Transformation of the uniform distribution

Question

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?

Thank you for your answer

Answer 1

No, the required integral is $$\begin{align}\mathsf P(-2\ln(X)\leq t) =&~ \mathsf P(X\geq\mathsf e^{-t/2})\\[1ex] =&~ \mathbf 1_{\exp(-t/2)\in[0;1]}\int_{\exp(-t/2)}^1\operatorname d x \\[1ex]=&~ (1-\mathsf e^{-t/2})~\mathbf 1_{t\in[0;\infty)}\end{align}$$

Answer 2

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 $$P(T\le t)=\int_{0}^{t}\frac{1}{2}e^{-T/2}dT$$ $$=-\left[e^{-T/2}\right]_{0}^t$$ $$=1-e^{-t/2}$$