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Let $Y\sim N(0,1)$ and $X$ be a real-valued random variable with distribution $\mu$ ($\mu$ is not abs. continuous with respect to Lebesgue measure) independent of $Y$. Define $Z=X+Y$

How can we explain the conditional distribution of $Z$ given $X=x$ in a measure-theoretic sound way? (Of course it should be $N(x,1)$). To me the (rigorous) notion of conditional probability distributions is something we can define once we have a joint probability distribution of $(Z,X)$ (see below). But to define a joint distribution of $(Z,X)$ many appeal to the fact that $Z$ given $X=x$ has a special form. So where do we start?


The following definition is from Pollard's 'A user's guide to measure theoretic probability' (p. 113).

Letting $(\mathbb{R}^2,\mathcal{B}(\mathbb{R}^2),P)$ be a measure space, and $T: \mathbb{R}^2\rightarrow \mathbb{R}$ be the projection $T:(z,x)\mapsto x$, the conditional distribution of $P$ given $T$ is the family of proability measures in $\mathbb{R}^2$ $$\{P_x,x\in \mathbb{R}\},$$ satisfying

  1. $P_x(\mathbb{R}\times x)=1$ (only gives mass to the line $\mathbb{R}\times x$, so of course we could think of the measure as a measure on $\mathbb{R}$)
  2. $P(A\times B)=\int_BP_x(A\times B) \, \mu(dx),$ where $\mu$ denotes the projection of $P$ to $\mathbb{R}$ (i.e. $\mu([a,b])=P(\mathbb{R}\times [a,b])$)

Edit: Thanks for the helpful comments, however, I was looking for a more general approach than provided in Conditional density of Sum of two independent and continuous random variables, and have edited my question so that $X$ need not be absolutely continuous.

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    Well, without specifying that $X$ and $Y$ are independent, the problem is ill defined, and after specifying that, then I think it is farily straightforward. – P. Quinton Mar 24 '23 at 12:41
  • Yes of course, I forgot! It might be trivial but I don't understand why – SoupMath Mar 24 '23 at 12:44
  • https://math.stackexchange.com/questions/473790/conditional-density-of-sum-of-two-independent-and-continuous-random-variables – Derek Allums Mar 24 '23 at 12:55
  • Thank you for the link. It was not quite what I was looking for, and I have edited my question accordingly – SoupMath Mar 24 '23 at 17:42

1 Answers1

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There's no need to study the joint distribution of $Z$ and $X$. Instead, apply the measure-theoretic result:

If $X$ and $Y$ are independent, then $E[g(X,Y)\mid X=x]=\phi(x)$ where $\phi$ is defined by $\phi(x):=E[g(x,Y)]$.

(Formally, the result asserts that $\phi(X)$ is a version of the conditional expectation $E[g(X,Y)\mid X]$. One proof is here.)

In your situation, you are looking at $g(X,Y):=I(X+Y\in A)$ for $A$ a Borel set. Then $$\phi(x):=E[g(x,Y)]=E[I(x+Y\in A)]=P(x+Y\in A),$$ and $P(X+Y\in A\mid X=x)=\phi(x)$ is the probability that a $N(x,1)$ variable lies in set $A$, as anticipated.

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