Let $X$ be a non-negative random variable with cummulative distribution function $F_X$. What does $E[X] = \int_0^\infty x\,dF_X(x)$ mean?

The definition that I have for the expectation for a postitive random variable is: $E[X] = \sup\{ E[Y]:Y \text{ is a simple function}, 0 \leq Y \leq X \}$.

And for a simple r.v. we have $E[Y]=\sum_I^ma_iP(A_i)$

I don't see how we got form the definition to $E[X] = \int_0^\infty x\,dF_X(x)$

  • 1
    $\begingroup$ In your definition of $E[X]$ in terms of $E[Y]$, how do you define $E[Y]$? $\endgroup$
    – J.G.
    Mar 3, 2020 at 21:11
  • $\begingroup$ It's in the third line with a prob. space in the back ground (so $A_i$ is an event). And $a_i$ a real number $\endgroup$ Mar 3, 2020 at 21:16
  • $\begingroup$ Can you see how those two definitions are exactly the same when $X$ is simple? $\endgroup$ Mar 3, 2020 at 21:16
  • $\begingroup$ @BrianMoehring Well kind of I only don't understand where $dF_X(x)$ comes from $\endgroup$ Mar 3, 2020 at 21:21

1 Answer 1


Your definition of $E[X]$ is actually the integral with respect to the measure $P$. That is $$E[X] = \int_\Omega X \,dP$$ A well known formula for computing this is $$\int_\Omega X \,dP = \int_{\mathbb{R}} x \,dP_X(x)$$ Where $P_X$ is the distribution of $X$, in other words $P_X(A)=P(X\in A)$. Now the notation $\int_\mathbb{R} x \,dF_X(x)$ refers to the Lebesgue-Stieltjes integral associated with the distribution function $F_X$. It turns out by uniqueness of Lebesgue-Stieltjes measures, that the Lebesgue-Stieltjes measure of $F_X$ is in fact the probability distribution $P_X$.

Therefore the two integrals $\int x \,dF_X(x)$ and $\int x \,dP_X(x)$ are integrals with respect to the same measure, and therefore identical.


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