Proof of Cauchy-Schwarz inequality for conditional expectations

Question

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.

Answer 1

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 $$ \mathbb E[XY|\mathcal A]\leq \mathbb E[X^2|\mathcal A]^{1/2}\mathbb E[Y^2|\mathcal A]^{1/2}. $$

Answer 2

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.