I'm stuck on the following:
Consider a random variable $X$ whose probability mass function is given by:
$$
p(x)=
\begin{cases}
0.1,\quad &x=-3\\
0.2, &x=0\\
0.3, &x=2.2\\
0.1, &x=3\\
0.3, &x=4\\
0,&\text{otherwise}
\end{cases}
$$
Let $F(x)$ be the corresponding cdf. Find $E(F(X))$.
Thanks.
Thanks for the edits, Stefan. This is not a homework problem. I'm studying for my P1 exam after being out of school for some time.
So far, I have:
$$
F(x)=
\begin{cases}
0,\quad &x<-3\\
0.1, &-3<=x<0\\
0.3, &0<=x<2.2\\
0.6, &2.2<=x<3\\
0.7, &3<=x<4\\
1,& 4<=x
\end{cases}
$$
Best Answer
Compare $$ \mathbb E(F(X))=\sum_xp(x)F(x)=\sum_xp(x)\sum_{y\leqslant x}p(y) $$ with $$ 1=\sum_xp(x)\cdot\sum_yp(y)=2\sum_xp(x)\sum_{y\leqslant x}p(y)-\sum_xp(x)^2 $$ to deduce that $$ \mathbb E(F(X))=\frac12\left(1+\alpha\right),\qquad\text{with}\ \alpha=\underline{\ \ \ \ \ \ \ \ }. $$