# Compound of Exponential and Inverse Gamma Parameter

I am trying the prove the following:

Show that an exponential random variable such that the inverse of the parameter is gamma-distributed is Pareto-distributed. More precisely, show that if $$X | M = m \sim \mathrm{Exp}(m)$$ with $M^{−1} \sim \Gamma(p, a)$, then $X$ has a (translated) Pareto distribution.

I found a very similar problem proved already on stackexchange here: compound of gamma and exponential distribution

Except in my problem the pdf of the inverse gamma distributed variable $X$ is given by

$$f_X(x) = \frac {\beta^{\alpha}}{\Gamma (\alpha)} x^{-\alpha -1} e^{\frac{-\beta} {x}}$$

and the pdf of the compound exponential variable $Y$ is given by

$$f_{Y|X}(y|x) = Xe^{-Xy}$$

and thus

$$f_Y(y) = \int_{x=0}^\infty f_{Y \mid X}(y \mid x) f_X(x) \, \mathsf dx$$

I tried to follow analogous steps to the other problem, but the algebra just doesn't seem to work out. I can't seem to manipulate the integral to factor out the translated Pareto distribution and form the kernel of a known distribution within the integrand. Any help would be greatly appreciated.

The key is that the choice of parametrization has to match the type of distribution involved: or, more specifically, if the exponential distribution is parametrized by rate, then the parameter should be gamma distributed; if the exponential distribution is parametrized by scale, the parameter should be inverse gamma distributed (which is the same as saying that the inverse of the parameter is gamma distributed). So for example: if $$X \mid M \sim \operatorname{Exponential}(\mu = M),$$ and $$\frac{1}{M} \sim \operatorname{Gamma}(a,b),$$ then the marginal distribution will be Pareto, since the conditional density of $X$ given $M$ is $$f_{X \mid M}(x) = \frac{1}{M} e^{-x/M},$$ and the density of $M$ is inverse gamma, which also has a term of the form $e^{-b/m}$.
Let's go through this explicitly. Let $U = 1/M \sim \operatorname{Gamma}(a,b)$, so that $$f_U(u) = \frac{b^a u^{a-1} e^{-bu}}{\Gamma(a)}, \quad u > 0.$$ Then the inverse transformation gives $$f_M(m) = \frac{1}{m^2} f_U(1/m) = \frac{b^a e^{-b/m}}{m^{a+1} \Gamma(a)}, \quad m > 0,$$ which is inverse gamma. Now the marginal distribution of $X$ is \begin{align*} f_X(x) &= \int_{m=0}^\infty f_{X \mid M}(x;m) f_M(m) \, dm \\ &= \int_{m=0}^\infty \frac{1}{m}e^{-x/m} \cdot \frac{b^a e^{-b/m}}{m^{a+1} \Gamma(a)} \, dm \\ &= \frac{b^a}{\Gamma(a)} \int_{m=0}^\infty m^{-a-2} e^{-(b+x)/m} \, dm \\ &= \frac{b^a}{\Gamma(a)} \int_{u=\infty}^0 u^{a+2} e^{-(b+x)u} (-u^{-2}) \, du \\ &= \frac{b^a}{\Gamma(a)} \cdot \frac{\Gamma(a+1)}{(b+x)^{a+1}} \int_{u=0}^\infty \frac{(b+x)^{a+1} u^a e^{-(b+x)u}}{\Gamma(a+1)} \, du \\ &= \frac{ab^a}{(b+x)^{a+1}}, \quad x > 0.\end{align*} This is a (shifted) Pareto, as claimed.
What we should note is that the computation is much more straightforward if we had just stuck to the gamma distribution in the first place, rather than involving the inverse gamma. That is to say, if we had simply integrated with respect to $u$ to begin with.
• Sorry for digging up an old post, but in the third last line, your bounds are $\infty$ to $0$. How did that change back to $0$ to $\infty$ in the line immediately after? Commented Mar 24, 2022 at 11:43
• @EthanMark From Line 3 to Line 4, we used the substitution $$u = 1/m, \quad du = -1/m^2 \, dm,$$ and this introduces a negative sign, so that from Line 4 to Line 5, we eliminate the negative sign by reversing the direction of integration; e.g., $$\int_{x=a}^b -f(x) \, dx = \int_{x=b}^a f(x) \, dx.$$ Commented Mar 25, 2022 at 1:40