By definition of conditional expectation,
[tex]\mathrm E\left[ X \mid Y = y \right] = \displaystyle \sum_x x\, p_{X\mid Y}(x \mid y) = \sum_x x \frac{p_{X,Y}(x,y)}{p_Y(y)}[/tex]
where [tex]p_Y(y)[/tex] is the marginal probability mass function (pmƒ) of Y. We find this by summing up all joint probabilities over the possible values of X :
[tex]p_Y(y) = \displaystyle \sum_x p_{X,Y}(x,y) = p_{X,Y}(1,y) + p_{X,Y}(2,y) + p_{X,Y}(3,y)[/tex]
Consult the given values of the joint pmƒ. The values you actually list are unclear… if p(1, 1) = 1, then all other probabilities should be 0, so you must mean something else.
To get around this, I'll just make up a valid joint pmƒ of my own, and show you how to get what you need **using this pmƒ**.
[tex]\begin{array}{c|c|c|c} & A=1 & A=2 & A=3 \\ ---&---&---&--- \\ B=1 & 0.3 & 0.1 & 0.05 \\ B=2 & 0.02 & 0.02 & 0.01 \\B=3 & 0.25 & 0 & 0.25 \end{array}[/tex]
The marginal pmƒ for B is
[tex]p_B(b) = \begin{cases} p_{A,B}(1,1)+p_{A,B}(2,1)+p_{A,B}(3,1) = 0.45 & \text{if }b = 1 \\ p_{A,B}(2,1)+p_{A,B}(2,2)+p_{A,B}(3,1) = 0.05 & \text{if }b = 2 \\ p_{A,B}(3,1)+p_{A,B}(3,2)+p_{A,B}(3,3)=0.5 & \text{if }b = 3\\0&\text{otherwise}\end{cases}[/tex]
Now get the conditional expectation:
[tex]\mathrm E\left[A \mid B = b\right] = \begin{cases}\dfrac{1\cdot p_{A,B}(1,1)+2\cdot p_{A,B}(2,1)+3\cdot p_{A,B}(3,1)}{p_B(1)} \approx 1.44 & \text{if }b = 1 \\\\\dfrac{1\cdot p_{A,B}(1,2)+2\cdot p_{A,B}(2,2)+3\cdot p_{A,B}(3,2)}{p_B(2)} = 1.8 & \text{if }b = 2\\\\\dfrac{1\cdot p_{A,B}(1,3)+2\cdot p_{A,B}(2,3)+3\cdot p_{A,B}(3,3)}{p_B(3)} = 2 & \text{if }b = 3\end{cases}[/tex]
If A and B are independent, then [tex]p_{A,B}(a,b)=p_A(a)p_B(b)[/tex] and it follows that E[A | B] = E[A]. Compute the expectation of A :
[tex]\mathrm E\left[A\right] = \displaystyle \sum_a a \, p_{A,B}(a,b)[/tex]
[tex]\mathrm E[A] = \begin{cases}1\cdot p_{A,B}(1,1)+2\cdot p_{A,B}(2,1)+3\cdot p_{A,B}(3,1) = 0.65 & \text{if }b = 1 \\ 1\cdot p_{A,B}(1,2)+2\cdot p_{A,B}(2,2)+3\cdot p_{A,B}(3,2) = 0.9 & \text{if }b = 2\\ 1\cdot p_{A,B}(1,3)+2\cdot p_{A,B}(2,3)+3\cdot p_{A,B}(3,3) = 1 & \text{if }b = 3\end{cases}[/tex]
But E[A | B] ≠ E[A], so A and B are not independent.
