# Does this uniquely define a conditional expectation?

I know how we define conditional expectations in probability theory without measure theory (i.e. with density functions on R). My question is about defining it on general probability spaces using measure theory.

Assume a probability space $(\Omega, \sigma, P)$ If we want to define a conditional expectation $E(X|G)$ where G is a subset of Omega (i.e. Y is in sigma), then there is no problem whatsoever.

But What if instead we want to find $E(X|Y=y)$ where y is a single value of the random variable Y, so that it will have measure zero? then we cant calculate the conditional expectation the standard way.

So I came up with the following formula:

$$\lim_{\epsilon\to 0}\int x dP(x|Y_\epsilon)$$

where $Y_\epsilon=\{\omega : |Y(\omega)-y|<\epsilon\}$.

Does this correctly and uniquely define the conditional expectation of X given a measure zero realization of Y?

• In your second paragraph, I think you mean "If we want to define a conditional expectation $E[X|G]$ where [$G$ is an event with $P[G]>0$] then there is no problem whatsoever." Is this what you mean? Or are you trying to condition given a sigma-algebra (in which case $G$ would be a collection of sets in $\Omega$, not a single set). – Michael Dec 17 '17 at 16:14
• Why reinvent the wheel and not turn to the very concise definition of $E(X\mid Y)$? Then $E(X\mid Y)=g(Y)$ for some measurable $g$ and one knows that one can choose $E(X\mid Y=y)=g(y)$. – Did Dec 17 '17 at 16:22
• PS: The way you define is similar to the the way conditional expectation is usually taught in most (non-measure theory) probability classes: Define $P[X\leq x | Y_{\epsilon}]$ and then take $\epsilon\rightarrow 0$. This defines conditional CDF, and you can define conditional expectation from this. This gives great intuition but there are some problems that require the formal measure theory stuff for precision. One problem is, what happens if $P[Y_{\epsilon}]=0$ for $0 \leq \epsilon < 1$? There are some deeper examples of problems but I cannot find on a simple google search. – Michael Dec 17 '17 at 16:29

Another problem arises if $\lim_{\epsilon\rightarrow 0} P[X\leq 1| Y_{\epsilon}]$ does not exist. For example, let $Y$ be a nonnegative random variable, let $y=0$, and assume $P[Y=0]=0$. Suppose $(X,Y)$ satisfies the following for $k \in \{1, 2, 3, ...\}$:
If $k$ is even: \begin{align} P\left[X \leq 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (1/2)\frac{1}{2^{k^2}} \\ P\left[X > 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (1/2)\frac{1}{2^{k^2}} \end{align}
If $k$ is odd: \begin{align} P\left[X \leq 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (3/4)\frac{1}{2^{k^2}} \\ P\left[X > 1, Y \in \left[\frac{1}{k+1}, \frac{1}{k}\right)\right] &= (1/4)\frac{1}{2^{k^2}} \end{align}
Then $Y_{1/k} = \{\omega : 0 \leq Y < 1/k\}$ and since $1/2^{k^2}$ decreases so rapidly, we have for large $k$ that $\sum_{i=k}^{\infty} 1/2^{i^2} \approx 1/2^{k^2}$ and so $$P[Y_{1/k}] \approx P\left[Y \in \left(\frac{1}{k+1}, \frac{1}{k}\right]\right] = \frac{1}{2^{k^2}}$$ and \begin{align} P[X \leq 1, Y_{1/k}] &\approx (1/2)\frac{1}{2^{k^2}} \quad, \mbox{if $k$ even} \\ P[X \leq 1, Y_{1/k}] &\approx (3/4)\frac{1}{2^{k^2}} \quad, \mbox{if $k$ odd} \end{align}
So $$\lim_{m\rightarrow\infty} P[X\leq 1 | Y_{1/(2m)}] = 1/2$$ but $$\lim_{m\rightarrow\infty} P[X \leq 1 | Y_{1/(2m+1)}] = 3/4$$