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Does $\int E_{Y\lvert X}[Y\lvert X=x] p(y\lvert x)dy = E_{Y\lvert X}[Y\lvert X=x]$

In other words is the conditional expectation of the conditional expectation equal to the conditional expectation?

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2 Answers 2

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We know that there exists a measurable map $\varphi$ such that almost surely, $E[Y\vert X]=\varphi(X)$. So $$ E[E[Y\vert X]\vert X]=E[\varphi(X)\vert X]=\varphi(X)=E[Y\vert X]. $$

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For random variables $X,Y$ (defined on same probability space) we have under suitable conditions the random variable $\mathbb E[Y\mid X]$. It is measurable wrt $\sigma(X)$ and can be written as $f(X)$ where $f:\mathbb R\to\mathbb R$ is a Borel-measurable function. Then $f(x)=\mathbb E[Y\mid X=x]$ and consequently:$$\int\mathbb E[Y\mid X=x]p(y|x)dy=\int f(x)p(y|x)dy=f(x)\int p(y|x)dy=f(x)=\mathbb E[Y\mid X=x]$$

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