# Chebyshev's Inequality for conditional expectation

This is based on Durrett's 5.1.3

Prove Chebyshev's inequality. If $a > 0$ then $$\mathbb{P}(\lvert X \rvert \geq a | \mathcal{F}) \leq a^{-2}\mathbb{E}(X^2 | \mathcal{F})$$

First, I need to establish $X^2 \in L^1(\Omega, \Sigma, \mathbb{P})$, so the inequality is possible to have any meaning (otherwise functions are not defined). And i suppose $X \in L^1$, so the left side is defined.

But, following $L^1 \subseteq L^2$? I can't deduce anything about $X^2$.

Should I just suppose $X^2 \in L^1(\Omega, \Sigma, \mathbb{P})$? Or Durrett works in $L^2$?

We just solve problems from the book, so did during my fast-forward search I missed this assumption?

• There's a general formula: If $\Omega$ has finite measure and $X$ is in $L^q(\Omega)$, then for $1\leq p<q<\infty$ we have $\lVert X\rVert_p\leq P(\Omega)^{\frac{q-p}{pq}} \lVert X\rVert_q$ which implies $L^q(\Omega)\subset L^p(\Omega)$. Does my comment help a little bit? – Fakemistake Feb 15 '18 at 10:14
• Not really, I know about this inclusion. But in this specific question, no assumption is made regarding $X^2$ - this is what bothers me. – dEmigOd Feb 15 '18 at 10:22
• In the chebyshev inequality you have to assume $X\in L^2(\Omega)$, which implies ...? – Fakemistake Feb 15 '18 at 10:27
• I don't think this is true. "Naked" Chebyshev need not even $X \in L^1$, cause then rhs is infinite, but the inequality is good. – dEmigOd Feb 15 '18 at 10:53

Notice that: \begin{align} a\mathbf{1}_{|X|\geq a}\leq |X| \end{align} Squaring preserves the inequality since both sides are positive: \begin{align} a^2\mathbf{1}_{|X|\geq a}\leq |X|^2=X^2 \end{align} where $(\mathbf{1}_{|X|\geq a})^2=\mathbf{1}_{|X|\geq a}$ is used. By monotonicity and the linearity of the conditional expectation we have: \begin{align} a^2\E[\mathbf{1}_{|X|\geq a} \mid\mathcal F]\leq \E[X^2\mid \mathcal F] \end{align} Dividing both sides with $a^2$ yields: \begin{align} \P(|X|\geq a\mid\mathcal F)\leq a^{-2}\E[X^2\mid\mathcal F] \end{align}
• "Trivially integrable" only in case of $\lvert X \rvert$ - measurable. But just taking Conditional expectation on some arbitrary r.v. could be undefined (this is the main concern)? – dEmigOd Feb 25 '18 at 8:04