# Inverse Binomial Expected value

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 $$X \sim B(n,p)$$

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

• I wonder whether such general result exists!
– hola
Jun 27, 2014 at 14:44
• What is $1/x$ when $x=0$? Jun 27, 2014 at 14:50
• @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. Jun 27, 2014 at 15:17
• In that case, the expectation is obviously infinite, since there is positive probability that $x=0$. Jun 27, 2014 at 15:36
• It takes only a single infinite case to make the expectation infinite!
– whuber
Jun 27, 2014 at 15:36

Since $\Pr[X = 0] = p^n > 0$, what is $${\rm E}\left[\frac{1}{X}\right] = \sum_{i=0}^n \frac{1}{i} \Pr\left[X = i\right] = ?$$
• Check your expression carefully. Should $i$ start from $0$? Doesn't it make the whole expression meaningless?