3
$\begingroup$

In general, for a given random variable $X : \Omega \rightarrow \mathbb{R}$, the expectation is defined as $$ E[X] := \int_\Omega X dP, $$ where $P$ is a probability measure on $\Omega$. For discrete RVs, there's a simpler definition for the expectation, namely $$ E[X] := \sum_{x \in X(\Omega)}x P[X=x]. $$ Obviously the first definition should coincide with the second one if $X$ is discrete. If $X$ is simple (i.e. takes on only finitely many values), the result follows immediately from the definition of the Lebesgue integral on simple functions. However, if $X$ takes on (countable) infinitely many values (e.g. $X$ is a geometrically distributed), then it's not clear to me how one arrives at the 2nd definition, given the 1st one. In that case we get $$E[X] = \sup\{ \int_\Omega Y dP \mid \text{$Y$ is simple and $Y\leq X$} \},$$ but I couldn't see why this is equivalent to the infinite sum.

$\endgroup$
2
$\begingroup$

If $Y$ is non-negative, discrete, and integrable, let $\{y_n\}_{n=1}^{+\infty}$ the values taken by $Y$. Define $S_N:=\sum_{n=1}^Ny_n\chi_{Y^{-1}\{y_n\}}$: it's a simple function, and $0\leq S_N\leq Y$. So $E[Y]\geq \sum_{n=1}^Ny_nP(Y=y_n)$ for all $N$, which gives $E[Y]\geq \sum_{n=1}^{+\infty}y_nP(Y=y_n)$. We can apply monotone convergence theorem to get the other direction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.