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Given any event $A$ and $[0,1]$-valued random variable $X$ the conditional probability $P(A|X)$ is a random variable uniquely defined (outside a set of measure zero) by the requirement that

$$\displaystyle \int_{S}P(A|X) d P = P(S\cap A)$$

for every $X$-measurable set $S$, i.e $S = X^{-1}(U)$ for some measurable $U$.

In particular for $S$ the whole probability space we have

$$P(A)=\displaystyle \int P(A|X) d P $$

The discrete version of the above is as follows: Suppose Y is discrete random variable taking vales $y_1,y_2,\ldots y_n$ we can write

$P(A) = \sum_i P(A,Y=y_i) = \sum_i P(A | Y=y_i)P(Y=y_i)$.

meaning we integrate with respect to $X$

The notation suggests that if $Y$ is a discretisation of $X$, i.e $Y$ is $X$ rounded to the nearest multiple of $1/n$, and we let the discretisation become very fine then the right-hand-side should tend to

$$\sum_i P(A | Y=y_i)P(Y=y_i) \to \int P(A|X) dX$$

Here $dX$ means we integrate with respect to the measure induced by $X$. i.e $\mu_X(A) = P(X \in A)$.

But this is different from the definition of the conditional probability. Should I find this surprising? Where has my intuition for the meaning of these objects broken down?

Daron
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    It is not at all clear what you mean by "we integrate with respect to $X$", and how $\int P(A|X) dX$ is defined. I think you are mixing up the undergraduate notions of conditional probabilities (which are numbers) and 'modern' conditional probability (which is a function measurable with respect to some sub $\sigma$ algebra). – copper.hat Jun 27 '23 at 22:52
  • @copper.hat I mean the measure induced by the random variable $X$. I'll edit it. – Daron Jun 28 '23 at 06:40
  • Remember that $P(A|X)$ is a $\sigma(X)$ measurable function that 'best approximates' $P$ on the sub $\sigma$-algebra $\sigma(X)$. In some sense it is a 'projection' of $1_A$ onto $\sigma(X)$ (this can be made precise in a Hilbert space, see https://math.stackexchange.com/questions/2870000/what-is-the-intuition-behind-conditional-expectation-in-a-measure-theoretic-trea?rq=1 for example). (cont.) – copper.hat Jul 01 '23 at 20:45
  • A more standard approach would be to consider the conditional probability as defining a measure $\mu(B)= \int_B XdP$ where $B \in \sigma(X)$ (the latter being the important part). Since $\mu \ll P$, Radon Nikodym gives the existence of a $\sigma(X)$ measurable ${d\mu \over dP}$ such that $\mu(B)= \int_B {d\mu \over dP} dP$ and so $P(A|X) = {d\mu \over dP}$. – copper.hat Jul 01 '23 at 20:45
  • I suspect that trying to approximate $\sigma(X)$ by a sequence of finite $\sigma$-algebras is a much more difficult route. – copper.hat Jul 01 '23 at 20:47

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