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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 $f(y)$?

StubbornAtom
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HaneenSu
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1 Answers1

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

Kavi Rama Murthy
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  • 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 '19 at 23:38
  • I have edited my answer. – Kavi Rama Murthy Feb 22 '19 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 '19 at 00:31