about PDF of normal distribution function

Question

Definition: X is said to have a normal distribution if its PDF is given by $f_X(x)=\frac{1}{\sqrt{2 \pi \sigma}} e^{\frac{(-(x-\mu)^2}{2 \sigma^2}}$

How do people come up with this, and how am I supposed to remember it?

Answer 1

You can start by remembering the following function.$$f(x) = e^{\frac{-x^2}{2}}$$

You know that for any continuous distribution: $$z(x) = \int_{-\infty}^{\infty}f(x)dx = 1$$ If you integrate $f(x)$ and try to check if it integrates to 1:

$$\int_{-\infty}^{\infty}f(x)dx\int_{-\infty}^{\infty}f(x)dx = \int_{-\infty}^{\infty}f(x)dx\int_{-\infty}^{\infty}f(y)dy$$

$$z^2(x) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)f(y)dxdy$$ $$z^2(x) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^\frac{-(x^2 + y^2)}{2}dxdy$$

From here, you can use polar coordinates and get: $$x^2 + y^2 = r^2$$ $$x = r\cos\theta, y = r\sin\theta$$ $$z^2(x) = \int_{0}^{2\pi}\int_{0}^{\infty}e^\frac{-r^2}{2}rdrd\theta = \int_{0}^{2\pi}1d\theta = 2\pi$$ $$z(x) = \sqrt{2\pi}$$

Since your initial function has to integrate to 1, you can transform f(x) into: $$f(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$$ This is the normal distribution with $\mu=0$ and $\sigma^2 = 1$. If you want to generalize, you can use linearity: $$ Y = \sigma X + \mu $$ $$ P(Y < y) = P(\sigma X + \mu < y) = P(X < \frac{y - \mu}{\sigma}) = F(\frac{y - \mu}{\sigma})$$

$$f(y) = F'(y)$$ $$f(y) = \frac{1}{\sigma}f(\frac{y - \mu}{\sigma})$$ $$f(y) = \frac{1}{\sqrt{2\pi}\sigma}e^\frac{-(x-\mu)^2}{2\sigma^2}$$

It might be tedious to go over this over and over again but I think understanding where it comes from might help remembering it on long term.