I need to proof that the Pareto random variable is a mixture of the Gamma and Exponential distribution but need help with one part of the proof.

Consider $X$ being Exponential with parameter $\lambda$ and $\Lambda$ being Gamma with parameters $\alpha$ and $\beta$. So we can say the mixture distribution of $X$ is

$$ \begin{align} f_{X|\alpha, \beta} &= \int_0^\infty \lambda e^{-\lambda x} \cdot \frac{1}{\Gamma (\alpha) \beta ^ \alpha} \lambda^{\alpha - 1} e^{-\frac \lambda \beta} d\lambda\\ &=\frac {1} {\Gamma (\alpha) \beta ^ \alpha} \int_0^\infty \lambda ^\alpha e^{-\lambda x - \lambda \frac 1 \beta} d\lambda\\ &=\frac{\Gamma (\alpha +1)}{\Gamma (\alpha) \beta^\alpha} \int_0^\infty \frac{\lambda ^\alpha e^{-\lambda x - \lambda \frac 1 \beta}}{e^{-\lambda}\lambda^\alpha} d\lambda\\&= \frac{\alpha}{\beta^\alpha}\int_0^\infty e^{\lambda \frac{\beta - x -1}{\beta}} d\lambda\\ &=\frac{\alpha}{\beta^\alpha}\begin{bmatrix} \frac{\beta}{\beta - x -1}e^{\lambda\frac{\beta - x - 1}{\beta}} d\lambda\end{bmatrix}^\infty _0\\&=?\\ &=\frac{\alpha}{\beta^\alpha}\begin{bmatrix}\frac{\beta}{\beta x+1}\end{bmatrix}^{\alpha +1} \end{align} $$

  • $\begingroup$ I think the equality between second and third row is not correct. $\endgroup$ Commented Jan 31, 2014 at 15:28
  • $\begingroup$ Hello @JuhoKokkala, any idea how to fix it? Well, I used the identity $\Gamma (\alpha +1) = \int_0^\infty e^{-\lambda}\lambda^\alpha dt$ $\endgroup$ Commented Jan 31, 2014 at 15:30
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    $\begingroup$ But that identity does not imply that dividing the integrand by $e^{-\lambda}\lambda^\alpha$ and multiplying the result by $\Gamma(\alpha+1)$ cancel out - note that $\int (f(x) / g(x)) dx$ is not generally equal to $\int f(x) dx~/ \int g(x) dx$. $\endgroup$ Commented Jan 31, 2014 at 15:57
  • $\begingroup$ could you propose an answer and help me with the solution and proof? $\endgroup$ Commented Jan 31, 2014 at 16:07
  • $\begingroup$ Sure! I have solved it successfully. Thank you for your help! $\endgroup$ Commented Feb 2, 2014 at 6:58

1 Answer 1


The solution to this question was recently provided on the mathematics exchange

compound of gamma and exponential distribution

  • $\begingroup$ Thank you! I shall preserve this link because I used it to prove the above. $\endgroup$ Commented Feb 2, 2014 at 6:57

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