How to transform gaussian(normal) distribution to uniform distribution?

Solution 1:

\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 $0 \le x\le 1,$ $$ \Pr(\Phi(Z) \le x) = \Pr( Z \le \Phi^{-1}(x)) = \Phi(\Phi^{-1}(x)) = x. $$

Solution 2:

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 $t \geq 3$.