# Holder conditional inequality

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.

• Proving it this way is a little cumbersome and maybe not very intuitive. You could also prove it using regular conditional distributions. See Theorem 6.60 in math.swansonsite.com/19s6245notes.pdf for a proof like you mentioned. See p. 111 in that same link for a proof using regular conditional distributions. Commented May 2, 2019 at 18:33
• @JasonSwanson there is a shorter proof than the link you gave, I've posted an answer. Commented Nov 16, 2019 at 23:36

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 $$\mathbb{E}(|XY||\mathcal{F})\leq U_{0}V_{0}.$$

• Thank you for this nice proof. At the end you get $\mathbb{E}(|XY||\mathcal{F})\leq U_{\epsilon}V_{\epsilon}$ almost surely, for all $\epsilon>0$. To finish I believe we should take $\epsilon=1/n$ for $n=1,2\dots$, and use the fact that a countable union of null sets is null, to obtain $\mathbb{E}(|XY||\mathcal{F})\leq U_{1/n}V_{1/n}$ for each $n$ almost surely (interchanging order). Commented Oct 26, 2021 at 8:34

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.

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

(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 $$\mathbb{P}(A)=0$$.

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 $$LHS\leq \mathbb{E}(\mathbb{1}_{G})^{1/p}\mathbb{E}(\mathbb{1}_{G})^{1/q}=\mathbb{E}(\mathbb{1}_{G})\leq\mathbb{E}(\mathbb{1}_{A})\ \text{as desired.}$$

• This is a nice argument, but your application of the standard Hölder inequality assumes that $\mathbb E\left[|X|^p\mathbb 1_G/U^p\right] < \infty$ and $\mathbb E\left[|Y|^q\mathbb 1_G/V^q\right] < \infty$. Is this obvious? Commented Aug 18, 2020 at 22:31
• @DFord hi Ford, it's been a long time and thanks for your words. This is not that obvious but indeed, as you suggested, is important. I guess this is also why I came up with another proof, you could glance above, the accepted one. In that proof, we do not need to worry about if $U$ and $V$ are $0$. Commented Aug 19, 2020 at 17:48