# Find a transformation $G(Y )$ such that, if $Y$ has a uniform distribution on the interval $(0, 1), G(Y )$ has the same distribution as $X$.

Find a transformation $G(Y)$ such that, if $Y$ has a uniform distribution on the interval $(0, 1), G(Y )$ has the same distribution as $X$ where $$F_X(x)\begin{cases} 1-e^{-x^2}\quad;x>0\\ 0\qquad\;\;\;\;\;\;;\text{otherwise} \end{cases}$$

I am finding this part of Probability really challenging. I am hoping that someone can either confirm my result, or point out flaws.

Attempt:

I have a theorem which states: Let $X$ be a continuous rv with monotonic increasing cdf $F(x)$. Let $Y = F (X)$. Then $Y$ is uniformly distributed on $[0, 1].$

Since our function is monotone we can set $y=1-e^{-x^2}$. Then $x=\sqrt{-\ln(1-y)}.$

Now to find our domain see that when $x=0$ the $F_X(x)=0$ and when $x=\infty$ then $F_X(x)=1$

So, I'm to sure this is correct but it seems to me that if we have the pdf all we need to do is find the inverse. This does not make sense to me though because if we have the pdf the process is much more complicated

• The function $G$ which you have implicitly produced is right. May 10, 2016 at 5:40
• So, when we have the cdf all we need to due is find the inverse? That is all. Seems too easy . . . for many functions anyway. May 10, 2016 at 5:42
• Yes, easy, at least when we can find an explicit formula for the inverse. This is one way to simulate a nicely behaved distribution. There are many other ways, big important subject. May 10, 2016 at 5:45