:00

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