# Proof of Cauchy-Schwarz inequality for conditional expectations

I'm taking a wild guess and say that this is probably a question with a trivial answer (being declared as an exercise almost everywhere), but as I am short on time and my mind being absolutely fried, I would appreciate any help. Feel free to close/delete this question if I might have skipped over a post haphazardly.

As you can read in the title, I am interested in proving the Cauchy-Schwarz inequality for conditional expectations:

$$\mathbb{E}(XY|\mathcal{A})^2 \leq \mathbb{E}(X^2|\mathcal{A})\mathbb{E}(Y^2|\mathcal{A})$$

for all $$X,Y \in \mathcal{L}^2(\mathbb{P})$$ for some probability measure $$\mathbb{P}$$ and a sub-$$\sigma$$-algebra $$\mathcal{A} \subseteq \mathcal{F}$$.

I have already tried to approach this question by using the regular CS inequality in conjunction with the contraction property of conditional expectations, but that string of inequalities breaks down at the last step. I've also tried to substitute the conditional expectations in the regular inequality, but that also led nowhere.

I am almost surely certain that this is a one line proof, but from studying all day I can't really see it. I have a semester worth of measure theory and stochastics, so feel free to comment/post any approach within reasonable distance of this skill level. Thank you for your attention.

• There are square roots missing on the right side. Jul 20, 2021 at 18:19
• Ah yes, my scripture does miss the square/square root. Thank you. Jul 20, 2021 at 18:23

One can quite easily modify one of the standard proofs of the classical Cauchy-Schwarz inequality. Since $$(X-Y)^2\geq 0$$ we have $$0\leq\mathbb E[(X-Y)^2|\mathcal A]=\mathbb E[X^2|\mathcal A]+\mathbb E[Y^2|\mathcal A]-2\mathbb E[XY|\mathcal A].$$ If we replace $$X$$ by $$\frac{(\mathbb E[Y^2|\mathcal A]+\epsilon)^{1/4}}{(\mathbb E[X^2|\mathcal A]+\epsilon)^{1/4}}X$$ and $$Y$$ by $$\frac{(\mathbb E[X^2|\mathcal A]+\epsilon)^{1/4}}{(\mathbb E[Y^2|\mathcal A]+\epsilon)^{1/4}}Y$$ and note the factors in front fo $$X$$ and $$Y$$ are $$\mathcal A$$-measurable, we get $$\mathbb E[XY|\mathcal A]\leq \frac 1 2\frac{\mathbb E[X^2|\mathcal A](\mathbb E[Y^2|\mathcal A]+\epsilon)^{1/2}}{(\mathbb E[X^2|\mathcal A]+\epsilon)^{1/2}}+\frac 1 2\frac{\mathbb E[Y^2|\mathcal A](\mathbb E[X^2|\mathcal A]+\epsilon)^{1/2}}{(\mathbb E[Y^2|\mathcal A]+\epsilon)^{1/2}}.$$ Letting $$\epsilon\searrow 0$$, we obtain $$\mathbb E[XY|\mathcal A]\leq \mathbb E[X^2|\mathcal A]^{1/2}\mathbb E[Y^2|\mathcal A]^{1/2}.$$

• Oh, not as trivial as I thought! Thank you, that is quite a neat argument. Just out of curiosity: Could one also eliminate the use of $\epsilon$ alltogether by inspecting the edge cases seperately? Maybe with indicator functions that filter all zeroes out and proving both statements seperately? Jul 20, 2021 at 18:51
• Maybe that could work, but I'd expect it to be rather cumbersome. In the case of the usual Cauchy-Schwarz inequality, if one of the factors on the RHS is zero, then it is quite easy to see that the LHS is zero. But in thise case if one factor on the RHS just vanishes at some points, it's not immediately clear that the LHS also vanishes at these points (except for some set of measure zero). That's why I wanted to avoid these nuisances altogether. Jul 20, 2021 at 22:31
• I have just noticed: Shouldn't the exponents on the fractions be $\frac{1}{4}$ instead of $\frac{1}{2}$? Because they are squared when we replace the random variables with the new terms. Mar 21, 2022 at 21:49

Consider a probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$. Given a random vector $$\textbf{X}=(X, Y)$$ and a $$\sigma$$ field $$\mathcal{A} \subset \mathcal{F}$$, a theorem says that the regular conditional probability exist.

Regular conditional probability is a random measure $$P(\omega, A)$$ such that $$P(\omega, \cdot)$$ is a probability in $$(\mathbb{R}^2, \mathcal{B}\{\mathbb{R}^2\})$$, and $$P(\cdot, A) \in \mathcal{A}$$ for every $$A \in \mathcal{B}\{\mathbb{R}^2\}$$, and $$E[h(\textbf{X})|\mathcal{A}](\omega)=\int_{\mathbb{R}^2} h(x,y) P(\omega,dxdy)$$ $$a.s.$$ for every Borel measurable function $$h \in \mathcal{B}\{\mathbb{R}^2\}$$.

Then set $$h_0(x,y)=xy, h_1(x,y)=x^2, h_2(x,y)=y^2$$, we have $$E[h_i(X,Y)|\mathcal{A}](\omega)= \int _{\mathbb{R}^2} h_i(x,y) P(\omega,dxdy)$$ $$a.s.$$. For every $$\omega$$, $$P(\omega, \cdot)$$ is a probability, so Cauchy’s inequality holds: $$\int _{\mathbb{R}^2} xy P(\omega,dxdy)^2 \le \int _{\mathbb{R}^2} x^2 P(\omega,dxdy) \int _{\mathbb{R}^2} y^2 P(\omega,dxdy) .$$ So we have $$E[XY|\mathcal{A}]^2 \le E[X^2|\mathcal{A}]E[Y^2|\mathcal{A}] \quad a.s.$$ This method is also available for proofing Holder’s and Minkowski’s inequality for conditional expectation.

• Thank you for your answer! I know this is a late response, but I'm writing a digital version of my lecture notes and would try to keep the number of theorems we didn't go over (see e.g. your result about existence of conditional expectation of random vectors) as low as possible to make the scripture pretty much self-contained. That makes MaoWao's answer "better". Sorry, but you will still get my upvote anyway! Aug 4, 2021 at 9:34
• @TheOutrageousZ Thank you. It’s right to choose the answer that helps you most, of course. Aug 4, 2021 at 11:57