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Let $Y=X^2$. Find the denisity $f_Y$ of $Y$.

This is my approach $\rightarrow Y=X^2 \rightarrow Y=\sqrt{X}$ then

$F_Y(Y)=P(Y \leq X)= P(Y \leq X^2) \rightarrow P(\sqrt{Y} \leq X)$.

Or am I supposed to use this formula for density:

$$\frac{1}{2\sqrt{2\pi}\sigma}e^{-\left(\frac{x-\mu}{2\sigma}\right)^2}?$$

2 Answers 2

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Hint : when $X$ is standard, you can extend it to the non-standard

We define the cdf of $Y$ as $$F_Y(y)=P(Y \leq y)$$

First of all, $F_Y(y)= 0$ if $y<0$, because $Y=X^2$ is non-negative.

We assume $y \geq 0$, and we have $$F_Y(y)=P(Y \leq y)=P(X^2 \leq y)=P(\sqrt{X^2} \leq \sqrt{y})=P(|X| \leq \sqrt{y})$$

This can be rewritten as $$P(|X| \leq \sqrt{y})=P(-\sqrt{y}\leq X\leq \sqrt{y})=P( X\leq \sqrt{y})-P( X\leq -\sqrt{y})=F_X(\sqrt{y})-F_X(-\sqrt{y})$$

Using the symmetry of $X$, we have $F_X(-\sqrt{y})=1-F_X(\sqrt{y})$

Thus, $$F_Y(y)=2F_X(\sqrt{y})-1$$

To get the density $f_Y$, we derive $F_y$ wrt $y$, $$f_Y(y)=\frac{f_X(\sqrt{y})}{\sqrt{y}}$$

where $f_X$ is the density function of $X$

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    Oh okay, i see , man I was missing a lot of steps then – 2017-02-24
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    You gotta use it at some point, in your result, you do not use that $f_X(x)=f_X(-x)$ – 2017-02-24
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You found the accumulate distribution, but not the density. For all $y$, we have $F_Y(y)=\mathbb{P}(Y \leq y)=\mathbb{P}(X^2 \leq y)$. If $y<0$, then $\mathbb{P}(X^2 \leq y)=0$. If $y \geq 0$, $\mathbb{P}(X^2 \leq y)=\mathbb{P}(\left |{X}\right | \leq \sqrt[]{y})=F_X(\sqrt[]{y})-F_X(-\sqrt[]{y})$. Then, we derive, and we obtain $f_Y(y)=\frac{{\partial F_Y(y)}}{{\partial y}}=\frac{f_X(\sqrt[]{y})}{2\sqrt[]{y}}+\frac{f_X(-\sqrt[]{y})}{2\sqrt[]{y}}$ for all $y\geq 0$.

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    So I messed up completely then – 2017-02-24
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    My proof is the same to the previously, but I didn't suppose symmetry in the distribution. If $X$ is normal standard, both answers are correct. – 2017-02-24