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Is there a conditional Markov inequality? I mean, assume that $X$ is a random variable on the probability space $(\Omega, \mathcal{A}, \mathbb{P})$, and $X\geqslant 0$. Is then

$$\mathbb{P}(X\geqslant a \mid \mathcal{F})\leqslant \frac{1}{a}\mathbb{E}(X\mid \mathcal{F}),$$

with $\mathcal{F}\subseteq \mathcal{A}$ and $a> 0$?

I tried to prove that by looking at the inequality $a\mathbb{1}_{X\geqslant a}\mathbb{1}_A\leqslant X\mathbb{1}_A$ for all $A\in \mathcal{F}$. If I use the expectation of this inequality, the desired result follows. Is this correct?

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1 Answer 1

Yes, that's the conditional Markow inequality and your proof is fine (at least for $a>0$; for $a=0$ the expression $\frac{1}{a}$ doesn't make sense at all). There is a (slight) generalisation:

$$\mathbb{P}(X>Y \mid \mathcal{F}) \leq \frac{\mathbb{E}(X \mid \mathcal{F})}{Y}$$

for $Y \in L^1(\mathcal{F}), Y>0$.

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In your generalized version, what does divide by Y imply? Is there a pointer that you can give for this general version? –  TenaliRaman Nov 19 '12 at 21:23
    
What do you mean by "what does divide by Y imply?"? We have $Y>0$, so it's no problem to divide by Y...? (And the proof for this generalisation is exactly the same.) –  saz Nov 20 '12 at 16:20
    
I didn't mean that, I probably didn't write my question correctly. I am just asking a very basic question here :-). When you say $Y \in L^1(\mathcal{F})$, the $L^1$ here is the Lebesgue space right, so it is a function correct? So, I think I understand what $\mathbb{P}(X > Y|\mathcal{F})$ means, but I don't understand the RHS. –  TenaliRaman Nov 20 '12 at 16:46
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Ah, okay, i see. $Y$ is a random variable, yes. The point is that $\mathbb{P}(X>Y|\mathcal{F})$ is not a constant, but also a random variable $\Omega \ni w \mapsto \mathbb{P}(X>Y|\mathcal{F})(w)$! So the equation means that $\mathbb{P}(X>Y|\mathcal{F})(w) \leq \frac{\mathbb{E}(X|\mathcal{F})(w)}{Y(w)}$ for $\mathbb{P}$-almost all $w \in \Omega$. –  saz Nov 20 '12 at 16:58
    
"The point is that $\mathbb{P}(X > Y|\mathcal{F})$ is not a constant, but also a random variable" --> This is going to take me several days to digest properly :-). Thanks for the clarification! –  TenaliRaman Nov 20 '12 at 17:06

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