TASK
Consider a probability space $\Omega$ with four elements $a,b,c,d$. Define $\sigma$-algebra $F$ on $\Omega$-collection of subsets of $\Omega$.
The actual question
Given Probability Measure
- $\mathbb P(\{a\})=\frac 16$
- $\mathbb P(\{b\})=\frac 13$
- $\mathbb P(\{c\})=\frac 14$
- $\mathbb P(\{d\})=\frac 14$
We define two random variables, $X$ and $Y$ as follows:
$$ X(a)=1, \quad X(b)=1, \; \quad X(c)=-1, \quad X(d)=-1, $$ $$ Y(a)=1, \quad Y(b)=-1, \quad Y(c)=1, \quad Y(d)=-1,$$
a. Determine $\mathbb{E}[Y|X]$ and verify the Partial Averaging Property
b. Do the same for $\mathbb{E}[Z|X]$, set $Z=X+Y$
с. Show that $\mathbb{E}[Z|X]$-$\mathbb{E}[Y|X]$=$X$
My Thoughts
a.
According to partial averaging property $$\boxed{\int_{[a,b]} \mathbb{E}[Y|X(w)]d\mathbb{P}(w) = \int_{[a,b]} Y(w)d\mathbb{P}(w)}\\ \text{and} \\\boxed{\int_{[c,d]} \mathbb{E}[Y|X(w)]d\mathbb{P}(w) = \int_{[c,d]} Y(w)d\mathbb{P}(w)}$$
- So for $\int_{[a,b]}, Y(w)d\mathbb{P}(w)= Y(a)\mathbb{P}(a)+Y(b)\mathbb{P}(b)=\frac{-1}6$
$\mathbb{E}[Y|X](a)=-1/6*2/1=-1/3$, $\mathbb{E}[Y|X](b)=-1/6*2/1=\frac{-1}3$
$\int_{[a,b]} \mathbb{E}[Y|X(w)]d\mathbb{P}(w) = \mathbb{E}[Y|X](a)\mathbb{P}(a)+\mathbb{E}[Y|X](b)\mathbb{P}(b)=\frac{-1}6$ - For $\int_{[c,d]} Y(w)d\mathbb{P}(w)$= $Y(c)\mathbb{P}(c)+Y(d)\mathbb{P}(d)=0$
$\mathbb{E}[Y|X](c)$=$\mathbb{E}[Y|X](d)=0$
$\int_{[c,d]} \mathbb{E}[Y|X(w)]d\mathbb{P}(w) = \mathbb{E}[Y|X](c)\mathbb{P}(c)+\mathbb{E}[Y|X](d)\mathbb{P}(d)=0$
So averaging property is verified.
b.
If I understand the logic right we have $Z(a)=2, Z(b)=0, Z(c)=0, Z(d)=-2$ and and can calculate $\mathbb{E}[Z|X]$ the same way, but I feel that I am wrong.
c.
I assume that it is necessary to use properties of conditional expectation here. I started like this :
$\mathbb{E}[Z|X]-\mathbb{E}[Y|X]=\mathbb{E}[Z-[Y|X](linearity)=\mathbb{E}[Z-Y]$ (take out known)
Can somebody be so kind as to check my thoughts and help me with c. part? Thanks a lot!