Gaussian distribution — Jemoka Knowledge Base

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 \left( \frac{x-\mu}{\sigma}\right) \end{equation}

We can’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} \left(\mu, \Sigma\right) \end{equation} then

\begin{equation} \mathbb{E}\left[z\right] = \mu \end{equation}

\begin{align} \text{Cov}\left(z\right) &= \mathbb{E}\left[\left(z-\mu\right) \left(z - \mu\right)^{T}\right] = \Sigma \\ &= \mathbb{E}\left[zz^{T}\right] - \left(Ez\right)\left(Ez\right)^{T} \end{align}

\begin{equation} p\left(z\right) = \frac{1}{\left(2\pi\right)^{\frac{|z|}{2}}|\Sigma|^{\frac{1}{2}}} \exp \left(-\frac{1}{2} \left(x-\mu\right)^{T} \Sigma^{-1} \left(x - \mu\right)\right) \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<x) \implies P\left(\frac{X-\mu}{\theta} < \frac{x-\mu}{\theta}\right) \implies P\left(Z< \frac{x-\mu}{\theta}\right) = \Phi\left(\frac{x-\mu}{\theta}\right) \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} \left(\mqty[\mu_{a}\\ \mu_{b}], \mqty(A & C \\ C^{\top} & B)\right) \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} &= \mu_a + CB^{-1}\left(b - \mu_{b}\right) \\ \sigma_{a|b} &= 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}