# Independent gamma and normal distribution

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?

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 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 $t\to\exp((t/\theta)y)/\theta.$ – whuber Nov 30 '13 at 19:31