# Fubini's theorem and uncorrelated/independent random variables

Suppose $X$ and $Y$ are two real-valued random variables, and $f:\mathbb{R}^2\to \mathbb{R}$ is Borel measurable.

I was wondering if $X$ and $Y$ being uncorrelated or independent implies that $$\mathrm{E}_{X,Y} f(X,Y) = \mathrm{E}_X [\mathrm{E}_Y f(X,Y)] = \mathrm{E}_Y [\mathrm{E}_X f(X,Y)]?$$ Thanks and regards!

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The hard part of answering your question will be figuring out what your notation means. –  Michael Hardy Feb 11 '12 at 21:08
I might be abusing notations, but I am not sure. Could you let me know what you think? –  Tim Feb 11 '12 at 21:10
I deleted the comment and made it an answer, sorry. The random variables being merely uncorrelated is not sufficient. It is not hard to construct a counterexample from whatever your favorite uncorrelated not-independent random variables are. –  Chris Janjigian Feb 11 '12 at 21:20
@Chris: Thanks! I was wondering what causes ambiguity to you and/or Michael in my notations? –  Tim Feb 13 '12 at 5:13
Generally $E_x$ refers to the conditional expectation given that $X = x$ rather than the expectation with respect to the marginal law of $X$. –  Chris Janjigian Feb 13 '12 at 5:21

Independence is equivalent to the measure on the product space being the product measure. This is actually just a restatement of the definitions: $\mu_{X_1,...,X_n}(E_1,..,E_n) = P((X_1,...,X_n) \in (E_1,...,E_n) = \prod P(X_i \in E_i) = \prod \mu_i (E_i) = \mu_1 \times ... \times \mu_n (E_1 \times ... \times E_n)$

This is sufficient for Fubini's Theorem (modulo existence of integrals).

edit: Looking back at this, I suppose I should have been clearer since I may have misinterpreted your question. This is unambiguously true for product measures and could be true depending on what you mean by $E_X$ and $E_Y$ more generally. Let's just move down to the absolutely continuous case to make this a bit clearer: In general, we have $f_{X,Y}(x,y) = f_{X|Y}(x|y) f_Y (y)$ (this does generalize beyond the absolutely continuous case, but the notation is cumbersome). If your question does not involve conditioning then find any random variable for which the conditional probability $f_{X|Y} (x|y)$ does not equal the marginal $f_X (x)$. This is how I interpreted your original question, but that is probably not what you meant.

In light of your comment I will give a standard counterexample. Let $B = 1$ or $0$ each with probability $\frac{1}{2}$ and $D = 1$ or $-1$ each with probability $\frac{1}{2}$. Then $A = BD$ is uncorrelated with $B$, but $E A^2 B$ = $\frac{1}{2} \neq E_A E_B A^2 B = E_A A^2 E_B B = \frac{1}{4}$ where I am interpreting $E_B$ to mean the integral with respect to the marginal distribution of $B$.

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+1. Thanks! (1) What does "modulo existence of integrals" mean? (2) How about uncorrelatedness? –  Tim Feb 11 '12 at 21:20
Fubini's theorem is only a theorem when the integrand is absolutely integrable or positive. I'm reasonably certain that the variables being uncorrelated is not enough (as I said above if I recall it isn't hard to construct a counterexample). I'll play with it a bit and see if I come up with one. –  Chris Janjigian Feb 11 '12 at 21:23
Thanks for the edit! (1) My questions do not involve conditioning, just joint distribution and marginal distribution. Is this how you interpreted my questions? (2) I wonder what "If your question does not involve conditioning then find any random variable for which the conditional probability fX|Y(x|y) does not equal the marginal fX(x)" is for? –  Tim Feb 11 '12 at 22:01
Yes, that is how I interpreted it originally. I'll add a counterexample to that to my answer then. That comment was for a way to find a counterexample, but I'll give an easier construction. –  Chris Janjigian Feb 11 '12 at 22:06
Thanks! (1) In the counterexample, do you mean $E e^{AB} \neq E_A [E_B e^{AB}]$ instead? (2) Maybe you have explained, but what does "modulo existence of integrals" mean? –  Tim Feb 11 '12 at 22:33