# A certain expectation of a function of independent gammas

Suppose that $$Y_1...,Y_n$$ are independant gamma random variables: $$Y_i \sim \Gamma(\alpha_i,\beta_i)$$, with density $$f_i(t) = \frac{t^{\alpha_i-1} e^{-\frac{x}{\beta}}}{\Gamma(\alpha) \beta^{\alpha}}$$.

Suppose that $$a_1,...a_m$$ and $$b_1,...,b_m$$ are strictly positives reals.

I want to compute:

$$\mathbb E \left(\prod\limits_{j=1}^m \left(b_j + Y_1 +...+ Y_n\right)^{-a_j}\right)$$

My first attemp was expanding the function inside the expectation into a multivairate taylor serie in $$(Y_1,...,Y_n)$$ and then use the fact that $$\mathbb E(Y_i^k)$$ is known. But it is quite inneficient.

Maybe from the structure of the function we could expand in a 'power serie' with non-integer powers (which would probably depend on $$a_1,...,a_m$$), and the serie would converge more efficiently (since $$\mathbb E(Y_i^a)$$ is known for any $$a$$, even $$a$$ that are not integers).

How can i do that ?

• Is there a reason to expect a closed-form (or at least simpler) expression? Even when $n = m = 1$ the answer does not seem to be simple. Jan 15, 2021 at 17:53
• For the record: $n = m = 1$ leads to some hypergeometric rubbish, and for $n = 1$ and $m = 2$ Mathematica is not able find a closed-form expression. Jan 15, 2021 at 18:00
• I am pretty sure that a closed-form expression will not be posible, but a good convergent serie yes. For n=m=1, this expectation is in essence the laplace transform of the product of two gammas, which can be expressed by some confluent hypergeometric functions. For n,m > 1, i dont know yet but i have hope haha.
– lrnv
Jan 15, 2021 at 18:02
• Btw, could you share how you did it with mathematica ?
– lrnv
Jan 15, 2021 at 18:07
• By the way, I wonder why someone downvoted the question, it seems perfectly reasonable. Jan 15, 2021 at 18:12