Let $Z = X + Y$ where $X \sim N\left(\mu, \sigma^2 \right)$ and $Y \sim \Gamma\left(k, \theta \right)$ using this parametrization of the Gamma distribution. Also assume $X$ and $Y$ are independent. Then what is the distribution (pdf) of $Z$?

I thought this question looked straightforward, but it doesn't seem so. For example, I have tried using the convolution formulas here, but can't seem to find a closed form expression for the integral. I have also tried multiplying the moment generating functions (mgfs) of $X$ and $Y$, but it does not seem to match up to any known mgfs.

Is it possible to find a closed-form solution for the distribution of $Z$? If not, how about some sort of approximation?


1 Answer 1


So $X$ and $Y$ are independent, with $$ f_X(x) = \frac{e^{-(x-\mu)^2/2\sigma^2}}{\sqrt{2\pi}\sigma} \, , $$ and $$ f_Y(y) = \frac{y^{k-1}e^{-y/\theta}}{\theta^k\Gamma(k)} I_{(0,\infty)}(y) \, . $$ If $Z=X+Y$, then $$ f_Z(z)=\int_{-\infty}^\infty f_X(z-y) f_Y(y)\,dy \, . $$ Completing the square in the exponent and looking for the terms involving $y$, we find that the necessary integration in $y$ is $$ A(z)=\int_0^\infty \frac{y^{k-1} e^{-(y-z-\mu+\sigma^2/\theta)^2/2\sigma^2}}{\sqrt{2\pi}\sigma} \,dy \, , $$ which is available in special cases. For example, if $k$ is an odd integer, then $$ A(z) = \frac{1}{2} \mathrm{E}[U^{k-1}] \, , $$ in which $$ U\sim\mathrm{N}\left( z-\mu+\frac{\sigma^2}{\theta}, \sigma^2\right) \, . $$ Please, check the algebra and group the remaining terms to get the final expression of $f_Z(z)$ depending on $A(z)$.

  • 1
    $\begingroup$ +1 You can, with relative ease, obtain expressions for all natural numbers $k$ in the usual way by replacing $y^{k-1}/(\theta^k\Gamma(k))$ by the Maclaurin series of $t\to\exp((t/\theta)y)/\theta.$ $\endgroup$
    – whuber
    Commented Nov 30, 2013 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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