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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?

Community
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Don
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    $P(n/(X_1+...+X_n) \geq a) = 1$ 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

3 Answers3

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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 $$f_{1/T}(x) = \frac{\lambda^n e^{-\lambda/x}}{x^{n+1} \Gamma(n)}, \quad x > 0.$$

heropup
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  • Thank you for your answer. Is $f_T$ the $n-$fold convolution of $f_{X_i}$? – Don Apr 25 '15 at 11:55
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    @Don See the following post: http://math.stackexchange.com/questions/655302/gamma-distribution-out-of-sum-of-exponential-random-variables/655315#655315 – heropup Apr 25 '15 at 16:30
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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.

MBW
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I also agree with the math, but per the technical point, the cited 'n' here for the Gamma-provided solution is, more precisely, 'n/2', relating back to the Exponential case. This is, in accord, with the opening statement:

I know that the sum $X_1 + ... + X_n$ has Gamma distribution with parameters $2n$...

So, importantly, the bias correction factor becomes n/(n-2) (and, not n/(n-1)). Support this for assertion (which is important in small sample scenarios) can currently be found in Wikipedia on the Exponential distribution citing this as the appropriate correction factor for the MLE.

Note: Having worked in correcting small sample estimation of parameters, for various distributions including the Exponential, even the bias-corrected estimate remains noisy, in my opinion.

AJKOER
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