Question

Given \(\int_{C}\left[1 / \pi\left(1+x^{2}\right)\right] d x\), where \(C \subset \mathcal{C}=\\{x:-\infty

Short answer

The integral function \(\frac{1}{\pi\left(1+x^{2}\right)}\) fulfills the necessary conditions for a probability density function, as it is nonnegative and its integral over the range equals to 1. Thus, it can serve as a probability set function for a random variable \(X\) whose space is \(\mathcal{C}\).

Step-by-step solution

Check for non-negativity

Firstly, check if the function is non-negative for all \(x \in \mathcal{C}\): \(1 / \pi\left(1+x^{2}\right)\) is always greater than zero for \(x \in R\). Hence, this function fulfills the non-negativity condition.

Calculate the integral

Calculate the integral of the function over the given range. \[\int_{-\infty}^{\infty}\frac{1}{\pi\left(1+x^{2}\right)} d x\] Recognize this as the integral of the standard Cauchy distribution, which is known to equal 1.

Prove as a Probability Set Function

Since the function satisfies the two necessary conditions for a probability density function - i.e. it is nonnegative and its integral over the range equals to 1 - we can conclude that the given integral could serve as a probability set function of a random variable \(X\) whose space is \(\mathcal{C}\).