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The answer to this question gives a replacement rule for conditional probability. But how do you prove this? I tried integrating both sides w.r.t. $P_X$ and fiddling around but it didn't get me anywhere.

Rule: Let $X$ and $Y$ be as above random variables, and let $\xi:\mathbb{R}^2\to\mathbb{R}$ be $(\mathcal{B}(\mathbb{R}^2),\mathcal{B}(\mathbb{R}))$-measurable. Then $$ P(\xi(X,Y)\in D\mid X=x)=P(\xi(x,Y)\in D\mid X=x),\quad D\in\mathcal{B}(\mathbb{R}), $$ holds for $P_X$-a.a. $x$. This is saying that "conditional on $X=x$ we may replace $X$ by $x$".

[EDIT]

Managed to prove this without $Y$, for measurable $g$ we have: $\int_C \mathbf{P} (g (X) \in B \mid X = x) P_X \left( \text{d} x \right) =\mathbf{P} (g (X) \in B, X \in C) = \int_C {\bf 1}_{g (x) \in B} P_X \left( \text{d} x \right)$. Also $\int_C \mathbf{P} (g (x) \in B \mid X = x) P_X \left( \text{d} x \right) = \int_C {\bf 1}_{g (x) \in B} P_X \left( \text{d} x \right)$.

But this no longer works with $Y$ as $g(X,Y)$ is no longer a function of $X$.

[EDIT2] Many thanks to Stefan Hansen who posted an answer here to the same question.

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simonzack
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1 Answers1

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  • Probability of an event is the expectation of an indicator function that the event occurs, and
  • $\mathsf E(g(X,Y)\mid X=x) = \mathsf E(g(x,Y)\mid X=x)$

Thus: $$\begin{align} \mathsf P(\xi(X,Y)\in D\mid X=x) & =\mathsf E(\mathbf 1_{\xi(X,Y)\in D}\mid X=x) \\ & = \mathsf E(\mathbf 1_{\xi(x,Y)\in D}\mid X=x) \\ & = \mathsf P(\xi(x, Y)\in D\mid X=x) \end{align}$$

Graham Kemp
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  • Why is the second equality true? It looks like a kind of replacement rule for conditional expectation, do you mind explaining that? – simonzack May 05 '15 at 12:19