Find $Y=f(X)$ such that $Y \sim \text{Uniform}(-1,1)$.

Question

If $X_1,X_2\sim \text{Normal} (0,1)$, then find $Y=f(X)$ such that $Y \sim \text{Uniform}(-1,1)$.

I solve problems where transformation is given and I need to find the distribution. But here I need to find the transformation. I have no idea how to proceed. Please help.

Answer 1

The idea of this classical homework is to exhibit a usual function $f$ such that $f(X_1,X_2)$ has the desired distribution (in this context, the CDF of a gaussian random variable is not considered as a usual function).

To do so, recall (or reprove) that the radius $R=\sqrt{X_1^2+X_2^2}$ of a planar standard normal random variable $(X_1,X_2)$ has density $r\mathrm e^{-r^2/2}$ on $r\geqslant0$. A consequence is that $\mathrm e^{-R^2/2}$ is uniform on $(0,1)$, hence a solution to the question asked is $$ f(x_1,x_2)=2\mathrm e^{-(x_1^2+x_2^2)/2}-1. $$

Answer 2

Why not use the probability integral transform? Note that if $$ F(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-u^2/2} du $$ then $F(X_i) \sim U(0,1)$. So you could take $f(x) = 2*F(x) - 1$.

Answer 3

The cumulative distribution function $F(X)$ for any random variable $X$ always have a $U(0,1)$ distribution. Hence, $2F(X)-1$ will always follow $U(-1,1)$ distribution.

Regards Nicola Bradshaw http://tutorteddy.com/site/free_statistics_help.php