0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ In general, if $A,B$ are sets, $\int_A f(x) 1_{B}\,dx=\int_{A\cap B}f(x)\,dx$. What is $(-\infty,\exp(-t/2))\cap (0,1)$? $\endgroup$ – πr8 Jul 9 '16 at 10:30
  • $\begingroup$ @πr8 hi, I'm a little bit confused, would this be $$(-\infty,\exp(-t/2))\cap (0,1)=(0,1)$$? $\endgroup$ – MarcE Jul 9 '16 at 11:40
  • 1
    $\begingroup$ @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 $t\geq 0$ . $\endgroup$ – Graham Kemp Jul 9 '16 at 12:16
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $f_T=1\times |\frac{\mathrm{d}X}{\mathrm{d}T}|$? $\endgroup$ – MarcE Jul 9 '16 at 11:43
  • $\begingroup$ I am not familiar with CDFs. My first equation is the transformation formula in terms of PDFs. $\endgroup$ – velut luna Jul 9 '16 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.