Homework help finding pdf's of y given pdf's of x - stuck

Question

If anyone can give me the steps on how to find pdf$\,'$s of $y$ given $x$.

Let X be a continuous random variable with probability density function given by $$ {\rm f}\left(x\right) =\left\lbrace\begin{array}{lcl} {1 \over x^{2}} & \mbox{if} & x \geq 1 \\[2mm] 0&&\mbox{otherwise} \end{array}\right. $$ A Random Variable $Y$ is given by $$ Y = \left\lbrace\begin{array}{lclrcl} 2x & \mbox{if} &\quad X & \geq & 2 \\[2mm] x^{2} & \mbox{if} &\quad X & < & 2 \end{array}\right. $$
Find the probability density function of $Y$.

I would need to take the cumulative density function of ${\rm f}\left(x\right)$ with limits of $Y$ ?. Then derivative of that function or ?. Any help to get on the right track much appreciated.

Answer 1

The formula is $f_Y(y)=\begin{cases} f_X(g^{-1}(y)) \left| \frac{d}{dy} g^{-1}(y) \right| ,y\in \mathcal Y \\ 0,\text{else} \end {cases}$

First build the inverse function of $Y=g(x)$ (with definition set)

Then insert $g^{-1}(y)$ in f(x). Keep in mind, that $f(x) =0$, if $x \leq 1$.

Then calculate $\left| \frac{d}{dy} g^{-1}(y) \right|$.

greetings,

calculus

Answer 2

One strategy is to find the cumulative distribution function $F_Y(y)$ of $Y$, and then differentiate. It is clear that $F_Y(y)=0$ if $y\le 1$.

We are told that $Y=X^2$ if $X\lt 2$. So if $1\lt y\lt 4$, we have $$F_Y(y)=\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(X\le \sqrt{y})=\int_1^{\sqrt{y}} \frac{1}{x^2}\,dx.$$ If we just want to find the density, we can use the Fundamental Theorem of Calculus to find the derivative without calculating the integral. However, in this case the integration is easy.

If $y\ge 4$, the calculation is similar. We get $F_Y(y)=\Pr(Y\le y)=\Pr(X\le y/2)=\int_1^{y/2} \frac{1}{x^2}\,dx$.