Ever wonder why the way we normalize a value for looking up the standard normal table works?

For example, suppose we know a random variable \(X\) has a normal distribution with mean \(\mu\) and standard deviation \(\sigma\), and we want to find \(\mathbb{P}(X \le L)\).

As usual, the procedure is to normalize the value \(L\) into \(\displaystyle M = \frac{L - \mu}{\sigma}\) and then look up \(M\) from the standard normal table. This works. But why?

Technically the reason is simple. It's just a change of variable in the underlying integral.

How? Remember \(\mathbb{P}(X \le L)\) is the cumulative distribution function of the normal distribution?

In particular, by definition \[\begin{aligned} \mathbb{P}(X \le L) &= \frac{1}{\sqrt{2\pi}\sigma} \int_{-\infty}^{L} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx \\ \end{aligned}\] Let \(\displaystyle y = \frac{x-\mu}{\sigma}\) and \(\displaystyle M = \frac{L-\mu}{\sigma}\) \[\begin{aligned} \frac{dy}{dx} &= \frac{1}{\sigma} \\ dy &= \frac{dx}{\sigma} \\ \end{aligned}\] Substituting variable with \(y\), which leads to the necessity of adjusting the bound with \(M\), we turn the above integral into: \[\begin{aligned} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{M} e^{-\frac{y^2}{2}} \, dy \\ \end{aligned}\] but that's equivalent to \(\mathbb{P}(Y \le M)\) under the standard normal distribution! That's why it works.

# posted by rot13(Unafba Pune) @ 12:41 AM