3
$\begingroup$

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

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

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

1 Answer 1

2
$\begingroup$

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

compound of gamma and exponential distribution

$\endgroup$
1
  • $\begingroup$ Thank you! I shall preserve this link because I used it to prove the above. $\endgroup$ Feb 2, 2014 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.