C Tedious results

C.1 Normal distribution

The normal distribution with parameters \(\mu\) and \(\sigma^2 > 0\) has pdf \[f(x) = \frac{1}{\sqrt{2\pi} \sigma} \exp \left[ -\frac{(x-\mu)^2}{2 \sigma^2} \right]\] It is clear \(f(x)>0\) for all \(x\), but we should confirm that this pdf integrates to one. Unfortunately this integral is one of those that either you know how to solve or you waste a huge amount of time with.

First, we will perform a substitution of \(z=\frac{x-\mu}{\sigma}\) and therefore \(dz = dx/\sigma\) and we are interested in showing that \[\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-z^2/2} \, dz = 1\] or equivalently that \[\underbrace{\int_{-\infty}^{\infty} e^{-z^2/2} \, dz}_{I} = \sqrt{2\pi}\] or that \[I^2 = 2\pi\] We will consider this \(I^2\) value as \[\begin{aligned} I^2 &= \int_{-\infty}^{\infty} e^{-z^2/2} \, dz \; \int_{-\infty}^{\infty} e^{-w^2/2} \, dw \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-z^2/2} \,e^{-w^2/2} \, dz\,dw \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\frac{1}{2}(z^2 + w^2)} \, dz\,dw \\ \end{aligned}\]

Because \(z^2 + w^2\) is the equation of a circle in the \(z,w\) plane, we think to convert to polar coordinates with the subsitution \(z=r \cos \theta\) and \(w= r \sin \theta\) and therefore \(z^2 + w^2 = r^2\) and \(dz\,dw = r\, dr\, d\theta\) therefore \[\begin{aligned} I^2 &= \int_0^{2\pi} \int_0^{\infty} e^{-\frac{1}{2} r^2} \,r \, dr \, d\theta \end{aligned}\] Finally this integral can be attacked using common methods and we’ll do a \(u\)-substitution with \(u=r^2/2\) and therefore \(du=r\,dr\) and we have

\[\begin{aligned} I^2 &= \int_0^{2\pi} \int_0^{\infty} e^{-u} \, du \, d\theta \\ &= \int_0^{2\pi} 1 \, d\theta \\ &= 2 \pi \end{aligned}\]