Finding pdf of $\frac1X$ given pdf of $X$ is $f_X(x)= \frac{24}{x^4}I_{(2,∞)}(x)$

Question

Consider a random variable $X$ having pdf

$$f_X(x) = \frac{24}{x^4}I_{(2,∞)}(x).$$

(a) Give the pdf of $Y= \frac{1}X$.

(b) Give the value of $E(Y) =E(\frac1X)$.

$f_X(x) = \frac{24}{x^4}I_{(2,∞)}(x)$ $\Rightarrow$ $F_X(x) = \int\frac{24}{x^4}dx=-8x^{-3}$

From graphing $Y = \frac1X$ when $x\in(2,\infty)$ then $y\in(0,\frac12)$

$F_Y(y)=P(Y\leq y)$

$=P(\frac1X \leq y)$

$=P(X \geq \frac1y)$

$=1-P(X \leq \frac1y)$

$=1-F_X(\frac1y)$

$=1-\int_2^{\frac1y}\frac{24}{x^4}dx$

$=1-\frac{-8}{x^3}|_{2}^{\frac1y}$

$=8y^3$

$\Rightarrow f_Y(y)=24y^2$

So the PDF I obtained is

$$F_Y(y) = \begin{cases} {1} & \text{$y \geq \frac12$} \\ 8y^3 & \text{$0 \lt y \lt \frac12$} \\{0} & \text{$y \leq 0$}\end{cases}$$

I have that $E(Y) = \int_0^{\frac12}y(24y^2)dy=\frac38$

Did I do these correctly?

Answer 1

A way to check is to say $$ \mathbb{E}[Y] = \mathbb{E}[1/X] = \int_2^\infty \frac{24}{x^4}\frac{dx}{x} = 24 \int_2^\infty x^{-5}dx = \left. \frac{-6}{x^4}\right|_2^\infty = \frac{6}{16} = \frac{3}{8} $$ so you seem to be correct.