# Expectation of inverse of sum of random variables

Let $X_i$'s ($i=1,..,n$) be i.i.d. random variables with mean $\mu$ and variance $\sigma^2$.

Is there a method that can be used to compute $\mathbb{E}[1/(X_1+...+X_n)]$?

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No. The information is severely insufficient and the answer is "expectation is undefined" more often than not. –  fedja Feb 13 '13 at 20:39
Since you do not specify the distribution of the $X_i$, I assume it is not known. Then the answer is no. –  André Nicolas Feb 13 '13 at 20:39

Assuming the expectation does exist, and further assuming $X$ to be positive random variables: $$\mathbb{E}\left(\frac{1}{X_1+\cdots+X_n}\right) = \mathbb{E}\left( \int_0^\infty \exp\left(-t (X_1+\cdots+X_n) \right) \mathrm{d}t \right)$$ Interchanging the integral over $t$ with expectation: $$\mathbb{E}\left( \int_0^\infty \exp\left(-t (X_1+\cdots+X_n) \right) \mathrm{d}t \right) = \int_0^\infty \mathbb{E}\left(\exp\left(-t (X_1+\cdots+X_n) \right) \right) \mathrm{d}t$$ Using iid property: $$\int_0^\infty \mathbb{E}\left(\exp\left(-t (X_1+\cdots+X_n) \right) \right) \mathrm{d}t = \int_0^\infty \mathbb{E}\left(\exp\left(-t X \right) \right)^n \mathrm{d}t$$ So should you know the Laplace generating function $\mathcal{L}_X(t) = \mathbb{E}\left(\mathrm{e}^{-t X} \right)$ we have: $$\mathbb{E}\left(\frac{1}{X_1+\cdots+X_n}\right) = \int_0^\infty \mathcal{L}_X(t)^n \mathrm{d} t$$
As others have commented, you need more information. However, you do have an estimate (using convexity): assuming the random variables are positive, $$E\left[ \frac{1}{X_1 + \ldots + X_n} \right] \ge \frac{1}{E[X_1 + \ldots + X_n]} = \frac{1}{n\mu}$$