Given the density function, $f_{X}\left(x\right)=\dfrac{1}{4\sqrt{x}}$, for $1<x<9$, find the density function for $Y=\dfrac{1}{X}$.
I solved $f_{Y}\left(y\right)=\dfrac{\dfrac{1}{y}\sqrt{\dfrac{1}{y}}}{4}$. How do I find the range of $y$?
Thank you.
On
Correct so far.
The support for your $f_Y(y)$ shall be $1\leq 1/y\leq 9$, which is to say, $1/9\leq y\leq 1$.
$\begin{align}f_Y(y) &= \lvert\mathrm d(1/y)/\mathrm d y\rvert\, f_Y(1/y)\\[1ex]&=(1/y^2)\cdot(\sqrt{y~}/4)\,\mathbf 1_{1\leqslant 1/y\leqslant 9}\\[1ex]&=(1/4y\sqrt{y~})\,\mathbf 1_{1/9\leqslant y\leqslant 1}\\[1ex] &=\begin{cases}\dfrac{1}{4y\sqrt{y~}}&:&\dfrac 19\leqslant y\leqslant 1\\0&:&\text{elsewhere}\end{cases}\end{align}$
The range of a function is the image of the domain: and taking the support as the domain, and for this function is simply $[f_Y(1) \ldots f_Y(1/9)]$, ie $[1/4\ldots27/4]$
Since $1 < x < 9$, you have $\frac{1}{9} < y < 1$. So the density of $Y$ is defined on that domain. Since the density function you have computed is monotonic, you can just plug in the extreme values of the domain and see that $1/4 < f_Y(y) < 27/4$. This is the range.