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Let $(\Omega,\mathcal F,\mu)$ be a probability space. It is well-known that if $\mathcal A$ is a sub-$\sigma$-algebra of $\mathcal F$, $p\geqslant 1$ and $X$ is an element of $\mathbb L^p$ which takes real values, then $$\tag{1} \lVert\mathbb E[X\mid\mathcal A]\rVert_p\leqslant \lVert X\rVert_p.$$

We now consider for $p\gt1$ the weak $\mathbb L^p$-space, denoted by $\mathbb L^{p,\infty}$, which consists of all the real valued random variables such that $\lVert X\rVert_{p,\infty}^*:= \left(\sup_{t\gt 0 }t^p\mu\{|X|\gt t\}\right)^{1/p} $ is finite. It is known that $\lVert \cdot \rVert_{p,\infty}^*$ is not a norm because the triangle inequality fails. However, defining $$\lVert X\rVert_{p,\infty}:=\sup\left\{\mu(A)^{1/p-1}\mathbb E[| X|\cdot\mathbb 1_{A} ], A\in\mathcal F,\mu(A)\gt 0 \right\},$$ then $\lVert X\rVert_{p,\infty}$ is a norm and $$\tag{2} \lVert X\rVert_{p,\infty}^*\leqslant \lVert X\rVert_{p,\infty}\leqslant \frac p{p-1}\lVert X\rVert_{p,\infty}^*.$$


Questions:

  1. Does the inequality $$\lVert\mathbb E[X\mid\mathcal A]\rVert_{p,\infty} \leqslant \lVert X\rVert_{p,\infty}$$ hold in general?
  2. If not, is there a norm $N_p$ on $\mathbb L^{p,\infty}$ which is equivalent to $\lVert\cdot\rVert_{p,\infty} $ and for which the inequality $$N_p\left(\mathbb E[X\mid\mathcal A]\right)\leqslant N_p(X) $$ takes place?

Motivations. In general, subadditive sequences have good properties. If $T\colon\Omega\to\Omega$ is a measure preserving map and $\mathcal M$ is a sub-$\sigma$-algebra of $\mathcal F$ such that $T \mathcal M\subset \mathcal M$, then defining $S_n(f):=\sum_{j=0}^{n-1} f\circ T^i$ ($f$ is $\mathcal M$-measurable), the sequence $\left(\lVert\mathbb E\left[S_n(f)\mid \mathcal M\right]\rVert_p\right)_{n\geqslant 1} $ is subadditive for any $p\geqslant 1$. A similar result for weak $\mathbb L^p$ norms would be nice.


For the first question, we can show that $$t\mu\{\mathbb E[X\mid\mathcal A]>t\}^{1/p} \leqslant \frac 1{\mu\{\mathbb E[X\mid\mathcal A]>t\}^{1-1/p}}\mathbb E[f\mathbb{1}\{\mathbb E[X\mid\mathcal A]>t\} ],$$ hence $$\lVert \mathbb E[X\mid\mathcal A]\rVert_{p ,\infty}^*\leqslant \lVert f\rVert_{p,\infty},$$ and it follows from (2) that $$\lVert \mathbb E[X\mid\mathcal A]\rVert_{p ,\infty}\leqslant \frac p{p-1} \lVert f\rVert_{p,\infty},$$ so the question is actually whether we can get rid of the factor $\frac p{p-1}$.

We can try to show this when $f$ is a simple function, but I fail to see how to apply efficiently convexity inequalities.

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    $\begingroup$ en.wikipedia.org/wiki/Interpolation_space#Real_interpolation $\endgroup$ Jan 28, 2015 at 20:32
  • $\begingroup$ @AlexanderShamov Indeed, it seems that my question is related to interpolation of Banach spaces. However, I fail to see how to apply one of the methods of interpolation in the link. For example, the connexion of the $\lVert \cdot\rVert_{p ,\infty}$ with the obtained norm in the interpolated space is not clear (to me). Do you have a precise idea in how to apply the theorem? Anyway I will study the book by Bennett and Sharpley. $\endgroup$ Jan 28, 2015 at 22:10
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    $\begingroup$ If I'm not mistaken, in the case $X_0 = L^1$, $X_1 = L^\infty$, $\theta = 1 - \frac{1}{p}$, the functional $K(x, t, X_0, X_1)$ defined in that Wikipedia article coincides with $\sup_{\mu[A] = t} \mathsf{E} |x| \mathsf{1}[A]$. Which makes the $\Vert \cdot \Vert_{p,\infty}$ norm exactly equal to the $K(\theta,\infty)$ interpolation norm. $\endgroup$ Jan 29, 2015 at 15:42
  • $\begingroup$ You are right and i can conclude from Theorem 1.11 301 of Bennett and Shapley. Maybe you could post this as an answer. $\endgroup$ Jan 29, 2015 at 16:51
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    $\begingroup$ The Calderon-Mitiagin interpolation theorem says that all rearrangement invariant function spaces are exact interpolation spaces between $L_1$ and $L_\infty$. This is Theorem 2.2 in Bennett-Sharpley and can be found in many other books as well. $\endgroup$ Mar 30, 2015 at 0:05

2 Answers 2

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Yes. You can modify your definition of the norm as: $$\|X \|_{p,\infty}=\sup [(\int_\Omega \phi )^{1/p-1}\mathbb{E}(|X|\phi), \text{ } 0\leq \phi \leq 1] $$ Then $$\|\mathbb{E}(X|\mathcal{A}) \|_{p,\infty}=\sup [(\int_\Omega \mathbb{E}(\phi|\mathcal{A}))^{1/p-1}\mathbb{E}(|X|\mathbb{E}(\phi|\mathcal{A})), 0\leq \phi \leq 1] $$

And we can conclude because $0\leq \phi \leq 1 \Rightarrow 0\leq \mathbb{E}(\phi|\mathcal{A})\leq 1$

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my typesetting was so bad it wasn't worth posting my answer but the suggestion was to apply neyman pearson lemma to show that you can replace sup over sets A with only sets of form A=(X>C), then compute norm of E(X|A). I did assume a r.v.s had continuous distributions if convenient.

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  • $\begingroup$ Thanks for your reply. Could you include the statement of Neyman Pearson's lemma? $\endgroup$ Jan 27, 2015 at 18:10
  • $\begingroup$ en.wikipedia.org/wiki/Neyman%E2%80%93Pearson_lemma in this case take likelihood ratio proportional to X, then it says maximize E(X; A) over P(A) = c by choosing A = (X > A), which is the set where the likelihood ratio, X, is large. $\endgroup$
    – mike
    Jan 27, 2015 at 18:50
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    $\begingroup$ Could you please give more details? For the moment I fail to see the connection between likelihood ratios and the context of my question. $\endgroup$ Jan 28, 2015 at 16:04

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