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I expect this should be a basic property of regular conditional densities/stochastic kernels, but somehow I am having trouble verifying this.

Suppose we have random variables $X$ and $Y$ with (smooth) joint probability density $p(x,y)$ and (smooth) conditional density for X given Y, namely $p(x|y)$. Then, is it true that: $$p(x|y \in A) = \frac{1}{\mathbb{P}(Y \in A)} \int_A p(x|y) p(y) dy $$

pseudocydonia
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1 Answers1

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I would write something like

$$p(x \mid y \in A)=\dfrac{\int\limits_{y \in A} p(x,y) \,dy}{\int\limits_{z}\int\limits_{y \in A} p(z,y) \,dy\, dz}$$ providing that the denominator is positive, where $z$ ranges over all possible values of $x$

The numerator is pretty much your $\int\limits_A p(x\mid y) \,p(y) \,dy $ and the denominator your $\mathbb{P}(Y \in A)$, so your expression works too.

Henry
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  • What if $A=\Big{\frac{1}{n}\Big}{n=1}^{\infty}$? Would this conditional density be $$p(x|y\in A)=\frac{\sum{n=1}^{\infty}p(x,1/n)}{\sum_{n=1}^{\infty}p_Y(1/n)}$$ where $p_Y(y)=\int_{\mathbb{R}}p(x,y)dx$ – Matthew H. Mar 07 '21 at 15:45
  • @MatthewPilling Yes-ish: it looks as if your $p(x,y)$ is neither a density not a mass function. If it is a density then you could end up with $\frac00$ – Henry Mar 07 '21 at 17:13