This is Durrett $3^{rd}$ Example 1.4.2, I understand most of it but I am stuck in the end. This example is designed to show that $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)$ does not necessarily imply $X$ and $Y$ are independent.
To show this, he provided a table of discrete joint distribution of $X$ and $Y$ as follows:
where $a>0$, $b>0$ and $c\geq 0$, and $2a+2b+c=1$.
This gives me the first question:
I understand he is making the total probability being $1$, and each single probabilities non-negative. Then we only need $a,b,c\geq 0$, but why he made $a,b>0$ and $c\geq0$ instead?
It is easy to see $\mathbb{E}(XY)=0=\mathbb{E}(X)\mathbb{E}(Y)$ because $X$ and $Y$ are symmetrically distributed about $0$, so $\mathbb{E}X=0$ and $\mathbb{E}Y=0$ which imply that $\mathbb{E}X\mathbb{E}Y=0$ and the table is arranged such that $xy\mathbb{P}_{X,Y}(x,y)=0$ for all $x,y$, so $\mathbb{E}(XY)=0$.
However, to prove the dependence, he said
They are not independent since $$\mathbb{P}(X=1, Y=1)=0\neq ab=\mathbb{P}(X=1)\mathbb{P}(Y=1).$$
This gives me the second confusion:
How did he know the probability of $X$ and $Y$ themselves? The table is joint distribution, right?
Did he actually refer that $\mathbb{P}(X)$ is the column when $Y=0$? But then by the same argument $\mathbb{P}(Y)$ is the column when $X=0$. Then we need $2a+c=1$ and $2b+c=1$, which implies $a=b$, but he does not require this.
Is there a way to retrieve "single distribution" from the discrete joint distribution table?
Additionally, it will really appreciated if anyone could teach me how to compute $\mathbb{E}(Y|X)$ given this table. Thank you!
Thank you so much in advance!
Given a joint distribution, we can compute the marginal distribution.
$$P(X=1) = P(X=1, Y=-1) + P(X=1, Y=0)+P(X=1, Y=1)=0+a+0=a$$
which is the sum of the first row.
$$P(Y=1) = P(X=1, Y=1) + P(X=0, Y=1) + P(X=-1, Y=1)=0+b+0=b$$
In general $$P(X=x) = \sum_y P(X=x, Y=y)$$
and $$P(Y=y) = \sum_x P(X=x, Y=y)$$
Since $$P(X=1, Y=1)=0$$ which was read off from the upper left corner, the condition that $a>0, b>0$ are stated to ensure that the product $ab$ can't be equal to $0$, which established the validity of the counter example.
$$E(Y|X=1)=1\cdot Pr(Y=1|X=1) + 0\cdot Pr(Y=0|X=1) - 1\cdot Pr(Y=-1|X=1)=0$$
Similarly
$$E(Y|X=-1)=1\cdot Pr(Y=1|X=-1) + 0\cdot Pr(Y=0|X=-1) - 1\cdot Pr(Y=-1|X=-1)=0$$
\begin{align}E(Y|X=0)&=1\cdot Pr(Y=1|X=0) + 0\cdot Pr(Y=0|X=0) - 1\cdot Pr(Y=-1|X=0) \\ &= \frac{b}{2b+c}+0 - \frac{b}{2b+c}\\&=0\end{align}
That is $E(Y|X)=0$ which actually can be read off since each row is symmetrical.