Whats the formula for the probability density function of skewed normal distribution

The formula for the probability density function of a standard normal distribution that isn't skewed is: $$P(x) = \frac{1}{\sqrt{2π}}e^{-(x^2 / 2)}$$ where, $π = 3.14, e = 2.718$.

What if it is skewed left or right like this: What is the formula for the standard normal distribution after it's been skewed.

If anyone is wondering I'm asking because I need the formula for a program I'm writing where certain 'effects' are strong in the beginning and weaker towards the end or weak at the beginning and stronger towards the end. If this doesn't make sense just forget it.

Solution 1:

The skewed normal distribution is defined as follows:

Let $\phi(x) := \frac{1}{\sqrt{2\pi}} e^{-\frac 1 2 x^2}$ be the PDF of the standard normal distribution.

Let $\Phi(x) := \int_{-\infty}^x \phi(t) dt$ the corressponding CDF.

Then we can define the PDF of the skewed normal distribution with a skewnessparameter $a$ as

$$f_a(x) = \frac{\phi(x)\Phi(ax)}{\Phi(0)} \qquad (*)$$

You can show that $\int_{-\infty}^{\infty} f_a(x) dx = 1 \forall a\in \mathbb R$

If you look at plots you will easily see, that $f_a$ is left skewed for positive $a$ and right skewed for negative $a$.

If you want to shift the distribution about $u$ for adjusting the mean, you can just use $x \mapsto f_a(x-u)$. If you want to adjust the variance you can just use $x \mapsto \frac{1}{s} f_a\left(\frac{x}{s}\right)$ (or a combination of both).

These are the plots of $\color{blue}{f_0}, \color{green}{f_1},\color{red}{f_3}, \color{#0aa}{f_{15}}$:

$(*):$ This idea works also for other distributions, in your case you can replace $\phi,\Phi$ with the correspoding PDF/CDF of the cauchy, student-t, slash etc...