Standard Normal Distribution in a Nutshell

Given a Gaussian function that depicts the well-known bell-shaped curve.

$$p(x)=e^{-x^{2}/2},\quad x\in (-\infty ,\infty )$$

We can compute the integral.

$$\int _{-\infty }^{\infty }p(x)\,dx=\int _{-\infty }^{\infty }e^{-x^{2}/2}\,dx={\sqrt {2\pi \,}}.$$

Dividing the Gaussian function by this factor scales the area under the curve to one which gives us the probability density function of the standard normal distribution.

$$\varphi (x)={\frac {1}{\sqrt {2\pi \,}}}p(x)={\frac {1}{\sqrt {2\pi \,}}}e^{-x^{2}/2}$$

One can see that including the factor of \(\frac{1}{2}\) in the Gaussian function gives us the established definition of the standard normal distribution having \(\mu = 0\) and \(\sigma = 1\).

If one wonders why the normalization constant includes \(\pi\) in the denominator, the answer is a bit more involved but the short version is that when computing the Gaussian Integral the variables are transformed to polar coordinates.

I stumbled over this derivation while peeking into The Principles of Deep Learning Theory.

The text is rather dense, but appears to be a wortwhile read.

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