The probability density function of a random variable $X$ is given by
$$f_X (x) =\begin{cases}
\frac{5}{x^2} & x > 5\\
0 & \text{otherwise}\\
\end{cases}
$$
Find the probability density function of $Y = 4-3X $. Don't forget to state the support of the p.d.f (set of values $y$ for which $y > 0$).
Going to solve this using this formula from my textbook
$$f_Y(y)\frac{f_X(h^{-1}(y))}{|h'(h^{-1}(y))|}$$
Let $h^{-1}(y) = 4 – 3y$.
Then $h(y) = \dfrac{4-y}{3}$ and $h'(h^{-1}(y)) = 0-3y^{1-1} = -3$
$$\dfrac{\dfrac{5}{(4-3y)^2}}{|-3|} = \dfrac{5/3}{(4-3y)^2}$$
with support
$$\frac{4-y}{3} > 5 \iff4 – y > 15 \iff y < -11$$
Answer is
$$\dfrac{15}{(4-y)^2}$$
which is different from mine. Why?
If I let $h^{-1}(y) = \frac{4-y}{3}$,
then
$$\frac{5 \cdot 9}{(4-y)^2} \cdot \frac{1}{\left|h'\left(\frac{4-y}{3}\right)\right|} = \frac{45}{(4-y)^2} \cdot \frac{1}{\frac{1}{3}} = \frac{45 \cdot 3}{(4-y)^3}$$
which is again, not the same as the solution.
I understand it for the most part except I don't get the answer.
I still don't get how $$h'(h^{-1}(y)) = h'\left(\frac{4-y}{3}\right) = -3$$
Best Answer
You're given $y = 4-3x$. Let this $y=h(x)=4-3x$. Then $x=h^{-1}(y)=\dfrac{4-y}{3}$. So by the given formula,$$f_Y(y)=\frac{f_X(h^{-1}(y))}{|h'(h^{-1}(y))|}=\frac{\dfrac{5}{\left(\frac{4-y}{3}\right)^2}}{|-3|}=\frac{15}{(4-y)^2}.$$ Note that $h'$ is constant, since $Y$ is linear: $$h'(\color{blue}{x})=-3\,\,\forall x\implies h'(\color{blue}{h^{-1}(y)})=-3.$$ On the other hand, if you've noticed, a shorthand for this when $Y=\alpha X+\beta$, $$f_Y(y)=\dfrac{f_X(\dfrac{y-\beta}{\alpha})}{|\alpha|}$$ for $\alpha\ne0.$