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Let $X_1, X_2, \cdots, X_n \sim Gamma(\alpha, \beta)$. How do we compute $E\left(\cfrac{\sum_1^n X_i}{(\prod_1^n X_i)^{1/n}}\right)$ ?

I am stuck on how to compute this expectation. I know that $\cfrac{1}{X_i}$ follows inverse-gamma but how to we handle $\cfrac{1}{(X_i)^{1/n}}$ and how do we separate the sum and the product ?

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    $\begingroup$ 1. is this for some subject? 2. the title incorrectly describes what you have. It's not over a product of inverse gamma. The X's are gammas, and it's not simply a product but a geometric mean, so a correct description would be "over the geometric mean of gammas" or "times a geometric mean of inverse gammas" (though it would be important to notice that there are common indices in the numerator and denominator). 3. Did you omit any conditions (such as independence of the X's)? $\endgroup$ – Glen_b Oct 23 '17 at 7:22
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This is rather straightforward (when the $X_i$'s are independent): \begin{align*}\mathbb{E}\left(\cfrac{\sum_{i=1}^n X_i}{(\prod_{j=1}^n X_j)^{1/n}}\right) &= \sum_{i=1}^n \mathbb{E}\left(\cfrac{ X_i}{(\prod_{j=1}^n X_j)^{1/n}}\right)\\ &= \sum_{i=1}^n \mathbb{E}[X_i^{1-1/n}]\times \mathbb{E}\left(\cfrac{1}{(\prod_{j\ne i} X_j)^{1/n}}\right)\\ &= \sum_{i=1}^n \mathbb{E}[X_i^{1-1/n}]\times \prod_{j\ne i}\mathbb{E}\left[X_j^{-1/n}\right]\\ &= n\mathbb{E}[X_1^{1-1/n}]\mathbb{E}\left[X_1^{-1/n}\right]^{n-1}\\ &= n\times \beta^{1/n-1}\dfrac{\Gamma(\alpha+1-1/n)}{\Gamma(\alpha)}\times\left[\beta^{1/n}\dfrac{\Gamma(\alpha-1/n)}{\Gamma(\alpha)}\right]^{n-1}\\ &=n\times \beta^{1/n-1+(n-1)/n}\times\dfrac{\Gamma(\alpha+1-1/n)\Gamma(\alpha-1/n)^{n-1}}{\Gamma(\alpha)^n}\\ &=n\times \dfrac{(\alpha-1/n)\Gamma(\alpha-1/n)^{n}}{\Gamma(\alpha)^n}\end{align*} If considering the expectation of the ratio of the arithmetic mean to the geometric mean, i.e., when divding the above by $n$, one obtains a ratio that converges to $1$ with $\alpha\to\infty$ enter image description here

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    $\begingroup$ Nothing in the question says any of the variables are independent (and if they are independent, the question looks like a bookwork exercise). How did you take the right hand side of the first line to the second line? $\endgroup$ – Glen_b Oct 23 '17 at 7:15
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    $\begingroup$ @Glen_b: Granted. If the $X_i$'s are dependent the question has no unique solution. Hence the default assumption that they are independent when writing$$X_1, X_2, \cdots, X_n \sim Gamma(\alpha, \beta)$$instead of $$X_1, X_2, \cdots, X_n \stackrel{\text{iid}}{\sim} Gamma(\alpha, \beta)$$ $\endgroup$ – Xi'an Oct 23 '17 at 7:59
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    $\begingroup$ Neat that the $\beta$ parameter cancels out and does not appear in the solution $\endgroup$ – wolfies Oct 23 '17 at 13:05
  • $\begingroup$ @wolfies: it follows from the fact that $\beta$ is a scale parameter, hence cancels top and bottom. $\endgroup$ – Xi'an Oct 23 '17 at 14:39

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