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Suppose $X_1,\ldots,X_n$ is a random sample from the $\Gamma(k,\lambda)$ distribution where $\lambda$ is unknown and $k$ is a positive integer and known. How can I find $$E\left[\frac{\sum_{i=1}^n X_i^2}{(\sum_{i=1}^n X_i)^2}\right] \>?$$

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I think the formula is much more readable without the \displaystyle's, so I removed them. In particular, the title took up too much space on the front page when it used them. However, I suppose if you really want to add it back into the body of the question only, that would be okay. – Zev Chonoles Apr 29 '12 at 2:36
I think the $\lambda$ will cancel from the numerator and the denominator, so the answer will not depend on $\lambda$. – Michael Hardy Apr 29 '12 at 2:39
up vote 14 down vote accepted

I will assume $X_i$ be independent.

Using identity, for $a>0$, $$ \frac{1}{a^2} = \int_0^\infty t \mathrm{e}^{-a t} \mathrm{d} t $$ Rewrite the expectation as follows: $$ \begin{eqnarray} \mathbb{E}\left( \frac{\sum_{i=1}^n X_i^2}{\left(\sum_{i=1}^n X_i \right)^2} \right) &=& \int_0^\infty t \cdot \mathbb{E}\left(\sum_{i=1}^n X_i^2 \cdot \exp\left(-t \cdot \sum_{i=1}^n X_i\right) \right) \mathrm{d} t \\ &=& n \int_0^\infty t \cdot \mathbb{E}\left(X_1^2 \cdot \exp\left(-t \cdot \sum_{i=1}^n X_i\right) \right) \mathrm{d} t \\ &=& n \int_0^\infty t \cdot \left(\mathcal{M}_X(-t)\right)^{n-1} \cdot \mathcal{M}^{\prime\prime}_X(-t) \mathrm{d} t \\ &=& n \int_0^\infty t \cdot \left(1+\frac{t}{\lambda}\right)^{-k (n-1)} \cdot \frac{k(k+1)}{\lambda^2} \left( 1+ \frac{t}{\lambda}\right)^{-2-k} \mathrm{d} t \\ &\stackrel{\color\maroon{t \to \lambda u}}{=}& n k (k+1) \int_0^\infty \frac{t}{(1+t)^{k n +2}} \mathrm{d} t \\&\stackrel{t \to \frac{1-s}{s}}{=}& n k (k+1) \int_0^1 (1-s) s^{k n-1} \mathrm{d} s = n k (k+1) \left( \frac{1}{n k} - \frac{1}{n k+1} \right)\\&=& {\color\maroon{\frac{k+1}{k \cdot n +1}}} \end{eqnarray} $$ where $\mathcal{M}_X(t)$ denotes the moment generating function of the $\Gamma(k,\lambda)$ distribution.

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This is excellent. Maybe you could add an explanation about why the last integral is $1/(kn(kn+1))$ (possibly through the change of variable $s=1/(1+t)$ and a Beta identity). Upvoted anyway. – Did Apr 29 '12 at 8:48
@Didier Thanks for the feedback and upvote. I have expanded evaluation of the integral per your suggestion. It's better this way. – Sasha Apr 29 '12 at 14:56
(+1) That's clever, Sasha. – cardinal Apr 29 '12 at 16:04

Sasha gives a very nice answer. Here is an alternative using standard relationships between distributions. I, too, will assume that the $X_i$ are independent.

Write $\newcommand{\Beta}{\mathrm{B}}Y_i = \frac{X_i}{X_i + \sum_{i\neq j} X_j}$.

Then, $Y_i \sim \mathrm{Beta}(k,(n-1)k)$ since $X_i \sim \Gamma(k,\lambda)$ and $\sum_{i \neq j} X_j \sim \Gamma((n-1)k,\lambda)$ and the latter two are independent.

By linearity of expectation and the fact that each $Y_i$ has a common distribution, we have $$ \mathbb E\left(\frac{\sum_{i=1}^n X_i^2}{(\sum_{i=1}^n X_i)^2}\right) = n \mathbb E Y_1^2 \>. $$

But, $$ \mathbb E Y_1^2 = \frac{1}{\Beta(k,(n-1)k)} \int_0^1 y^2 y^{k-1} (1-y)^{nk-k-1} \,\mathrm dy = \frac{\Beta(k+2,(n-1)k)}{\Beta(k,(n-1)k)} = \frac{k+1}{n(nk + 1)} \>. $$

So, $$ \mathbb E\left(\frac{\sum_{i=1}^n X_i^2}{(\sum_{i=1}^n X_i)^2}\right) = \frac{k+1}{nk + 1} \>. $$

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Wow! No less than two good solutions to a same question... What is going on here? :-) – Did Apr 29 '12 at 16:27

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