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Find the PDF of $Y=X^2$ where $X\sim U(0, \theta)$

So I already know that $$F_X(x)=\cases {0 & $x\le0$ \\ \frac{x}{\theta} & $0<x<\theta$ \\ 1 & $x > \theta$}$$

My calculation for $f_Y(y)$:

$$F_Y(y) = F_{X^2}(y) = P(X^2\le y)=P(X\le \sqrt y)= F_X(\sqrt y) \\ = \cases {0 & $\sqrt y\le0$ \\ \frac{\sqrt y}{\theta} & $0<\sqrt y<\theta$ \\ 1 & $\sqrt y > \theta$}$$

To get the PDF from the CDF we evaluate the derivative:

$$\frac{d}{dy} F_Y(y) = \cases {\frac{1}{2\theta \sqrt y} & $0<\sqrt y<\theta$ \\ 0 & Otherwise}$$

Is that correct?

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  • $\begingroup$ I think there might be a mistake in your definition of $F_{X}(x)$. It should be $0$ after $\theta$ $\endgroup$ – Euler_Salter Dec 18 '16 at 15:21
  • $\begingroup$ Oh no sorry, that's the cdf, my mistake $\endgroup$ – Euler_Salter Dec 18 '16 at 15:22
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    $\begingroup$ yeah it all looks correct! $\endgroup$ – Euler_Salter Dec 18 '16 at 15:23
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Similarly, you can use a formula. Call $w(Y) = \sqrt{Y}$ the inverse transformation. Then you have $$f_{Y}(y) = f_{X}(w(y))\left|\frac{dw(y)}{dt}\right|$$ which indeed gives you $$f_{Y}(y) = f_{X}(\sqrt{y})\left|\frac{dw(y)}{dt}\right| = \frac{1}{\theta}\frac{1}{2\sqrt{y}} = \frac{1}{2\theta\sqrt{y}}$$ for $y$ in the domain of course

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