Let $X \thicksim U(-\pi/2, \pi/2)$ where $U$ denotes Uniform Distribution.

Then derive the probability density function of random variable $Y = \tan X$

First, since $X$ follows uniform distribution over the range of $(-\pi/2, \pi/2)$ there would be constant $1/\pi$ upon the given range as a probability.

Then by the transformation of $Y = \tan X$ the range mapped into $(-\infty, \infty)$ and the probability would be also become small satisfying $\int_{-\infty}^{\infty}\text{pdf}(Y) dy =1$

In this case, how could one find $\text{pdf}(Y)$ ? I want to think about a specific function that integrated from $-\infty$ to $\infty$ becomes 1 but can't imagine which would be proper.


1 Answer 1


Since $\tan(\cdot)$ is monotonic in $(-\pi/2,\pi/2)$, $$ \{ Y \leq y\} = \{X \leq \arctan(y)\}$$ So $$\mathbb{P}(Y\leq y) = \frac{1}{\pi}\arctan(y)$$ Differentiating both sides $$ f_Y(y) = \frac{\mathrm{d}}{\mathrm{d}y}\mathbb{P}(Y\leq y) = \frac{1}{\pi}\frac{1}{1+y^2}$$


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