What can we say about $\mathbb{E} \left(\frac{1}{\sum_{i=1}^N X_i}\right)$?

Asked Sep 18 '20 at 01:28

Active Sep 18 '20 at 07:13

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Let $X_i \sim \exp(\lambda)$ for $i= 1, ..., N$ and i.i.d. Hence $\mathbb{E}(X_i) = \frac{1}{\lambda}$. What can we say about $\mathbb{E} \left(\frac{1}{\sum_{i=1}^N X_i}\right)$?

I think that $\sum_{i=1}^N X_i \sim \exp(\lambda)$, but I am not sure we can say $\frac{1}{\sum_{i=1}^N X_i} \sim \exp(\lambda)$

edited Sep 18 '20 at 07:13

StubbornAtom

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asked Sep 18 '20 at 01:28

J.Doe

are the $X_i$, $I=1.\ldots,N$ independent? – Mittens Sep 18 '20 at 01:38
Yes, they are I.I.D – J.Doe Sep 18 '20 at 01:38
@JDoe: the sum of iid exponentials is a Gamma distribution here. That does not solve your problem just yet, but it corrects your first conjecture. – Mittens Sep 18 '20 at 01:41
It follows an inverse Gamma distribution. See https://en.wikipedia.org/wiki/Inverse-gamma_distribution – Alex Sep 18 '20 at 01:42
I am not familiar with the Gamma distribution yet. Are you up to build a full answer? – J.Doe Sep 18 '20 at 01:46
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Does this answer your question? Inverse of sum of exponential random variables (mean and variance) – Alex Sep 18 '20 at 02:15

What can we say about $\mathbb{E} \left(\frac{1}{\sum_{i=1}^N X_i}\right)$?

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