# $E(X)=\int_{0}^{\infty}P(X>y)dy$ for non-negative r.v. $X$

First off, I am aware of this answer: Show that $E(X)=\int_{0}^{\infty}P(X\ge x)dx$ for non-negative random variable $X$, but I want to know what is missing in my approach. I would calculate as follows $$E(X)=\int_{-\infty}^{\infty}xP(dx)=\int_{-\infty}^{\infty}\int_{0}^{x}dyP(dx)$$

where I use Fubini with $$0<y<x \text{ and}-\infty<x<\infty$$ giving $$0<y<\infty\text{ and } y<x<\infty$$ hence the above expression becomes $$\int_0^{\infty}\int_y^{\infty}P(dx)dy=\int_0^{\infty}P(X>y)dy$$

Now this is the right solution but nowhere did I (explicitly) use that $X$ is non-negative. Could someone point out what I am missing.

• What is your definition of $\int_0^xdy$ if $x<0$??? It arises in the RHS of your expression for $\mathbb EX$. – drhab Feb 12 '17 at 15:33
• @drhab Still $x$ I suppose. E.g. for $x=-2$: $\int_0^{-2}dy = -\int_{-2}^0dy=-[y]^0_{-2}=-(0+2)=-2$. – azureai Feb 12 '17 at 15:41
• In the second line you are saying to use Fubini with $0<y<x$, but no such $y$ exists if $x<0$. – drhab Feb 12 '17 at 15:47

Since you didn't assume $E|X|<\infty$ you need non-negativity to switch the order of integration. The version of Fubini for non-negative r.v. is sometimes called Tonelli's theorem.

Check out wikipedia's entry on Fubini's theorem and its counterexamples.