Question
If the pdf of $X$ is $f(x)=2 x e^{-x^{2}}, 0<x<\infty$, zero elsewhere, determine the pdf of $Y=X^{2}$.1.7.23. If the pdf of $X$ is $f(x)=2 x e^{-x^{2}}, 0<x<\infty$, zero elsewhere, determine the pdf of $Y=X^{2}$.
Step 1
The CDF, $F(x)$, is the integral of the pdf from $-\infty$ to $x$: $$F(x) = \int_{-\infty}^{x} f(t) dt = \int_{0}^{x} 2t e^{-t^{2}} dt.$$ This integral can be solved by substituting $u = t^2$, $du = 2t dt$. The integral then becomes: $$F(x) = \int_{0}^{x^2} e^{-u} Show more…
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