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Suppose that $X$ and $Y$ are both gamma distributed with shapes $a,b$ and unit scales / unit rates. To fix ideas, X has Laplace transform given by:

$$L_X(t) = \mathbb{E}(e^{-tX}) = (1 + t)^{-a}$$

How can we compute the Laplace transform of the independent product $Z=XY$, that is the integral:

$$L_{XY}(t) = \mathbb{E}(e^{-tXY}) = \mathbb{E}(L_Y(tX)) = \frac{1}{\Gamma(a)}\int\limits_{0}^{+\infty} (1+tx)^{-b} x^{a-1} e^{-x} \partial x \text{ ?}$$

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$$\frac{1}{\Gamma(a)}\int\limits_{0}^{\infty} (1+tx)^{-b} x^{a-1} e^{-x} \,dx=$$ $$\qquad=t^{-a} \, _1F_1\left(a;a-b+1;\frac{1}{t}\right)\frac{ \Gamma (b-a)}{\Gamma (b)}+t^{-b} \, _1F_1\left(b;b-a+1;\frac{1}{t}\right)\frac{\Gamma (a-b)}{\Gamma (a)}.$$

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  • $\begingroup$ According to en.wikipedia.org/wiki/Confluent_hypergeometric_function, could-nt it be framed as $t^{-a} U(a,a-b+1,\frac{1}{t})$, where U is the Tricomi confluent hypergeometric function ? $\endgroup$
    – lrnv
    Commented Dec 11, 2020 at 11:13
  • $\begingroup$ Yes it is ! Thanks a lot for pointing these functions to me ? $\endgroup$
    – lrnv
    Commented Dec 11, 2020 at 11:21

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