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Where does the extra $1_{X=x_n}$ come in discrete conditional expectation?

$$\mathbb{E}[Y|X=x]=\frac{\mathbb{E}[Y1_{X=x}]}{\mathbb{P}(X=x)}$$

But then I see that when $\Omega$ is partitioned into union of subsets, then

$$\mathbb{E}[Y|\sigma(X)]=\sum_n \frac{\mathbb{E}{[Y1_{X=x_n}]}}{\mathbb{P}(X=x_n)}\mathbb{1}_{X=x_n}$$

So where does this last $\mathbb{1}_{X=x_n}$.

https://www.stat.berkeley.edu/users/pitman/s205f02/lecture15.pdf

mavavilj
  • 7,270

1 Answers1

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If we write: $$f\left(x\right)=\mathbb{E}\left[Y\mid X=x\right]=\frac{\mathbb{E}Y\mathbf{1}_{X=x}}{P\left(X=x\right)}$$ then: $$f\left(X\right)=\mathbb{E}\left[Y\mid\sigma\left(X\right)\right]$$

For a fixed $\omega\in\Omega$ with $X\left(\omega\right)=x_{k}$ we get:

$$\begin{aligned}\left(\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}\right)\left(\omega\right) & =\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}\left(\omega\right)\\ & =\frac{\mathbb{E}Y\mathbf{1}_{X=x_{k}}}{P\left(X=x_{k}\right)}\\ & =\mathbb{E}\left[Y\mid X=x_{n}\right]\\ & =f\left(x_{n}\right)\\ & =f\left(X\left(\omega\right)\right) \end{aligned} $$

So the conclusion is that:

$$f\left(X\right)=\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}$$ or equivalently: $$\mathbb{E}\left[Y\mid\sigma\left(X\right)\right]=\sum_{n}\frac{\mathbb{E}Y\mathbf{1}_{X=x_{n}}}{P\left(X=x_{n}\right)}\mathbf{1}_{X=x_{n}}$$

drhab
  • 151,093
  • So you mean that it's meant to "extract out" that particular $x_k$ for which $X(\omega)=x_k$? Even if we decide to write the expectation as the "full sum". – mavavilj Nov 25 '19 at 14:52
  • Both sides are random variables on the same probability space and my answer shows that they give the same outcome if some $\omega\in\Omega$ is substituted. That means exactly that both sides are expressions for the same random variable so that the equality sign $=$ in between is justified. – drhab Nov 25 '19 at 14:57
  • Function $f$ must be looked at as some intermediator making things more easy to grasp. If I work with $f(x)$ then nothing goes wrong if I substitute random variable $X$ and get $f(X)$. That cannot be said if we use $\mathbb E[Y\mid X=x]$ for the same function. Then substitution of $X$ gives the weird expression $\mathbb E[Y\mid X=X]$ that makes things problematic. – drhab Nov 25 '19 at 15:02
  • I don't understand why they are the same. Is the indicator $=0$ for other than $x_k$? – mavavilj Nov 25 '19 at 15:15
  • $\mathbf1_{X=x_n}(\omega)=1\iff X(\omega)=x_n$ so if $X(\omega)=x_k$ then $\mathbf1_{X=x_n}(\omega)=1\iff x_k=x_n$. This will be the case iff $x_n=x_k$ or equivalently iff $n=k$ (provided that $x_1,x_2,\dots$ are distinct). – drhab Nov 25 '19 at 15:19
  • So the other indicators than that of $x_k$ are zero? – mavavilj Nov 25 '19 at 16:24
  • Yes, only one of them takes value $1$. – drhab Nov 25 '19 at 16:30