# Transformation of the uniform distribution

I struggle to understand the transformation of a random variable with uniform distribution. For example:

Let $X\sim \text{Uniform}(0,1)$ and $T=-2\ln(X)$ and I want to find the CDF of $T$, then I know that I can compute $$P(T\leq t)=P(-2\ln(X)\leq t)=P\left(X\leq e^\frac{-t}{2}\right)$$ $$=\int\limits_{-\infty}^{e^\frac{-t}{2}}\mathbb{1}_{\{0,1\}}\mathbb{d}t$$

But how do I compute this? How can I get a nice integral without the indicator function and the $\displaystyle e^\frac{-t}{2}$ as the upper bound?

• In general, if $A,B$ are sets, $\int_A f(x) 1_{B}\,dx=\int_{A\cap B}f(x)\,dx$. What is $(-\infty,\exp(-t/2))\cap (0,1)$? – πr8 Jul 9 '16 at 10:30
• @πr8 hi, I'm a little bit confused, would this be $$(-\infty,\exp(-t/2))\cap (0,1)=(0,1)$$? – MarcE Jul 9 '16 at 11:40
• @MarcE That's because $-2\ln(X)\leq t \iff X\geq \exp(-t/2)$ so you are looking for the interval $(\exp(-t/2);\infty)\cap (0;1)$, which is $(\exp(-t/2);1)$ iif $t\geq 0$ . – Graham Kemp Jul 9 '16 at 12:16
No, the required integral is \begin{align}\mathsf P(-2\ln(X)\leq t) =&~ \mathsf P(X\geq\mathsf e^{-t/2})\\[1ex] =&~ \mathbf 1_{\exp(-t/2)\in[0;1]}\int_{\exp(-t/2)}^1\operatorname d x \\[1ex]=&~ (1-\mathsf e^{-t/2})~\mathbf 1_{t\in[0;\infty)}\end{align}
The PDF of $T$ is $$P(T)=P(X)\left|\frac{dX}{dT}\right|$$ $$=1\times\left|\frac{dX}{dT}\right|$$ $$=\left|\frac{X}{2}\right|$$ $$=\frac{1}{2}e^{-T/2}$$ The CDF is $$P(T\le t)=\int_{0}^{t}\frac{1}{2}e^{-T/2}dT$$ $$=-\left[e^{-T/2}\right]_{0}^t$$ $$=1-e^{-t/2}$$
• Hi, thank you for the answer. Unfortunately I don't understand the first step. What is $|\frac{\mathrm{d}X}{\mathrm{d}T}|$? And why is the PDF $f_T=1\times |\frac{\mathrm{d}X}{\mathrm{d}T}|$? – MarcE Jul 9 '16 at 11:43