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?

  • $\begingroup$ 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? $\endgroup$ – Fakemistake Feb 15 '18 at 10:14
  • $\begingroup$ Not really, I know about this inclusion. But in this specific question, no assumption is made regarding $X^2$ - this is what bothers me. $\endgroup$ – dEmigOd Feb 15 '18 at 10:22
  • $\begingroup$ In the chebyshev inequality you have to assume $X\in L^2(\Omega)$, which implies ...? $\endgroup$ – Fakemistake Feb 15 '18 at 10:27
  • $\begingroup$ 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. $\endgroup$ – dEmigOd Feb 15 '18 at 10:53

$\newcommand{\E}{\mathbb E}\newcommand{\P}{\mathbb P}$I think as long as one talks about the conditional expectation, then one assumes indirectly the existence of it. Otherwise the question would not make sense. Notice that we don't need the integrability of $X$ to define $\P(|X|\geq a \mid\mathcal F)$. We need the integrability of $\mathbf{1}_{|X|\geq a}$ which is trivially integrable.

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}

  • $\begingroup$ "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)? $\endgroup$ – dEmigOd Feb 25 '18 at 8:04
  • $\begingroup$ @dEmigOd could you please reformulate the question? I don't get it. $\endgroup$ – Shashi Feb 25 '18 at 8:58

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