91Ó°ÊÓ

The random variable X has a continuous distribution for which the pdf is as shown below.

\(f\left( x \right) = \left\{ {\begin{aligned}{{}{}}{{e^{ - x}}}&{{\rm{for }}x > 0}\\0&{{\rm{otherwise}}}\end{aligned}} \right.\)

We know that:

\(\begin{aligned}{}F\left( x \right) &= \int\limits_0^x {f\left( x \right)dx} \\ &= \int\limits_0^x {{e^{ - x}}dx} \\ &= 1 - {e^{ - x}}.\end{aligned}\)

Then, the cumulative distribution function of X is \(F\left( x \right) = \left\{ {\begin{aligned}{{}{}}0&{{\rm{for }}x < 0}\\{1 - {e^{ - x}}}&{{\rm{for }}x > 0}\end{aligned}} \right..\)

Let the median of X be m. Then, we have:

\(\begin{aligned}{}\Pr \left( {X \le m} \right) = \frac{1}{2}\\\,\,\, \Rightarrow 1 - {e^{ - m}} = \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\, \Rightarrow {e^{ - m}} = \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow m = \ln 2\end{aligned}\)

Similarly:

\(\begin{aligned}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\Pr \left( {X \ge m} \right) = \frac{1}{2}\\ \Rightarrow 1 - \Pr \left( {X < m} \right) = \frac{1}{2}\\\,\,\,\,\, \Rightarrow 1 - \left( {1 - {e^{ - m}}} \right) = \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {e^{ - m}} &= \frac{1}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow m &= \ln 2.\end{aligned}\)

Thus, \(m = \ln 2\) satisfies both the conditions of a median. Hence, \(m = \ln 2\) is the median of X.