How to transform gaussian(normal) distribution to uniform distribution? I have gaussian distributed numbers with mean 0 and variance 0.2.
And I want to transform this distribution to uniform distribution [-3 3].
How can I transform gaussian distribution numbers to uniform distribution?
 A: \begin{align}
X & \sim N(0,0.2) = N\left( 0, \frac 1 5 \right) \\[10pt]
X\cdot\sqrt 5 &\sim N(0,1) \\[10pt]
\Phi(X\cdot\sqrt 5) & \sim \operatorname{Uniform}[0,1] \tag 1 \\[10pt]
-3 + 6 \Phi(X\cdot \sqrt 5) & \sim \operatorname{Uniform}[-3,3] 
\end{align}
A proof of line $(1)$ is as follows: Suppose $Z\sim N(0,1).$ Then for $0 \le x\le 1,$
$$
\Pr(\Phi(Z) \le x) = \Pr( Z \le \Phi^{-1}(x)) = \Phi(\Phi^{-1}(x)) = x.
$$
A: Let $X\sim \mathcal{N} (\mu, \sigma^2)$ have a normal distribution with mean $\mu=0$ and variance $\sigma^2 = 0.2$, which cumulative distribution function (CDF) is denoted by $\Phi_X$. The variable $Y = 6\Phi_X (X) - 3$ has a uniform distribution over $[{-3},3]$. In facts,
\begin{aligned}
\mathbb{P}(Y\leq t) &= \mathbb{P}\left(\Phi_X(X)\leq \frac{t+3}{6}\right) \\
&= \mathbb{P}\left(X\leq \Phi_X^{-1}\left(\frac{t+3}{6}\right)\right) \\
&= \Phi_X\left(\Phi_X^{-1}\left(\frac{t+3}{6}\right)\right) \\
&= \frac{t+3}{6} \, ,
\end{aligned}
if ${-3}\leq t \leq 3$, $\mathbb{P}(Y\leq t)=0$ if $t \leq {-3}$, and $\mathbb{P}(Y\leq t)=1$ if $t \geq 3$.
