Question: Let $T$ be an exponential random variable with parameter $\lambda$. Find a function $g$ such that $g(T)$ is a uniform random variable on $[0,1]$.
I used the density mapping theorem for single variable.
What I get after solving a differential equation is:
$$g(t)=C-e^{-\lambda t},$$
for some constant C. But I'm having trouble restricting the range of $g$ to be $[0,1]$. Can I just let $C=1$, so that $g$ is between $[0,1)?$ But now $1$ is not in the interval.
May I know what can I do?
Work:
Let $U=g(T)$. We want $g$ such that $$\bigg\rvert\frac{dg^{-1}(u)}{du}\bigg\rvert \lambda e^{-\lambda g^{-1}(u)}=1.$$
Solving this equation gives $g^{-1}(u)=\frac{1}{\lambda}\ln\left(\frac{1}{C-u}\right)$.
Best Answer
It's an immediate example of integral transform theorem thus the result is the CDF of $Y$
$$G=1-e^{-\lambda T}$$
You can verify this solution
$$\mathbb{P}(G\leq g)=\mathbb{P}(1-e^{-\lambda T}\leq g)=\dots=\mathbb{P}\left(T\leq -\frac{1}{\lambda}\log(1-g)\right)=$$
$$ =F_T \left( -\frac{1}{\lambda}\log(1-g)\right)=1-e^{-\lambda(-1/\lambda)\log(1-g)}=g $$