Question

Let \(X\) have the uniform pdf \(f_{X}(x)=\frac{1}{\pi}\), for \(-\frac{\pi}{2}

Short answer

The probability density function of \(Y=\tan X\) when \(X\) has uniform distribution in \(-\pi/2 < x < \pi/2\) is \(f_Y(y) = \frac{1}{\pi(1+y^2)}\) for \(-\infty < y < \infty\), which is the formula for a Cauchy distribution.

Step-by-step solution

Find inverse function

First, identify the inverse function, \(g^{-1}(y)\), which is the function that corresponds to \(x\) when given \(y\). Since \(y = \tan x\), so \(g^{-1}(y) = x = \arctan y\).

Find derivative

The next step is to find the derivative of the inverse function with respect to \(y\), \(\frac{d}{dy} g^{-1}(y)\). Taking the derivative of \(\arctan y\) yields \(\frac{1}{1 + y^2}\).

Substitute into the formula

Now we substitute these values into the formula for the transformation of random variables. The range of \(y\) will be from \(-\infty\) to \(\infty\) as \(\tan x\) approaches \(-\infty\) as \(x\) approaches \(-\frac{\pi}{2}\) from the right and \(\infty\) as \(x\) approaches \(\frac{\pi}{2}\) from the left. The result is \[f_Y(y) = \frac{1}{\pi} \cdot \left| \frac{1}{1+y^2} \right| = \frac{1}{\pi(1+y^2)} \] for \(-\infty < y < \infty\). This is the probability density function of the Cauchy distribution.

Simplify the solution

Finally, we don't need absolute value bars around the derivative as \(1/(1+y^2)\) is always positive. Thus the final form of the pdf of \(Y\) is \[f_Y(y) = \frac{1}{\pi(1+y^2)}, -\infty < y < \infty.\]