Finding the PDF of a transformed random variable

Question

Let $$ f_{X}(x) = \begin{cases} 4x, \qquad 0 \le x < \frac{1}{4} \\ \frac{8}{7}-\frac{4}{7}x, \qquad \frac{1}{4} \le x \le 2 \\ 0,\qquad \mathrm{else} \end{cases} $$

If $Y=X^2$, how can I find the PDF of Y? I don't believe I can partition the pieces because their y-components would not be equal.

Answer 1

As per the fact that your rv is non-negative, the transformation $Y=X^2$ is monotone. Thus simply use the well known formula

$$ \bbox[5px,border:2px solid black] { f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg| \qquad (1) } $$

leading immediately to

$$f_Y(y) = \begin{cases} 2, & \text{ $0\leq y<\frac{1}{16}$ } \\ \frac{4}{7\sqrt{y}}-\frac{2}{7}, & \text{ $\frac{1}{16} \leq y \leq 4$ }\\ 0, & \text{elsewhere }\\ \end{cases}$$

... without doing any integral but simply substituting $g^{-1}(y)$ and its derivative in formula (1)

Check:

$$\int_{-\infty}^{+\infty}f_Y(y)dy=2\times \frac{1}{16}+\int_{\frac{1}{16}}^{4}\Bigg[\frac{4}{7\sqrt{y}}-\frac{2}{7}\Bigg]dy=\frac{1}{8}+\frac{7}{8}=1$$

nice! :)

Answer 2

Note that $X \ge 0$. $$ F_Y(y) = \mathbb{P}[Y \le y] = \mathbb{P}[X \le \sqrt{y}] = F_X(\sqrt{y}). $$ Can you take it from here?

Answer 3

Use the CDF. For $t\in[0,16]$:

$$F_Y(t)=\Pr(Y\leq t)=\Pr(X^2\leq t)=\Pr(X\leq \sqrt{t})=\int\limits_0^{\sqrt{t}} f_X(x)dx$$

It is left to finish the integral and differentiate if you want the PDF.

Answer 4

Sorry Guys but there is something wrong in the text...

...the given density is a triangle with area $\frac{4\times 1}{2}=2$

thus the given $f_X(x)$ is not nice...