I would like to find the PDF of the random variable $Y=\sin(X)$ given the PDF of $X$:

$$f(x) = \frac{2x}{\pi^2} \text{ for } 0<x<\pi \text{ and } 0 \text{ otherwise}$$

Following the tips in the question here: Transformation of PDF

I found $$f(y)= \frac{1}{\sqrt{1-y^2}}\frac{2\sin^{-1}(y)}{\pi^2}$$

However, I am not sure about the range of $Y$. Is it correct to say that $Y$ is between $1$ and $-1$ for the above $f(y)$?


Since $\sin\, x$ takes values between $0$ and $1$ when $x$ is between $0$ adn $\pi$, the range for $y$ is $(0,1)$, not $(-1,1)$. For a proper application of the transformation formula you have to split the range $(0, \pi)$ of $x$ into $(0, \pi /2)$ and $(\pi /2, \pi)$.

  • $\begingroup$ Ops sorry yes I meant between 0 and 1. Is the solution overall correct? I am not sure if the solution itself is correct as the function Y=sin(x) is a many to one function. $\endgroup$ – HaneenSu Feb 22 at 23:38
  • $\begingroup$ I have edited my answer. $\endgroup$ – Kavi Rama Murthy Feb 22 at 23:50
  • $\begingroup$ Thanks, however, according to my understanding and as explained in the link in my question, it seems the function itself doesn't depend on the range. f(y)=f(x) dx/dy $\endgroup$ – HaneenSu Feb 23 at 0:31

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.