Problem source

The random variable $X$ can take the value $X=-1$, and also any value in the range $0\le X <\infty\,$. The distribution of $X$ is given by
\[
\P(X=-1) =m  \,, \ \ \ \ \ \ \ \P(0\le X\le x) = k(1-\e^{-x})\,,
\]
for any non-negative number $x$, where $k$ and $m$ are constants, and $m <\frac12\,$.
\begin{questionparts}
\item Find $k$ in terms of $m$. 
\item Show that $\E(X)= 1-2m\,$. 
\item Find, in terms of $m$, $\var (X)$ and the median value of $X$.
\item Given that 
\[
\int_0^\infty y^2 \e^{-y^2} \d y = \tfrac14 \sqrt{ \pi}\;,\]
find  $\E\big(\vert X \vert^{\frac12}\big)\,$ in terms of $m$.
\end{questionparts}

Solution source

\begin{questionparts}
\item We must have the total probability summing to $1$, therefore $1 =m + k$ (as $x \to \infty$) therefore $k = 1-m$.

\item $\,$ \begin{align*}
&& \E[X] &= \mathbb{P}(X=-1) \cdot (-1) + \int_0^{\infty} kx e^{-x} \d x \\
&&&= -m + (1-m) = 1-2m
\end{align*}

\item $\,$ \begin{align*}
&& \var[X] &= \E[X^2]-\E[X]^2 \\
&&&= \mathbb{P}(X=-1)\cdot(-1)^2 + (1-m)\int_0^{\infty} x^2e^{-x} \d x  - (1-2m)^2 \\
&&&= m + (1-m)(1+1^2) - (1-2m)^2 \\
&&&= 3-4m - 1+4m -4m^2 \\
&&&= 2(1-m^2)
\end{align*}

To find the median $q$, we need
\begin{align*}
&& \frac12 &= \mathbb{P}(X \leq q) \\
&&&= m + (1-m)(1-e^{-q}) \\
\Rightarrow && e^{-q} &= 1-\frac{\frac12-m}{1-m} \\
&&&= \frac{1-m - \frac12+m}{1-m} \\
&&&= \frac{1}{2(1-m)} \\
\Rightarrow && q &= \ln 2(1-m)
\end{align*}

\item $\,$ \begin{align*}
&& \E\left [|X|^{\frac12}\right] &= \mathbb{P}(X=-1) \cdot 1 + \int_0^{\infty} \sqrt{x} (1-m)e^{-x} \d x \\
&&&= m + (1-m)\int_0^\infty \sqrt{x} e^{-x} \d x \\
u^2 = x, \d x = 2u \d u : &&&= m + (1-m) \int_{u=0}^{u=\infty} u e^{-u^2} \cdot 2u \d u \\
&&&= m + 2(1-m) \int_0^{\infty} u^2 e^{-u^2} \d u \\
&&&= m + (1-m)\frac{\sqrt{\pi}}2
\end{align*}
\end{questionparts}