# Pdf of $\sin X$ when $X$ has the pdf $f(x)=\frac{2x}{\pi^2}1_{0<x<\pi}$

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

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 $$f(y)$$?

## 1 Answer

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 $$(\pi /2, \pi)$$.

• 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. – HaneenSu Feb 22 at 23:38
• I have edited my answer. – Kavi Rama Murthy Feb 22 at 23:50
• 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 – HaneenSu Feb 23 at 0:31