Chapter 8, Problem 1CRE

To determine

To find:

The value of $P\left(X\le 1\right)$.

Answer to Problem 1CRE

$P\left(X\le 1\right)=\frac{3}{4}$

Explanation of Solution

Given information:

A probability density function is given by $p\left(x\right)=\frac{1}{\pi \left(x^2+1\right)}$ on the interval $\left(-\infty ,\infty \right)$.

Formula used:

A probability density function $p\left(x\right)$ of a random variable $X$, satisfies the following condition.

$P\left(a\le X\le b\right)={\displaystyle {\int }_a^{b}p\left(x\right)}dx=F\left(b\right)-F\left(a\right)$ such that $F$ is an antiderivative of $p$.

The indefinite integral can be calculated as

${\displaystyle \int \frac{dx}{1+x^2}={\tan }^{-1}x+C}$

Calculation:

Let us consider the given probability density function.

$p\left(x\right)=\frac{1}{\pi \left(x^2+1\right)}$

As $p\left(x\right)$ is a probability density function for the random variable $X$, so

$\begin{array}{c}P\left(X\le 1\right)={\displaystyle {\int }_{-\infty }^{1}p\left(x\right)}dx\\ =\underset{A\to -\infty }{\lim }{\displaystyle {\int }_A^1\frac{1}{\pi \left(x^2+1\right)}}dx\\ =\frac{1}{\pi }\underset{A\to -\infty }{\lim }{\displaystyle {\int }_A^1\frac{1}{\left(x^2+1\right)}dx}\\ =\frac{1}{\pi }\underset{A\to -\infty }{\lim }{\left[{\tan }^{-1}\left(x\right)\right]}_A^1\\ =\frac{1}{\pi }\underset{A\to -\infty }{\lim }\left[{\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(A\right)\right]\\ =\frac{1}{\pi }\left[{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(\infty \right)\right]\\ =\frac{1}{\pi }\left[\frac{\pi }{4}+\frac{\pi }{2}\right]\\ =\frac{3}{4}\end{array}$

Therefore, value of $P\left(X\le 1\right)$ is $\frac{3}{4}$.