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Let $(\Omega, \mathcal{A},P)$ be a probability space and let $X,Y\colon \Omega \rightarrow \mathbb{R}$ be random variables. Furthermore, let $Y$ be $p$-integrable. Then why is the conditional expectation $\mathbb{E}[Y \vert X]$ again necessarily $p$-integrable?

Kind regards and thank you very much!

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By conditional Jensen's inequality and from the fact that $x\mapsto |x|^p$ is convex for $p\ge 1$, we find that $$ \left|\Bbb E\left[Y\big|X\right]\right|^p\le\Bbb E\left[|Y|^p\big|X\right],\quad\text{a.s.} $$ It follows that $$ \Bbb E\left[\left|\Bbb E\left[Y\big|X\right]\right|^p\right]\le \Bbb E\left[\Bbb E\left[|Y|^p\big|X\right]\right]=\Bbb E\left[|Y|^p\right]<\infty. $$

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  • $\begingroup$ Fantastic! Thanks. $\endgroup$ – Joker123 Jan 25 at 15:45
  • $\begingroup$ Does this also hold if $X$ and $Y$ map into $\mathbb{R}^n$? In this case, I can only see how it holds for $p = 2$. $\endgroup$ – Joker123 Jan 25 at 16:11
  • $\begingroup$ I can make sure that the Jensen's inequality holds for any convex(concave) function $\psi:\Bbb R^n \to \Bbb R$, but I cannot find a reference to give you ... But we can prove it componentwise: i.e. for $\mathbf{Y}=(Y_1,Y_2,\ldots , Y_n)\in L^p$, we find that each $Y_i\in L^p \Longrightarrow \Bbb E[Y_i |\mathbf{X}]\in L^p$ and thus $\Bbb E[\mathbf{Y}|\mathbf{X}]=(\Bbb E[Y_1|\mathbf{X}],\ldots,\Bbb E[Y_n|\mathbf{X}])\in L^p$. $\endgroup$ – Song Jan 25 at 16:31

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