Holder conditional inequality

Question

we consider, on a probability space $(\Omega,\mathcal{A},P)$, two random variable $X$ and $Y$ and let $\mathcal{H} \subset \mathcal{A}$ be a $\sigma$-algebra. Let $p,q>1$ such that $\frac{1}{p}+\frac{1}{q}=1.$ Prove :$$E[|XY||\mathcal{H}] \leq (E[|X|^p|\mathcal{H}])^{1/p}(E[|Y|^q|\mathcal{H}])^{1/q}$$

I tried to use the Holder inequality for integral, I mean if we have : $$\forall B \in \mathcal{H},\int_B|XY|dP \leq \int_B(E[|X|^p|\mathcal{H}])^{1/p}(E[|Y|^q|\mathcal{H}])^{1/q}dP$$ then the problem is solved.

So I am stuck in proving the integral inequality.

Answer 1

OH! I came up a similar but much shorter proof:

Let $\epsilon\geq 0$, define $U_{\epsilon}:=\Big[\mathbb{E}(|X|^{p}|\mathcal{F})+\epsilon\Big]^{1/p}$ in $L_{p}(\Omega,\mathcal{F},\mathbb{P})$ and $V_{\epsilon}:=\Big[\mathbb{E}(|Y|^{q}|\mathcal{F})+\epsilon\Big]^{1/q}$ in $L_{q}(\Omega,\mathcal{F},\mathbb{P})$. Then, for each $\epsilon>0$, both $U_{\epsilon}$ and $V_{\epsilon}$ are uniformly bounded by $0$ from below.

Recall from the proof of regular Holder that we have, for all $x,y\geq 0$, that $$\dfrac{x^{p}}{p}+\dfrac{y^{q}}{q}-xy\geq 0,$$ for which we consider the first two derivatives in $x$ of the function on the LHS above.

This implies that for all $\omega$ and $\epsilon>0$, we have $$\Big|\dfrac{X(\omega)Y(\omega)}{U_{\epsilon}(\omega)V_{\epsilon}(\omega)}\Big|\leq\dfrac{1}{p}\Big|\dfrac{X(\omega)}{U_{\epsilon}(\omega)}\Big|^{p}+\dfrac{1}{q}\Big|\dfrac{Y(\omega)}{V_{\epsilon}(\omega)}\Big|^{q}.$$

Since both $1/U_{\epsilon}$ and $1/V_{\epsilon}$ are uniformly bounded, the expectation of both sides conditional upon $\mathcal{F}$ is well defined, and then it follows from the monotonicity that for almost every $\omega$, we have $$\dfrac{\mathbb{E}(|XY||\mathcal{F})}{U_{\epsilon}V_{\epsilon}}\leq\dfrac{1}{p}\dfrac{U_{0}^{p}}{U_{\epsilon}^{q}}+\dfrac{1}{q}\dfrac{V_{0}^{q}}{V_{\epsilon}^{q}}\leq\dfrac{1}{p}+\dfrac{1}{q}=1.$$

To finish, multiply both side by $U_{\epsilon}V_{\epsilon}$, and take $\epsilon\searrow 0$, then we have the desired inequality $$\mathbb{E}(|XY||\mathcal{F})\leq U_{0}V_{0}.$$

Answer 2

This is a pretty long proof, and note that we always have a "pulling-out" property

If $X\in\mathcal{F}$, $\mathbb{E}|Y|<\infty$, and $\mathbb{E}|XY|<\infty$, then $\mathbb{E}(XY|\mathcal{F})=X\mathbb{E}(Y|\mathcal{F})$.

Also a terminology that

when I say a random variable $X\in\mathcal{F}$ where $\mathcal{F}$ is a $\sigma-$algebra, I mean $X$ is $\mathcal{F}-$measurable.

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Proof:

Firstly, note that since by hypothesis that $|X|^{p}$ and $|Y|^{q}$ are both integrable, the conditional expectation $\mathbb{E}\Big(|X|^{p}\Big|\mathcal{F}\Big)$ and $\mathbb{E}\Big(|Y|^{q}\Big|\mathcal{F}\Big)$ make sense. Also, this hypothesis immediately gives us the regular H\"older, and thus $|XY|$ is integrable, so $\mathbb{E}\Big(|XY|\Big|\mathcal{F}\Big)$ makes sense.

Set $U:=\mathbb{E}\Big(|X|^{p}\Big|\mathcal{F}\Big)^{1/p}$ and $V:=\mathbb{E}\Big(|Y|^{q}\Big|\mathcal{F}\Big)^{1/q}$. Note that by definition of conditional expectation, $U, V\in\mathcal{F}$.

As usually, we firstly consider the case where $U=0$ or $V=0$. Define the set $$A:=\{\omega\in\Omega:U(\omega)=0\},$$ which is clearly in $\mathcal{F}$, and thus $\mathbb{1}_{A}\in\mathcal{F}$. Also by hypothesis $|X|^{p}\in\mathcal{F}$, and thus $|X|^{p}\mathbb{1}_{A}\in\mathcal{F}$, and thus $|X|^{p}\mathbb{1}_{A}=\mathbb{E}(|X|^{p}\mathbb{1}_{A}|\mathcal{F}),$ but $\mathbb{1}_{A}\in\mathcal{F}$, so by the pulling-out property, we have $\mathbb{E}(|X|^{p}\mathbb{1}_{A}|\mathcal{F})=\mathbb{1}_{A}\mathbb{E}(|X|^{p}|\mathcal{F})$, but $\mathbb{E}(|X|^{p}|\mathcal{F})=U^{p}$, so combining these results and taking expectations yield us $$\mathbb{E}(|X|^{p}\mathbb{1}_{A})=\mathbb{E}\Big(\mathbb{E}(|X|^{p}\mathbb{1}_{A}|\mathcal{F})\Big)=\mathbb{E}\Big(\mathbb{1}_{A}\mathbb{E}(|X|^{p}|\mathcal{F})\Big)=\mathbb{E}(\mathbb{1}_{A}U^{p})=0,$$ which implies $|X|\mathbb{1}_{A}=0$ a.s, which implies, if using the pulling-out property again, that $$\mathbb{E}(|XY|\Big|\mathcal{F})\mathbb{1}_{A}=\mathbb{E}(|XY|\mathbb{1}_{A}\Big|\mathcal{F})=0.$$

