# Condition that a Function Be a Probability Density Function

## Problem 756

Let $c$ be a positive real number. Suppose that $X$ is a continuous random variable whose probability density function is given by
\begin{align*}
f(x) = \begin{cases}
\frac{1}{x^3} & \text{ if } x \geq c\\
0 & \text{ if } x < c. \end{cases} \end{align*} (a) Determine the value of $c$.

(b) Find the probability $P(X> 2c)$.

## Solution.

### Solution of (a)

As $f(x)$ is a probability density function of a continuous random variable $X$, its integral must sum to 1, that is,
$\int_{-\infty}^{\infty} f(x) dx = 1.$ (This follows from $1=P(X\in (-\infty, \infty)) = \int_{-\infty}^{\infty} f(x) dx$.)

As $f(x)=1/x^3$ when $x \geq c$ and $f(x) = 0$ when $x < c$, we have \begin{align*} \int_{-\infty}^{\infty} f(x) dx & = \int_{-\infty}^c f(x) dx + \int_{c}^{\infty} f(x) dx\6pt] &=0 + \int_{c}^{\infty} \frac{1}{x^3} dx\\[6pt] &=\left[\frac{x^{-2}}{-2}\right]_c^{\infty}\\[6pt] &= \frac{1}{2c^2}. \end{align*} Since this must be equal to 1, we obtain \[1 = \frac{1}{2c^2} and thus $c = \frac{1}{\sqrt{2}},$ as $c$ is positive.

### Solution of (b)

In part (a), we found $c = 1/ \sqrt{2}$, hence $2c=\sqrt{2}$.

As $f(x)$ is the probability density function of the continuous random variable $X$, we have
$P(X > \sqrt{2}) = \int_{\sqrt{2}}^\infty f(x) dx.$

So, the desired probability can be computed as follows.
\begin{align*}
P(X>\sqrt{2}) &= \int_{\sqrt{2}}^{\infty} \frac{1}{x^3} dx\\[6pt] &= \left[\frac{x^{-2}}{-2}\right]_{\sqrt{2}}^{\infty}\\[6pt] &= \frac{1}{4}.
\end{align*}

Let $X$ and $Y$ be geometric random variables with parameter $p$, with $0 \leq p \leq 1$. Assume that $X$...