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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: $$ F(x)= \begin{cases} 0,\quad &x<-3\\ 0.1, &-3<=x<0\\ 0.3, &0<=x<2.2\\ 0.6, &2.2<=x<3\\ 0.7, &3<=x<4\\ 1,& 4<=x \end{cases} $$

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If this is a homework problem, please add the homework tag. Can you find the values that the random variable $F(X)$ takes on? For example, if $X = 3$, what is the value of $F(3)$? You should be able to arrive at the conclusion that $F(X)$ is a discrete random variable taking on values in $[0,1]$ with various probabilities and then be able to compute the mean of $F(X)$. –  Dilip Sarwate Dec 19 '12 at 18:01
    
@LittleRee Can you edit your addition into the original question, rather than posting it as a comment? Comments don't format well. –  Jonathan Christensen Dec 19 '12 at 18:43

1 Answer 1

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 $$ \mathbb E(F(X))=\frac12\left(1+\alpha\right),\qquad\text{with}\ \alpha=\underline{\ \ \ \ \ \ \ \ }. $$

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