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I have obtained the pdf and its distribution seems to be a triangle. Now, I want to take the square of this pdf like

${f_Y(y)}^2$

How can I obtain the square of this pdf?

Do, I just need to simply multiply the two $f_Y(y)$?

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    I suggest you use correct random variable notation. Specify the density function of whatever random variable $X$ you have in mind. Then the density function of $X^2$ should not be difficult to find.2012-12-23

1 Answers 1

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If $f(x)$ is the pdf of $X$, then $f(x)^2$ is NOT the pdf of $X^2$. Write $F(x)$ as the cdf of $X$: $F(x)=P(X\leq x)$. Then if $G$ is the cdf of $Y:=X^2$, $G(y):=P(Y\leq y)=P(x^2\leq y)=P(-\sqrt{y}\leq x\leq \sqrt{y})$, and assuming $X$ has a continuous cdf, $G(y)=F(\sqrt{y})-F(-\sqrt{y})$, which gives $g(y)=G'(y)=\frac{1}{2\sqrt{y}}(f(\sqrt{y})+f(-\sqrt{y}))$

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    f((y)^o.5) is a triangular distribution.So, It has different formula for -1 to 0 and 0 to 1. When I calculate this pdf for just one range the answer comes out to be in i(iota) since g(y)=1/(y)^0.5.Now,What to do?2012-12-23