# Expectation of Sum of Gamma over Product of Inverse-Gamma

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 ?

• 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)? Commented Oct 23, 2017 at 7:22

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 $\alpha\to\infty$
• @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 $$X_1, X_2, \cdots, X_n \stackrel{\text{iid}}{\sim} Gamma(\alpha, \beta)$$ Commented Oct 23, 2017 at 7:59
• Neat that the $\beta$ parameter cancels out and does not appear in the solution Commented Oct 23, 2017 at 13:05
• @wolfies: it follows from the fact that $\beta$ is a scale parameter, hence cancels top and bottom. Commented Oct 23, 2017 at 14:39