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Let a pair of random variables $(X, Y)$ over a finite product space $\mathcal{X}\times \mathcal{Y}$ be given. The conditional expectation operator is defined as $$(T_Yf) (y):=\mathbb{E}[f(X)|Y=y],$$ where $f$ is a real-valued function acting on $\mathcal{X}$.

It is well known the the operator $T$ is contractive in $L^p$-norm for $p\geq 1$, i.e., $$||Tf||_p\leq ||f||_p,$$ which can be easily proved using convexity of the map $t\mapsto t^p$ and Jensen's inequality.

I am looking at the map $p\mapsto \frac{||Tf||_p}{||f||_p}$ for $p\geq 1$. I am trying to see if this map is monotonic or not. Any idea?

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  • $\begingroup$ Does this monotonicity break for an arbitrary contractive operator? $\endgroup$
    – Suvrit
    Mar 17, 2015 at 1:16
  • $\begingroup$ @suvrit, I focus only on conditional expectation operator. For arbitrary contractive operator, I guess, the monotonicity breaks. $\endgroup$
    – math-Student
    Mar 17, 2015 at 2:53
  • $\begingroup$ I asked, because it is not clear from the outset, where would the fact that $T$ is a conditional expectation operator play a role for monotonicity --- it seems to be used mostly for the contractivity. On the other hand, even non-contractive operators can lead to a monotonic ratio, so maybe your hypothesis does play a role... $\endgroup$
    – Suvrit
    Mar 17, 2015 at 3:11
  • $\begingroup$ By "over a finite product" you probably mean "with values in"? $\endgroup$
    – Jochen Wengenroth
    Mar 17, 2015 at 7:50

1 Answer 1

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What if $X=\{1,2\}$ and $Y=\{1,2,3\}$. Equip both $X$ and $Y$ with the uniform probability measure. Set $f(x,1)=1$, $f(x,2)=x$ and $f(x,3)=2$. Then $Tf(x,1)=1$; $Tf(x,2)=1.5$ and $Tf(x,3)=2$.

Notice that $\|f\|_1=\|Tf\|_1=1.5$ and $\|f\|_\infty=\|Tf\|_\infty=2$, but $\|f\|_2>\|Tf\|_2$.

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  • $\begingroup$ Thanks for your answer. Now we know the map $p\mapsto \frac{||Tf||_p}{||f||_p}$ is not monotonic. How about the map $p\mapsto \sup_{f\neq 0}\frac{||Tf||_p}{||f||_p}$? $\endgroup$
    – math-Student
    Mar 17, 2015 at 19:02

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