4
$\begingroup$

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?

1 Answers 1

2

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

  • 0
    @Dole Sorry I don't have a reference by hand.2018-11-18