f(x,y)=(x+y)/30 for x=0,1,2,3; y=0,1,2
here is the chart for the probability of f(x,y), i.e., P(0,0)=0, P(1,0)=1/30, P(1,1)=2/15 ...
0 1/30 1/10 1/5
1/30 2/15 3/10 8/15
1/10 3/10 3/5 1
Find E[2X-Y]
E[2X-Y]=2E[X]-E[Y]
g(x)=1/30*[(x+0)+(x+1)+(x+2)]=(x+1)/10
h(y)=1/30*[(y+0)+(y+1)+(y+2)+(y+3)]=(2y+3)/15
2E[X]=4
E[Y]=19/15
So E[2X-Y]=41/15,
But when I use the chart, the answer is 31/10. So I don't know why, can you tell me how will you gonna do to solve this problem? Thanks
$\displaystyle E(X) = (0) \cdot \left(\frac{3}{30} \right) + (1) \cdot \left(\frac{6}{30} \right) + (2) \cdot \left(\frac{9}{30} \right) + (3) \cdot \left(\frac{12}{30} \right) = 2$
$\displaystyle E(Y) = (0) \cdot \left(\frac{6}{30} \right) + (1) \cdot \left(\frac{10}{30} \right) + (2) \cdot \left(\frac{14}{30} \right) = \frac{38}{30}$.
So it looks OK.
Lets compute the maginal distributions, which are the last row and column of the following table:
$\displaystyle
\begin{array}{cccc|c}
0 & 1/30 & 1/10 & 1/5 & 1/3 \\
1/30 & 2/15 & 3/10 & 8/15 & 1 \\
1/10 & 3/10 & 3/5 & 1 & 2 \\
\hline
2/15 & 7/15 & 1 & 26/15 &10/3
\end{array}
$
The sums of the last row and column should both be 1 which should be the sum of all the elements in the table if this truly represents the joint distribution of (x,y), but they do not, they are $\displaystyle 10/3$.
So there is an error in your table.
.