0
$\begingroup$

Consider the conditional expectation of $x$ given $y$, $$ \mathbb{E}(x | y) $$ where $x \in X$ and $y \in Y$ where $X, Y$ are Hilbert spaces (possibly infinite dimensional).

Question : I am looking for conditions under which $$ \mathbb{E}(x | \cdot ) : Y \to X $$ is a continuous function.

I'll give some simple examples to illustrate the problem:

Example 1: Assume $X = Y = \mathbb{R}$, let $x$ have a gaussian prior with mean 0 and unit variance and let $y$ be gaussian with mean $x$ and unit variance. In this case we have $$ \mathbb{E}(x | y) = \frac{y}{2} $$ which is continuous.

Example 2: Assume $X = Y = \ell_2$. Let $x$ have a gaussian prior with $x_i \sim N(0, 1/i)$ and let $y_i ~ \sim N(x_i / i, 1 / i^3)$. In this case we have $$ [\mathbb{E}(x | y)]_i = \left( \frac{1}{i} i^3 \frac{1}{i} + i \right)^{-1} \frac{1}{i} i^3 y_i = \frac{i}{2} y_i $$ This is a linear operator, but unbounded, hence $\mathbb{E}(x | \cdot)$ is not continuous.

$\endgroup$
  • $\begingroup$ Your question is not well posed. Apparently, $x$ is a fixed random variable with values in $X$. But what is then $y\mapsto E(x|y)$? $\endgroup$ – Jochen Wengenroth Feb 13 '18 at 8:39
  • $\begingroup$ I was sloppy on the notation to save some space. To be more strict, the idea is that $X/Y$ are $\mathcal{X}/\mathcal{Y}$ valued random variables and hence $E(X|Y)$ is an $\mathcal{X}$ valued random variable which is a function of a $\mathcal{Y}$ valued random variable. Thus the function I'm looking for is thus more strictly $y \to E(X|Y=y)$. $\endgroup$ – Jonas Adler Feb 13 '18 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.