The same argument can be applied to $B:=\{\omega\in\Omega:V(\omega)=0\}$ and to conclude that $$\mathbb{E}(|XY|\Big|\mathcal{F})\mathbb{1}_{B}=0.$$

It therefore suffices to show that $$\mathbb{E}(|XY|\Big|\mathcal{F})\mathbb{1}_{H}\leq UV,$$ where $H:=\{\omega\in\Omega:U(\omega)>0,\ V(\omega)>0\}$.

To proceed, we need to prove a lemma

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(Lemma) if $U$ and $V$ are $\mathcal{F}-$measurable random variable, such that $\mathbb{E}(U\mathbb{1}_{A})\leq \mathbb{E}(V\mathbb{1}_{A})$ for all $A\in\mathcal{F}$, then $U\leq V$ a.s. If $\mathbb{E}(U\mathbb{1}_{A})=\mathbb{E}(V\mathbb{1}_{A})$ for all $A\in\mathcal{F}$, then $U=V$ a.s.

Proof of Lemma:

Indeed, by reversing the role of $U$ and $V$, we see that the second claim follows from the first one. To prove the first part, define $A:=\{\omega\in\Omega:U(\omega)> V(\omega)\}$ which is clearly in $\mathcal{F}$. Then, $$0\leq \mathbb{E}\Big[(U-V)\mathbb{1}_{A}]=\mathbb{E}(U\mathbb{1}_{A})-\mathbb{E}(V\mathbb{1}_{A})\leq 0,$$ which implies that $\mathbb{E}\Big[(U-V)\mathbb{1}_{A}\Big]=0$, but $(U-V)\mathbb{1}_{A}> 0$ which together implies that $$(U-V)\mathbb{1}_{A}=0\ \text{a.s.}$$ and thus $\mathbb{P}(A)=0$.

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We now want to use the above lemma to show that $$\dfrac{\mathbb{E}(|XY|\Big|\mathcal{F})\mathbb{1}_{H}}{UV}\leq 1\ \text{a.s.} $$

To show this, we need to show for all $A\in\mathcal{F}$, we have $$\mathbb{E}\Big[\dfrac{\mathbb{E}(|XY|\Big|\mathcal{F})\mathbb{1}_{H}}{UV}\mathbb{1}_{A}\Big]\leq\mathbb{E}(\mathbb{1}_{A}).$$

To evaluate the LHS, let $A\in\mathcal{F}$ and define $G:=A\cap H$, then using the pulling out property, we have $$LHS=\mathbb{E}\Big[\mathbb{E}\Big(\dfrac{|XY|}{UV}\mathbb{1}_{G}|\mathcal{F}\Big)\Big],$$ and then by the second criterion of conditional property, we have $$\mathbb{E}\Big[\mathbb{E}\Big(\dfrac{|XY|}{UV}\mathbb{1}_{G\cap A}|\mathcal{F}\Big)\Big]=\mathbb{E}\Big(\dfrac{|XY|}{UV}\mathbb{1}_{G}\Big)=\mathbb{E}\Big(\dfrac{|X|}{U}\mathbb{1}_{G}\cdot\dfrac{|Y|}{V}\mathbb{1}_{G}\Big),$$ now treat $U$ and $V$ as simply constant numbers and apply regular H\"older, we have $$\Big(\dfrac{|X|}{U}\mathbb{1}_{G}\cdot\dfrac{|Y|}{V}\mathbb{1}_{G}\Big)\leq\Big(\mathbb{E}\Big(\dfrac{|X|^{p}}{U^{p}}\mathbb{1}_{G}\Big)^{1/p}\Big(\mathbb{E}\Big(\dfrac{|Y|^{q}}{V^{q}}\mathbb{1}_{G}\Big)^{1/q},$$ use the second criterion of conditional property, we have $|X|^{p}=\mathbb{E}(|X|^{p}|\mathcal{F})$ and $|Y|^{p}=\mathbb{E}(|Y|^{p}|\mathcal{F}),$ so that $$\Big(\mathbb{E}\Big(\dfrac{|X|^{p}}{U^{p}}\mathbb{1}_{G}\Big)^{1/p}\Big(\mathbb{E}\Big(\dfrac{|Y|^{q}}{V^{q}}\mathbb{1}_{G}\Big)^{1/q}=\Big(\mathbb{E}\Big(\dfrac{\mathbb{E}(|X|^{p}|\mathcal{F})}{U^{p}}\mathbb{1}_{G}\Big)^{1/p}\Big(\mathbb{E}\Big(\dfrac{\mathbb{E}(|Y|^{p}|\mathcal{F})}{V^{q}}\mathbb{1}_{G}\Big)^{1/q},$$ but notice that the numerator is exactly the denominator, and thus we have $$LHS\leq \mathbb{E}(\mathbb{1}_{G})^{1/p}\mathbb{E}(\mathbb{1}_{G})^{1/q}=\mathbb{E}(\mathbb{1}_{G})\leq\mathbb{E}(\mathbb{1}_{A})\ \text{as desired.}$$