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$\mathbf {The \ Problem \ is}:$ If $X$ is a random variable with distribution function $F,$ then show that $E[F[X]]= \frac{1}{2} + \frac{1}{2}\sum_{x} P\{X=x\}.$

$\mathbf {My \ Approach }:$ Actually, if $P_X$ is the probability measure corresponding to $X$, then by Lebesgue Decomposition Theorem, $P_X=P_1+P_2$ where $P_1$ is absolutely continuous and $P_2$ is discrete .

Then $\frac{1}{2}$ is appearing after integrating $F_1$(distribution function corresponding to $P_1$) but how to bring $\frac{1}{2}\sum_{x} P\{X=x\}$ ?

A hint is very much required at this moment .....Thanks in advance .

Rabi Kumar Chakraborty
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1 Answers1

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This is false. If $x$ takes the values $-1$ and $+1$ with probability $\frac 1 2 $ each then $EF(X)=\frac {F(-1)+F(1)} {2}=\frac {\frac 1 2 +1} 2=\frac 3 4$ but $\frac 1 2+\frac 1 2 \sum P(X=x)=1$ for any discetre random variable.

Actually, $\int F(x)dF(x)+ \int F(x-)dF(x)=1$ for any distribution function $F$. And since $F(x)=F(x-) +P(X=x)$, this gives $2\int F(x)dF(x) =1+\int P(X=x)dF(x)$ which implies that $EF(X)=\frac 1 2 +\frac 1 2\sum_x (P(X=x))^{2}$.

Ref: Integration by parts for Lebesgue Steiltjes Integrals in Real and Abstract Analysis by Hewitt and Stromberg.

Kavi Rama Murthy
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