# Inverse of a mean, exponential distribution, expected value

Could you help me find the expected value of this random variable?

Let $X_1, X_2, ...$ be independent identically exponentially distributed with parameter $\lambda$ random variables.

What is the expected value of $\frac{n}{X_1 + ... + X_n}$?

I've read these questions https://math.stackexchange.com/questions/1246590/expectation-of-inverse-of-sum-of-random-variables-exponential-distribution and Expectation of inverse of sum of random variables but there's nothing helpful there.

I know that the sum $X_1 + ... + X_n$ has Gamma distribution with parameters $2n$, $\frac{n}{\lambda}$ but either way I have a problem because $\mathbb{E} (\frac{1}{X}) \neq \frac{1}{\mathbb{E}X}$.

Could you tell me what I can do with this?

• $P(n/(X_1+...+X_n) \geq a) = 1$ for $a \leq 0$. For $a>0$, $P(n/(X_1+...+X_n) \geq a) = P((X_1+...+X_n) \leq \frac{n}{a}).$ Since you know the distribution of $X_1+...+X_n$, this enables you to work out the distribution of $n/(X_1+...+X_n)$, and hence its expectation. – Frank Apr 24 '15 at 17:34

If $X_i \sim \operatorname{Exponential}(\lambda)$ are iid such that $$f_{X_i}(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ then $T = \sum_{i=1}^n X_i \sim \operatorname{Gamma}(n,\lambda)$ with $$f_T(x) = \frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(n)}, \quad x > 0.$$ Then it is easy to see that $$\operatorname{E}[n/T] = \int_{x=0}^\infty \frac{n}{x} f_T(x) \, dx = \frac{n\lambda}{n-1} \int_{x=0}^\infty \frac{\lambda^{n-1} x^{n-2} e^{-\lambda x}}{\Gamma(n-1)} \, dx = \frac{n\lambda}{n-1},$$ for $n > 1$, since the last integral is simply the integral of of a $\operatorname{Gamma}(n-1,\lambda)$ PDF and is equal to $1$.
The distribution of $1/T$ is inverse gamma; i.e., $$T^{-1} \sim \operatorname{InvGamma}(n,\lambda)$$ with $$f_{1/T}(x) = \frac{\lambda^n e^{-\lambda/x}}{x^{n+1} \Gamma(n)}, \quad x > 0.$$
• Thank you for your answer. Is $f_T$ the $n-$fold convolution of $f_{X_i}$? – Don Apr 25 '15 at 11:55
It may help to use the fact that $$\mathbb{E}\left[X\right] = \int_0^{\infty} \mathbb{P}(X \geq x) dx$$ for positive continuous variables so that $$\mathbb{E}\left[\frac{n}{X_1+\ldots+X_n}\right] = \int_0^{\infty} \mathbb{P}\left(\frac{n}{X_1+\ldots+X_n} \geq x\right) dx = \int_0^{\infty} \mathbb{P}\left(X_1+\ldots+X_n \leq \frac{n}{x}\right) dx$$ Now use the fact that the last c.d.f is the c.d.f of a Gamma distribution with the parameters you gave.