Conditional Expectation Question Suppose $p(0,0)=\frac{1}{18}$, $p(0,1)=\frac{3}{18}$, $p(1,0)=\frac{4}{18}$, $p(1,1)=\frac{3}{18}$, $p(2,0)=\frac{6}{18}$, $p(2,1)=\frac{1}{18}$. 

Does $E[X|Y]= E[X]$ and $E[Y]= E[Y|X]$?
Well, $E[X|Y]$ is a random variable and $E[X]$ is a number, so no...

But it is true that $E[X]=E[E[X|Y]]$ and   $E[Y]=E[E[Y|X]]$; a formula which holds more generally for jointly distributed random variables $X$ and $Y$.

Let's verify this for the random variables $X$ and $Y$ in your question. Here, $X$ refers to the first coordinate of your joint mass function and $Y$ refers to the second coordinate.

First let's  figure out what values $X$ takes and it's marginal distribution.

Looking at your joint mass function $p(x,y)$, we see that $X$ takes the values $0$, $1$, and $2$.

To find the marginal distribution of $X$ we need to calculate the probability that $X$ takes on a specific value:
$$\eqalign{
p_X(0)=P[X=0]&amp;=\textstyle P[X=0 ,Y=0]+P[X=0,Y=1]= {1\over 18}+{3\over18}={4\over 18}\cr
p_X(1)=P[X=1]&amp;=\textstyle P[X=1 ,Y=0]+P[X=1,Y=1]= {4\over 18}+{3\over18}={7\over 18}\cr
p_X(2)=P[X=2]&amp;=\textstyle P[X=2 ,Y=0]+P[X=2,Y=1]= {6\over 18}+{1\over18}={7\over 18}.
}
$$

As a check of our work, note $p_X(0)+p_X(1)+p_X(2)=1$, as it should.

While we have everything in hand let's find $E[X]$:
$$\tag{1}
E[X]=0\cdot{4\over18}+1\cdot{7\over18}+2\cdot{7\over18}={14\over18}.
$$

Now on to $Y$:

$Y$ takes the values $0$, and $1$.

The marginal distribution of $Y$ is:
$$\eqalign{
p_Y(0)=P[Y=0]&amp;=\textstyle P[Y=0 ,X=0]+P[Y=0,X=1]+P[Y=0,X=2]\cr &amp;= {1\over 18}+{4\over18}+ {6\over 18}\cr &amp; ={11\over18}\cr
p_Y(1)=P[Y=1]&amp;=\textstyle P[Y=1 ,X=0]+P[Y=1,X=1]+P[Y=0,X=2]\cr &amp;=  {3\over 18}+{3\over18}+ {1\over 18}\cr &amp; ={7\over18}.\cr }
$$

Note $p_Y(0)+p_Y(1) =1$, as it should.

The expectation of $Y$ is
$$
E[Y]=0\cdot{11\over18}+1\cdot{7\over18} ={7\over18}.
$$



Note that if you wrote the joint mass function in tabular form
$$
\matrix{ X\backslash Y &amp; 0&amp;1  \cr  0\ \ \ \ \ &amp;{1\over18}&amp;{3\over18}  \cr
                        1\ \ \ \ \ &amp;{4\over18} &amp;{3\over18} \cr
\cr  2\ \ \ \ \  &amp;{6\over18} &amp;{1\over18}   } 
$$
Then the marginal distribution of $X$ is given by row sums of the table and the marginal distribution of $Y$ is given by the column sums of the table.



Now on to the random variable $E[X|Y]$. Note that  $E[X|Y]$ is a function of the random variable $Y$. The value of  $E[X|Y]$  at $y$ is by definition $E[X|Y=y]$. The quantity $E[X|Y=y]$ is called the conditional expectation of $X$ given $Y$ and its value at $Y=y$ is defined to be
$$
E[X|Y=y] =\sum_x xP[X=x|Y=y]=\sum_x x{p(x,y)\over p_Y(y)}.
$$
Note also that to find, for example $P[X=0|Y=1]$, you can go to the table and take the $(0,1)$ entry (which is $p(0,1)$) and divide by the column sum for column 1 (this gives $p_Y(1)$).

We have

$$\eqalign{
\Bbb E[X|Y=0]&amp;= 0\cdot P[X=0|Y=0]+1\cdot P[X=1|Y=0]+2\cdot P[X=2|Y=0]\cr
             &amp;=0+{p(1,0)\over p_Y(0)}+{p(2,0)\over p_Y(0)}\cr
             &amp;={4/18\over11/18}+{6/18\over11/18}\cr
             &amp;={10\over11}.
} $$

$$\eqalign{
\Bbb E[X|Y=1]&amp;= 0\cdot P[X=0|Y=1]+1\cdot P[X=1|Y=1]+2\cdot P[X=2|Y=1]\cr
             &amp;=0+{p(1,1)\over p_Y(1)}+{p(2,1)\over p_Y(1)}\cr
             &amp;={3/18\over7/18}+{1/18\over7/18}\cr
             &amp;={4\over7}.
} $$

Now, since  $E[X|Y]$ is a function of the random variable $Y$:
$$\eqalign{
E[E[X|Y]]&amp;= \sum_y E[X|Y=y]\cdot P[Y=y]\cr
 &amp;=      E[X|Y=0]\cdot P[Y=0]+ E[X|Y=1]\cdot P[Y=1]                      \cr 
&amp;={10\over11}\cdot {11\over18}+{4\over7}\cdot{7\over18}={14\over18};
}
$$
the same value as found in $(1)$, as claimed.

I'll leave the verification that  $E[Y]=E[E[Y|X]]$ as an exercise.
Use the fact that $E[X] = E(E[X|Y])$ and $E[Y] = E(E[Y|X])$.