Problem source

Let $\F(x)$ be the cumulative distribution function of a random variable $X$, which satisfies $\F(a)=0$ and $\F(b)=1$,  where $a>0$.  Let
\[
\G(y) = \frac{\F(y)}{2-\F(y)}\;.
\]
Show that $\G(a)=0\,$, $\G(b)=1\,$ and that $\G'(y)\ge0\,$.
Show also that
\[
\frac12 \le \frac2{(2-\F(y))^2} \le 2\;.
\]
The random variable $Y$ has cumulative distribution function $\G(y)\,$. Show that
\[
{ \tfrac12} \,\E(X) \le \E(Y) \le 2 \E(X) \;,
\]
and that
 \[
\var(Y) \le 2\var(X) +\tfrac 74 \big(\E(X)\big)^2\;.
\]

Solution source

\begin{align*}
&& G(a) &= \frac{F(a)}{2-F(a)}\\
&&&= 0 \tag{$F(a)= 0$}\\
\\
&& G(b) &= \frac{F(b)}{2-F(b)} \\
&&&= \frac{1}{2-1} = 1 \tag{$F(b)=1$}\\
\\
&& G'(y) &= \frac{F'(y)(2-F(y))+F(y)F'(y)}{(2-F(y))^2} \\
&&&= \frac{2F'(y)}{(2-F(y))^2} \geq 0 \tag{$F'(y) \geq 0$}
\end{align*}

\begin{align*}
&& 0 \leq F(y)\leq1\\
\Leftrightarrow&& 1\leq 2-F(y) \leq 2\\
\Leftrightarrow &&1 \leq (2-F(y))^2 \leq 4\\
\Leftrightarrow && 1 \geq \frac{1}{(2-F(y))^2} \geq \frac14 \\
\Leftrightarrow && 2 \geq \frac{2}{(2-F(y))^2} \geq\frac12
\end{align*}

\begin{align*}
&& \mathbb{E}(Y) &= \int_a^b y G'(y) \d y \\
&&&=  \int_a^b y F'(y) \underbrace{\frac{2}{(2-F(y))^2}}_{\in [\frac12, 2]} \d y \\
&&&\leq 2 \E[X] \\
&&&\geq \frac12 \E[X]\\
\\
&& \E[Y^2] &\leq 2\E[X^2] \\
&& \E[Y^2] &\geq \frac12\E[X^2] \\
\\
\Rightarrow && \var[Y] &= \E[Y^2]-\E[Y]^2 \\
&&& \leq 2 \E[X^2] - (\tfrac12\E[X])^2 \\
&&&= 2 \var[X] + \tfrac74(\E[X])^2
\end{align*}