A calculus of the absurd

21.6 Transformations of continuous random variables

Imagine we have a random variable \(X\), with probability density function

\begin{equation} f_X(x) = \begin{cases} 4x^3 & 0 < x \leqq 1 \\ 0 & \text {otherwise} \end {cases} \end{equation}

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

\begin{equation} Y = \frac {1}{X^4} \end{equation}

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

\begin{align} F_X(x) & = \begin{cases} 0 & x \leqq 0 \\ \int _0^{x} 4x^3 dx & 0 < x \leqq 1 \\ 1 & x > 1 \end {cases} \\ & = \begin{cases} 0 & x \leqq 0 \\ x^4 & 0 < x \leqq 1 \\ 1 & x > 1 \end {cases} \end{align}

Then, we can find the cumulative probability function for \(Y\) in terms of the cumulative probability function for \(X\). From the definition of the cumulative probability function we know that

\begin{equation} F_Y(y) = P(Y \leqq y) \end{equation}

Then as \(Y=\frac {1}{X^4}\) we can substitute \(X\) for \(Y\)

\begin{equation} P(Y \leqq y) = P(\frac {1}{X^4} < y) \end{equation}

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

\begin{align} P\left (Y \leqq y\right ) & = P\left (\frac {1}{X^4} < y\right ) \\ & = P\left (X^4 > \frac {1}{y}\right ) \\ & = P\left (X > \sqrt [4]{\frac {1}{y}}\right ) \\ & = 1- P\left (X < \sqrt [4]{\frac {1}{y}}\right ) \end{align}

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 \(F_X(x)\).

\begin{align} P\left (Y \leqq y\right ) & = 1- P\left (X < \sqrt [4]{\frac {1}{y}}\right ) \\ & = 1- F_X\left (\sqrt [4]{\frac {1}{y}}\right ) \\ & = \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} \\ & = \begin{cases} 1 & x \leqq 0 \\ 1- \frac {1}{y} & 0 < x \leqq 1 \\ 0 & x > 1 \end {cases} \end{align}

We also need to rewrite bounds in terms of \(y\), rather than \(x\). As \(Y=\frac {1}{X^4}\) we can write

\begin{equation} X = Y^{-\frac {1}{4}} \end{equation}

And thus that

\begin{align} P\left (Y \leqq y\right ) & = \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} \\ & = \begin{cases} 1 & y \geqq \infty \\ 1- \frac {1}{y} & y < \infty \text { and } y \geqq 1 \\ 0 & y < 1 \end {cases} \\ & = \begin{cases} 1- \frac {1}{y} & y \geqq 1 \\ 0 & y < 1 \end {cases} \end{align}

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

As we now have the cumulative probability function for \(Y\), the final step is to differentiate to get the probability density function.

\begin{align} \frac {d}{dx} \left [P\left (Y \leqq y\right )\right ] & = \begin{cases} \frac {d}{dx} \left [1- \frac {1}{y}\right ] & y \geqq 1 \\ 0 & y < 1 \end {cases} \\ & = \begin{cases} \frac {d}{dx} \left [ - y^{-1} \right ] & y \geqq 1 \\ 0 & y < 1 \end {cases} \\ & = \begin{cases} y^{-2} & y \geqq 1 \\ 0 & y < 1 \end {cases} \\ & = f_Y(y) \end{align}