40 Method of Transformations

Suppose we wish to find the distribution funciton for the random variable \(Y\) that is either a strictly increasing or strictly decreasing function (Such a function is sure to have an inverse, whereas a function like \(Y=X^2\) does not have an inverse).. If we know the distribution of \(X\), we may use the following to determine the cdf of \(Y\).

\[\begin{aligned} P(Y\leq y) &= P(h(X)\leq y) \\ &= P(h^{-1}(h(y))\leq h^{-1}(x)) \\ &= P(Y\leq h^{-1}(x)) \\ \Rightarrow F_Y(y) &= F_X(h^{-1}(x)) \end{aligned}\]

The pdf can now be found by taking the deriviative of the cdf.

\[\begin{aligned} f_Y(y) &= \frac{d(F_Y(y))}{d y} \\ &= \frac{d F_Y(h^{-1}(y))}{d y} \\ &= f_X(h^{-1}(y))\frac{d(h^{-1}(y))}{dy} \end{aligned}\]

40.1 Example: Cauchy Distribution

Let \(X\) have the Uniform pdf

\[f(x)=\left\{ \begin{array}{ll} \frac{1}{\pi}, & \frac{-\pi}{2}<x<\frac{\pi}{2}\\ 0 & otherwise \end{array}\right. \]

Let \(Y = \tan (X)\). The pdf of \(Y\) can be found as follows:

\[\begin{aligned} h(x) &= \tan (x) \\ \Rightarrow h^{-1}(x) &= \tan^{-1}(x) \\ \Rightarrow \frac{d h^{-1}(x)}{d x} &= \frac{1}{1+x^2} \\ \Rightarrow f_Y(y) &= f_X(h^{-1}(y))\frac{d(h^{-1}(y))}{d y} \\ &= f_X(tan^{-1}(x))\frac{1}{1+y^2} \\ &= \frac{1}{\pi}\frac{1}{1+y^2} \\ &= \frac{1}{\pi(1+y^2)} \end{aligned}\]

The domain of \(Y\) is transformed

\[ \frac{-\pi}{2} < x < \frac{-\pi}{2} \\ \Rightarrow \tan\big(\frac{-\pi}{2}\big) < \tan(x) < \tan\big(\frac{-\pi}{2}\big) \\ \Rightarrow -\infty < y < \infty \]

Thus the pdf of the \(Y\), known as the Cauchy distribution, is

\[f_Y(y) = \frac{1}{\pi(1+y^2)},\ \ -\infty<y<\infty \]