## Math

The formula of normal distribution is

$$$$e^{ ( (x - \mu) / \sqrt{2} \sigma )^2 }$$$$

where $\mu$ controls the “center” or “peak” of the distribution and $\sigma$ tells us how “wide” or “disperse” the distribution is.

To understand the distribution, we take some limits.

### $x = \mu$

First of all, when $x = \mu$ we have

$$e^0 = 1.$$

Notice the argument of the exponential is some squared value and can not be negative. This condition gives us the peak value.

### $x=\mu-a$ and $x=\mu + a$

For $x=\mu-a$, we have

$$e^{ ( (a) / \sqrt{2} \sigma )^2 }.$$

For $x=\mu + a$, we have

$$e^{ ( (a) / \sqrt{2} \sigma )^2 }$$

which is exactly the same as the previous case.

The distribution is symmetric around $x=\mu$.

### $x=\pm \infty$

We have 0 for both cases.

## Tricks

### Integral

We integrate distributions a lot. For Gaussian distribution, it is quite helpful to remember the following identity.

$$\int_{-\infty}^\infty e^ {- x^2} dx = \sqrt{\pi}.$$

It tells us that for $\mu=0$ and $\sigma=1/\sqrt{2}$, the area under the distribution is $\sqrt{\pi}$.

Hey, it is time to ask the question. Where the hell is the circle?

### Error Function

The error function is defined as

$$\mathrm{erf}(x) = \frac{1}{\sqrt{\pi}}\int_{-x}^{x} e^{ - t^2} dt$$

Obviously, the coefficient $\frac{1}{\sqrt{\pi}}$ normalizes the function to be within $[-1,1]$.

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