Sum of gamma and normal random variable

Question

If $X$ has an exponential distribution and $Y$ is normally distributed random variable, then what is the distribution of $Z=X+Y$?

Answer 1

The convolution between the pdfs of $X$ and $Y$, given by: $$ f_X(x) = \lambda e^{-\lambda x}\cdot\mathbb{1}_{x\geq 0},\qquad f_Y(x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ leads to: $$\begin{eqnarray*} f_{X+Y}(x) &=& \frac{\lambda}{\sqrt{2\pi\sigma^2}}\int_{0}^{+\infty}\exp\left(-\lambda t-\frac{(x-t-\mu)^2}{2\sigma^2}\right)\,dt\\&=&\frac{\lambda}{2}\cdot\exp\left(\frac{\lambda}{2}(2\mu+\lambda\sigma^2-2x)\right)\cdot\left(-1+\frac{1}{\sigma}+\text{Erfc}\left(\frac{\mu+\lambda \sigma^2-x}{\sigma\sqrt{2}}\right)\right)\end{eqnarray*}$$ where $\text{Erfc}$ is the complementary error function. It is a unimodal distribution quite close to the normal distribution with mean $\mu+\frac{1}{\lambda}$ and variance $\sigma^2+\frac{1}{\lambda^2}$, but with a non-zero skewness.