28.6 Transformations of continuous random variables

Imagine we have a random variable XX, with probability density function

fX(x)={4x30<x10otherwisef_{X}(x)=\begin{cases}4x^{3}&0<x\leqq 1\\ 0&\text{otherwise}\end{cases} (28.10)

and we want to find the probability density function for the random variable

Y=1X4Y=\frac{1}{X^{4}} (28.11)

The first thing to do is to find the cumulative probability function for XX, (by integrating)

FX(x)\displaystyle F_{X}(x) ={0x00x4x3𝑑x0<x11x>1\displaystyle=\begin{cases}0&x\leqq 0\\ \int_{0}^{x}4x^{3}dx&0<x\leqq 1\\ 1&x>1\end{cases} (28.12)
={0x0x40<x11x>1\displaystyle=\begin{cases}0&x\leqq 0\\ x^{4}&0<x\leqq 1\\ 1&x>1\end{cases} (28.13)

Then, we can find the cumulative probability function for YY in terms of the cumulative probability function for XX. From the definition of the cumulative probability function we know that

FY(y)=P(Yy)F_{Y}(y)=P(Y\leqq y) (28.14)

Then as Y=1X4Y=\frac{1}{X^{4}} we can substitute XX for YY

P(Yy)=P(1X4<y)P(Y\leqq y)=P(\frac{1}{X^{4}}<y) (28.15)

We can then manipulate this into some function of P(X<g(y))P(X<g(y)) (where g(y)g(y) is a function we need to determine).

P(Yy)\displaystyle P\left(Y\leqq y\right) =P(1X4<y)\displaystyle=P\left(\frac{1}{X^{4}}<y\right) (28.16)
=P(X4>1y)\displaystyle=P\left(X^{4}>\frac{1}{y}\right) (28.17)
=P(X>1y4)\displaystyle=P\left(X>\sqrt[4]{\frac{1}{y}}\right) (28.18)
=1P(X<1y4)\displaystyle=1-P\left(X<\sqrt[4]{\frac{1}{y}}\right) (28.19)

Note that in the second step we flipped the inequality because we took the reciprocal of both functions, and the reciprocal function makes bigger values smaller (and vice-versa) so to keep the inequality true, we had to flip the signs. From here, we plug into FX(x)F_{X}(x).

P(Yy)\displaystyle P\left(Y\leqq y\right) =1P(X<1y4)\displaystyle=1-P\left(X<\sqrt[4]{\frac{1}{y}}\right) (28.20)
=1FX(1y4)\displaystyle=1-F_{X}\left(\sqrt[4]{\frac{1}{y}}\right) (28.21)
={10x01(1y4)40<x111x>1\displaystyle=\begin{cases}1-0&x\leqq 0\\ 1-\left(\sqrt[4]{\frac{1}{y}}\right)^{4}&0<x\leqq 1\\ 1-1&x>1\end{cases} (28.22)
={1x011y0<x10x>1\displaystyle=\begin{cases}1&x\leqq 0\\ 1-\frac{1}{y}&0<x\leqq 1\\ 0&x>1\end{cases} (28.23)

We also need to rewrite bounds in terms of yy, rather than xx. As Y=1X4Y=\frac{1}{X^{4}} we can write

X=Y14X=Y^{-\frac{1}{4}} (28.24)

And thus that

P(Yy)\displaystyle P\left(Y\leqq y\right) ={1y14011y0<y1410y14>1\displaystyle=\begin{cases}1&y^{-\frac{1}{4}}\leqq 0\\ 1-\frac{1}{y}&0<y^{-\frac{1}{4}}\leqq 1\\ 0&y^{-\frac{1}{4}}>1\end{cases} (28.25)
={1y11yy< and y10y<1\displaystyle=\begin{cases}1&y\geqq\infty\\ 1-\frac{1}{y}&y<\infty\text{ and }y\geqq 1\\ 0&y<1\end{cases} (28.26)
={11yy10y<1\displaystyle=\begin{cases}1-\frac{1}{y}&y\geqq 1\\ 0&y<1\end{cases} (28.27)

Note that here it is assumed that 10=\frac{1}{0}=\infty (it makes it nice and easy to work with the bounds).

As we now have the cumulative probability function for YY, the final step is to differentiate to get the probability density function.

ddx[P(Yy)]\displaystyle\frac{d}{dx}\left[P\left(Y\leqq y\right)\right] ={ddx[11y]y10y<1\displaystyle=\begin{cases}\frac{d}{dx}\left[1-\frac{1}{y}\right]&y\geqq 1\\ 0&y<1\end{cases} (28.28)
={ddx[y1]y10y<1\displaystyle=\begin{cases}\frac{d}{dx}\left[-y^{-1}\right]&y\geqq 1\\ 0&y<1\end{cases} (28.29)
={y2y10y<1\displaystyle=\begin{cases}y^{-2}&y\geqq 1\\ 0&y<1\end{cases} (28.30)
=fY(y)\displaystyle=f_{Y}(y) (28.31)