---
title: "Gaussian distribution"
source: https://www.jemoka.com/posts/kbhgaussian_distribution/
---

constituents \(\mu\) the mean \(\sigma\) the variance requirements \begin{equation} X \sim N(\mu, \sigma^{2}) \end{equation}
Its PDF is:
\begin{equation} \mathcal{N}(x \mid \mu, \sigma^{2}) = \frac{1}{\sigma\sqrt{2\pi}} e^{ \frac{-(x-u)^{2}}{2 \sigma^{2}}} \end{equation}
where, \(\phi\) is the standard normal density function
Its CDF:
\begin{equation} F(x) = \Phi \qty( \frac{x-\mu}{\sigma}) \end{equation}
We can&rsquo;t integrate \(\Phi\) further. So we leave it as a special function.
And its expectations:
\(E(X) = \mu\)
\(Var(X) = \sigma^{2}\)
additional information multi-variant Gaussian density \begin{equation} z \sim \mathcal{N} \qty(\mu, \Sigma) \end{equation}
then
\begin{equation} \mathbb{E}\qty [z] = \mu \end{equation}
\begin{align} \text{Cov}\qty(z) &amp;= \mathbb{E}\qty [\qty(z-\mu) \qty(z - \mu)^{T}] = \Sigma \\ &amp;= \mathbb{E}\qty [zz^{T}] - \qty(Ez)\qty(Ez)^{T} \end{align}
\begin{equation} p\qty(z) = \frac{1}{\qty(2\pi)^{\frac{|z|}{2}}|\Sigma|^{\frac{1}{2}}} \exp \qty(-\frac{1}{2} \qty(x-\mu)^{T} \Sigma^{-1} \qty(x - \mu)) \end{equation}
linear transformations on Gaussian For some:
\begin{equation} Y = aX + b \end{equation}
where \(X \sim \mathcal{N}\)
We will end up with another normal \(Y \sim \mathcal{N}\) such that:
mean: \(au + b\) variance: \(a^{2}\sigma^{2}\) standard normal The standard normal is:
\begin{equation} Z=\mathcal{N}(0,1) \end{equation}
mean 0, variance 1. You can transform anything into a standard normal via the following linear transform:
transformation into standard normal \begin{equation} X \sim \mathcal{N}(\mu, \sigma^{2}) \end{equation}
and, we can shift it into a standard normal with:
\begin{equation} Z = \frac{X-\mu}{\sigma} \end{equation}
therefore, we can derive what the CDF of the normal distribution by shifting it back into the center:
\begin{equation} P(X&lt;x) \implies P\qty(\frac{X-\mu}{\theta} &lt; \frac{x-\mu}{\theta}) \implies P\qty(Z&lt; \frac{x-\mu}{\theta}) = \Phi\qty(\frac{x-\mu}{\theta}) \end{equation}
normal maximizes entropy no other random variable uses as little parameters to convey as much information
approximation of binomial distribution with normal distribution You can use a normal distribution to approximate binomial approximation. However, be aware of a continuity correction
adding Gaussian distributions for independent:
\begin{equation} X+Y \sim \mathcal{N}(\mu_{1}+\mu_{2}, \sigma_{1}^{2}+\sigma_{2}^{2}) \end{equation}
conditioning Gaussian distributions For distributions that follow Gaussian distributions, \(a, b\), we obtain:
\begin{align} \mqty[a \\ b] \sim \mathcal{N} \qty(\mqty[\mu_{a}\\ \mu_{b}], \mqty(A &amp; C \\ C^{\top} &amp; B)) \end{align}
meaning, each one can be marginalized as:
\begin{align} a \sim \mathcal{N}(\mu_{a}, A) \\ b \sim \mathcal{N}(\mu_{b}, B) \\ \end{align}
Conditioning works too with those terms, for \(a|b\):
\begin{align} \mu_{a|b} &amp;= \mu_a + CB^{-1}\qty(b - \mu_{b}) \\ \sigma_{a|b} &amp;= A - CB^{-1}C^{\top} \end{align}
standard normal density function This is a function used to model many Gaussian distributions.
\begin{equation} \phi(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^{2}}{2}} \end{equation}
This function is the CDF of the standard normal.
standard normal density function is also symmetric:
\begin{equation} \phi(a) = 1- \phi(a) \end{equation}
