if I have two r.v. X,Y. Are the following correct? If so, I'd love to know why.
- $\mathbb{E}[Y|Y] = Y $ ? (if Y is given, does that mean that its expected value becomes just the scalar it has become?)
- $\mathbb{E}[X+Y|Y] = Y+ \mathbb{E}[X|Y] $ (linearity of a scalar since Y is given?)
- $\mathbb{E}[(X-d)\cdot(Y-r)|Y] = (Y-r)\cdot\mathbb{E}[(X-d)|Y] $ , where r and d are scalars..
Thanks
Let $X,Y$ be random variables (on the same probability space) and let $X$ have finite expectation.
Let $f:\mathbb R\to\mathbb R$ be a Borel measurable function.
Then $f(Y)=\mathbb E[X\mid Y]$ if: $$\mathbb EX\mathbf1_B(Y)=\mathbb Ef(Y)\mathbf1_B(Y)\text{ for every Borel set }B\tag1$$
This is a sufficient condition and from $(1)$ it can be deduced that also:$$\mathbb EXg(Y)=\mathbb Ef(Y)g(Y)\text{ for every suitable function }g\tag2$$
Here "suitable" stands for Borel measurable and such that $g(Y)$ has finite expectation.
In the sequel of this answer $f(Y)$ stands for $\mathbb E[X\mid Y]$ so that $(1)$ and $(2)$ both are valid.
1) is correct. For this observe that $(1)$ is a true statement if $f=\mathsf{id}_\mathbb R$ and $X=Y$.
2) is correct. For this observe that: $$\mathbb E[X+Y]\mathbf1_B(Y)=\mathbb EX\mathbf1_B(Y)+\mathbb EY\mathbf1_B(Y)=\mathbb Ef(X)\mathbf1_B(Y)+\mathbb EY\mathbf1_B(Y)=\mathbb E[f(Y)+Y]\mathbf1_B(Y)$$
3) is correct if $Y$ has finite expectation.
It is easy to deduce that $\mathbb E[X-d\mid Y]=f(Y)-d$ and applying $(2)$ on the function $g$ prescribed by $y\mapsto y-r$ we find that:$$\mathbb E[(X-d)(Y-r)\mid Y]=(f(Y)-d)(Y-r)$$