How to calculate the pdf of sum of two independent random variables?

Question

If I have an Exponential random variable and also a Normal random variable that are independent, how do I calculate the pdf of their sum?

Answer 1

We use convolution.

Let $Z=X+Y$, where $X$ and $Y$ are independent,

then $$f_Z(z) = \int_{-\infty}^\infty f_X(t)f_Y(z-t) \, dt $$

Edit:

$$f_Z(z) = \int_{0}^\infty f_X(t)f_Y(z-t) \, dt $$

$$f_X(t)= \lambda \exp(-\lambda t)$$

$$f_Y(z-t) = \frac{1}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{(z-t-\mu)^2}{2\sigma^2}\right)$$

\begin{align}f_X(t) f_Y(z-t) &= \frac{\lambda}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{(z-t-\mu)^2+2\sigma^2\lambda t}{2\sigma^2}\right) \\ &= \frac{\lambda}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{(t-(z-\mu))^2+2\sigma^2\lambda t}{2\sigma^2}\right)\\ &=\frac{\lambda}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{t^2-2(z-\mu)t+(z-\mu)^2+2\sigma^2\lambda t}{2\sigma^2}\right)\\ &=\frac{\lambda}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{t^2-2(z-\mu-\sigma^2 \lambda)t+(z-\mu)^2}{2\sigma^2}\right)\\ &=\frac{\lambda}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{(t-(z-\mu-\sigma^2 \lambda))^2+(z-\mu)^2-(z-\mu-\sigma^2\lambda)^2}{2\sigma^2}\right)\\ &=\frac{\lambda}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{(t-(z-\mu-\sigma^2 \lambda))^2}{2\sigma^2}\right)\exp\left(- \frac{\sigma^2\lambda(2z-2\mu-\sigma^2\lambda)}{2\sigma^2}\right)\\ \end{align}

\begin{align} f_Z(z) &=\lambda \exp\left( \frac{\lambda(\sigma^2\lambda+ 2 \mu- 2z )}{2}\right) \int_0^\infty \frac{1}{\sqrt{2\pi \sigma^2}}\exp\left( -\frac{(t-(z-\mu-\sigma^2 \lambda))^2}{2\sigma^2}\right) \, dt \\ &= \lambda \exp\left( \frac{\lambda(\sigma^2\lambda+ 2 \mu- 2z )}{2}\right) \int_{-\frac{(z-\mu-\sigma^2\lambda)}{\sqrt{2}\sigma}}^\infty \frac{1}{\sqrt{\pi }}\exp\left( -s^2\right) \, ds \\ &= \frac{\lambda}{2} \exp\left( \frac{\lambda(\sigma^2\lambda+ 2 \mu- 2z )}{2}\right) \int_{-\frac{(z-\mu-\sigma^2\lambda)}{\sqrt{2}\sigma}}^\infty \frac{2}{\sqrt{\pi }}\exp\left( -s^2\right) \, ds \\ &= \frac{\lambda}{2} \exp\left( \frac{\lambda(\sigma^2\lambda+ 2 \mu- 2z )}{2}\right) \operatorname{erfc} \left( -\frac{(z-\mu-\sigma^2\lambda)}{\sqrt{2}\sigma}\right) \\ \end{align}