1

I realize questions regarding integrating the normal distribution are numerous, but I wasn't able to find an already answered question that helped me with this. The integral is:

\begin{align*} \int_{\mathbb{R}} e^{-ax}\,dP = \int^\infty_{-\infty} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{-(x-\mu)^2}{2\sigma^2}-ax}\, dx \end{align*} but where to take it from here is leaving me clueless. Maple tells me it should evaluate to $e^{\frac{1}{2}a(a\sigma^2-2\mu)}$, but how do I show this?

BallzofFury
  • 1,154

3 Answers3

1

Maybe I'm only stating the obvious here: a probability distribution integrates to 1.

Marc-Andre Seguin
  • 161
1

You need to complete the square in the exponentiated part and you will get the kernel of another normal distribution. That kernel can be normalized and integrated to 1. The left over will give you what Maple gave you.

Ben Elizabeth Ward
  • 131
1

\begin{align*} &\,\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}-ax\right)\mathrm{d}x\\=&\,\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu+a\sigma^2)^2}{2\sigma^2}-a\mu+\frac{a^2\sigma^2}{2}\right)\mathrm{d}x\\=&\,\exp\left(\frac{a^2\sigma^2}{2}-a\mu\right)\underbrace{\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu+a\sigma^2)^2}{2\sigma^2}\right)\mathrm{d}x}_{=1}.\end{align*}

The last integral is $1$ because the integrand is just the probability density function of $\mathcal{N}(\mu-a\sigma^2,\sigma^2)$.

triple_sec
  • 23,377
  • Very neat, I hadn't thought of making another normal distribution out of it! – BallzofFury Oct 16 '13 at 20:03
  • In fact, this formula just gives you the expected value of a lognormal distribution (whenever $a\neq 0$). If $\xi\sim\mathcal{N}(\mu,\sigma^2)$, then $-a\xi\sim\mathcal{N}(-a\mu,a^2\sigma^2)$, so that $\exp(-a\xi)$ is lognormal! – triple_sec Oct 16 '13 at 20:09