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 $c$.

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*}

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\end{align*}
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