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Denote $(\Omega,\mathcal{F},P)$ to be our probability space and $X:\Omega\to\mathbb{R}$ a random variable. Suppose we have a measurable convex function $f:\mathbb{R}\to\mathbb{R}$. From Jensen's inequality, we know that for all sub-sigma-algebra $\mathcal{G}\subset\mathcal{F}$: $$ E[f(X):\mathcal{G}] \geq f(E[X:\mathcal{G}]). $$ Now let us denote a partition of $\mathbb{R}$ by $\mathbb{R}^{(n)} = \{\mathbb{R}^{(n)}(k) \subset \mathbb{R}: k=1,..,n\}$ such that $P(X \in \mathbb{R}^{(n)}(k)) = 1/n$ for all $k =1,...,n$. Define a finer partition $\mathbb{R}^{(2n)} = \{\mathbb{R}^{(2n)}(k) \subset \mathbb{R}: k=1,..,2n\}$ such that each component in $\mathbb{R}^{(n)}$ is a proper superset of 2 components in $\mathbb{R}^{(2n)}$ and such that $P(X \in \mathbb{R}^{(2n)}(k)) = \frac{1}{2n}$ for all $k =1,...,2n$

Now define random variable $X^{(n)} = E[X:X\in\mathbb{R}^{(n)}]$. From Jensen's, we know that $$ E[f(X):X^{(n)}] \geq f(E[X:X^{(n)}]). $$ I want to show that as $n\to\infty$, the above holds with equality, i.e. I want to show that in the limit $$ \lim_{n\to\infty}E[f(X):X^{(2n)}] = \lim_{n\to\infty}f(E[X:X^{(2n)}]) = f(X) $$ almost surely. Is this possible??? Do I even need convexity?

Thank you.

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  • $\begingroup$ This may seem like a silly question but you use some notation that is a bit different to what I am used to. By $E[f(X):\mathcal G]$ do you mean "the expectation of $f(X)$ conditioned on the sigma algebra $\mathcal G$"? $\endgroup$
    – Therkel
    Oct 5, 2017 at 10:49
  • $\begingroup$ Would it not be enough to show that $\mathbb R^{(n)} \to \mathbb R$ for $n\to \infty$? $\endgroup$
    – Therkel
    Oct 5, 2017 at 10:56
  • $\begingroup$ Hi Therkel. Yes. I meant the expectation of f(X) conditioned on the sigma algebra $\mathcal{G}$. $\endgroup$
    – jerom
    Oct 5, 2017 at 18:46
  • $\begingroup$ Regarding $\mathbb{R}^{(n)}\to\mathbb{R}$, what does that mean exactly? $\endgroup$
    – jerom
    Oct 5, 2017 at 18:48
  • $\begingroup$ Essentially what I want to prove is that the random variables $E[f(X):X^{(n)}]$ and $f(E[X:X^{(n)}])$ both converge almost surely to f(X). I'm not sure how to do this. $\endgroup$
    – jerom
    Oct 5, 2017 at 18:55

1 Answer 1

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We may as well assume the first partition occurs for $n=1$, so we have partitions at $n=2^k$ for every $k\ge0$. Let $\mathcal F_k=\sigma\{[X\in A]:A\in\mathbb R^{(2^k)}\}$, the $\sigma$-field generated by these partitions. If we assume both $X$ and $f(X)$ are integrable, then $X^{(2^k)}=E[X|\mathcal F_k]$ and $Y^{(2^k)}=E[f(X)|\mathcal F_k]$ both define martingales. The martingale convergence theorem implies that $$X^{(2^k)}\to E[X|\mathcal F_\infty],\qquad Y^{(2^k)}\to E[f(X)|\mathcal F_\infty]$$ as $k\to\infty$ almost surely, where $\mathcal F_\infty:=\sigma\left(\bigcup_k\mathcal F_k\right)$. Thus, your desired result will follow for any continuous $f$ if we can show that $X$ is $\mathcal F_\infty$-measurable, no convexity required. (If $f$ is injective, this is if and only if.)

However, this isn't always true. Assume $X$ is uniformly distributed on $(0,1)$. Define $$\mathbb R^{(2^k)}(j)=\left[\frac{j-1}{2^{k+1}},\frac{j}{2^{k+1}}\right)\cup\left[\frac{j-1+2^k}{2^{k+1}},\frac{j+2^k}{2^{k+1}}\right)$$ for $2\le j\le2^k-1$, and $$\mathbb R^{(2^k)}(1)=(-\infty,2^{-{k+1}}),\qquad\mathbb R^{(2^k)}(2^k)=[1-2^{-(k+1)},\infty).$$ Note that $P(X\in\mathbb R^{(2^k)}(j))=2^{-k}$ for all $j$, so this partition meets your conditions. However, if $A\in\mathbb R^{(2^k)}$ for some $k$, then $A\cap[0,1)$ is closed under the map $\varphi:x\mapsto x+\frac12\mod1$, hence $[X\in B]\in\mathcal F_\infty$ implies $B\cap[0,1)$ is also closed under $\varphi$. In particular, this implies that $[\frac12\le X<1)\notin\mathcal F_\infty$, so $X$ is not $\mathcal F_\infty$-measurable.

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  • $\begingroup$ OK I see but I still don't understand why $[\frac{1}{2}\leq X < 1] \not\in \mathcal{F}_\infty$? $\endgroup$
    – jerom
    Oct 6, 2017 at 14:46
  • $\begingroup$ Let $B=[\frac12,1)$. This set is clearly not closed under the map $\varphi$ (e.g. $\frac34\in B$ but $\frac14\notin B$). As discussed, if $B\subset[0,1)$ is such that $[X\in B]\in\mathcal F_\infty$, then $B$ is closed under $\varphi$. $\endgroup$
    – Jason
    Oct 6, 2017 at 17:14

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