# What does $E[X] = \int_0^\infty x\,dF_X(x)$ mean?

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 $$E[X] = \int_0^\infty x\,dF_X(x)$$

• In your definition of $E[X]$ in terms of $E[Y]$, how do you define $E[Y]$?
– J.G.
Mar 3, 2020 at 21:11
• 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 Mar 3, 2020 at 21:16
• Can you see how those two definitions are exactly the same when $X$ is simple? Mar 3, 2020 at 21:16
• @BrianMoehring Well kind of I only don't understand where $dF_X(x)$ comes from Mar 3, 2020 at 21:21

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.