# Expectation of the product of two mean independent random variables

I have random variables $$a$$ and $$b$$, which are such that $$\mathbb{E}[a|b]=0$$. I am trying to compute $$E[ab]$$. Is the following correct? $$\mathbb{E}[ab]=\mathbb{E}[\mathbb{E}[ab|b]]=\mathbb{E}[b\mathbb{E}[a|b]]=0$$

Where I have used the law of iterated expectations. Thank you.

• Hi: The first equality is the law of iterated expectations. It's also called the tower property. Apr 25, 2022 at 11:26
• @markleeds Yes. Have I used it correctly? Apr 25, 2022 at 11:28
• You will need $E[ab]$ to exist: there will be examples where it does not even if $E[a \mid b]=0$, such as $b$ being a Cauchy random variable and independently $a=\pm1$ with equal probability Apr 25, 2022 at 12:01
• Hi Charles: You've used it correctly. Robertas explanation gives all the gory details in quite a nice way. My intuition is that, if you have a conditional expectation and then you take the expectation of that conditioned expectation, then you are kind of taking a mean over the whole conditional density ( so in a sense, averaging the condition ), so you get back the unconditioned expectation of that variable. ( the $b$ comes out ). I don't know if that makes sense to you but that's how I think of it. Apr 26, 2022 at 14:50

Long answer: I'll use notation $$X$$ and $$Y$$ for random variables and $$x$$ and $$y$$ for their values. Let's assume that $$(X,Y)$$ has continuous distribution (for discrete distribution the proof below is similar). Then
$$\mathbb{E}(X\cdot Y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x\cdot y\cdot f(x,y)\, dxdy = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x\cdot y\cdot f_X(x|y)\cdot f_Y(y)\, dxdy =$$ $$=\int_{-\infty}^{\infty}y\left(\int_{-\infty}^{\infty} x\cdot f_X(x|y)\,dx\right)f_Y(y)\, dy = \int_{-\infty}^{\infty}y\cdot \mathbb{E}(X|y)\cdot f_Y(y)\, dy = \mathbb{E}(Y\cdot\mathbb{E}(X|Y))$$ Explanation for the last quality:
1. Keep in mind that $$\mathbb{E}(X|y)$$ is not a random variable (it's conditional expectation of $$X$$ with condition $$Y=y$$). It depends on $$y$$, so it's a function of $$y$$.
2. But in the last integral we have $$y\cdot \mathbb{E}(X|y)$$, which is another function of $$y$$, let's say $$\varphi(y)$$.
3. $$\varphi(y)$$ is not a random variable, but $$\varphi(Y)=Y\cdot\mathbb{E}(X|Y)$$ is random variable and the expectation of that variable is $$\mathbb{E}\varphi(Y) =\int_{-\infty}^{\infty} \varphi(y)\cdot f_Y(y)\, dy = \int_{-\infty}^{\infty}y\cdot \mathbb{E}(X|y)\cdot f_Y(y)\, dy$$ So, if $$\mathbb{E}(X|Y)=0$$ then $$\mathbb{E}(X\cdot Y) = \mathbb{E}(Y\cdot\mathbb{E}(X|Y)) = \mathbb{E}(Y\cdot 0) = 0$$ Bonus Tip: This proof doesn't require $$X$$ and $$Y$$ to be independent.
• The OP doesn't assume $X$ and $Y$ have densities. Apr 25, 2022 at 13:07