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Let $(Y, \Sigma,\mu)$ be measure space and $X$ a Polish space endowed with its Borel $\sigma$-algebra. Suppose that $f:Y\times X\to \mathbb R$ is a Carathéodory function (i.e. continuous in $x\in X$ for each $y\in Y$, measurable and bounded by a $L^1$ function that does not depend on $x$). Let ${\Sigma}_0$ be sub $\sigma$-algebra of $\Sigma$, and let $g(\cdot,x)=E(f(\cdot,x)|\Sigma_0)$ denote the conditional expectation with respect to ${\Sigma}_0$. Does $g$ have a version that is a Carathéodory function as well?

Thanks!

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2 Answers 2

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Edit: The below answer is invalid, since $\Sigma_0$ is a sub sigma algebra of $Y$, not $X \times Y$.

I think the answer is no - take $X = Y = [0, 1]$, and $f(y, x) =y$. Pick some Borel set $E \subset [0, 1]$ such that $E$ and its complement have nonzero measure in every open interval, and let $\Sigma_0$ be the sigma algebra generated by the sets $[0, 1] \times E$ and $\mathcal B([0, 1]^2 \setminus [0, 1] \times E)$ where $\mathcal B$ denotes the restriction of the Borel sigma algebra to the given set.

Then $g(y, x) = \frac{1}{2}$ for $(y, x) \in [0, 1] \times E$ and $g(y, x) = y$ otherwise.

This has no modification that is continuous in $x$ for every $y$, since for every $y$ except for $y= \frac{1}{2},$ $g(y, \cdot)$ is essentially discontinuous everywhere.

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  • $\begingroup$ In the question, $\Sigma_0$ is a sub-$\sigma$-algebra of the $\sigma$-algebra on $Y$, not $X\times Y$. $\endgroup$ Nov 24 at 8:19
  • $\begingroup$ Oh, my mistake.. $\endgroup$
    – Nate River
    Nov 24 at 11:03
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Here is a positive answer for the case that $\Sigma_0$ is generated by a random variable with values in a Polish space, so that we can use regular conditional probabilities and for some kernel $\kappa:Y\to\Delta(Y)$, we can let $$\mathbb{E}(f(\cdot,x)|\Sigma_0)_y=\int f(,\cdot,x)~\mathrm d\kappa_y.$$ Then continuity follows from the assumption that there is a dominating integrable function and the dominated convergence theorem.

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