Expected Value, Expected Variance,covariance

[prodicalboxer]

Can someone help me with this problem?

The joint probability mass function of X and Y, p(i,j)=P{X=i,Y=j}, is given as follows
p(-1,-2)=1/9, p(-1,-1)=1/18, p(-1,0)=1/12, p(-1,1)=0,
p(0,-2)=1/12, p(0,-1)=1/9, p(0,0)=0, p(0,1)=1/8,
p(1,-2)=0, p(1,-1)=1/8, p(1,0)=1/4, p(1,1)=1/18,

a) Compute the E[X], Var(X), and Cov(X,Y)
b) Calculate P{X,Y=k} for k=-2,-1,0,1,2
c) Evaluate E[Y|X=k] for k=-1,0,1

here is what I attempted to do:
E[X]=E[X1] + E[X2]+....E[Xn]=np
Var(X)= E[X^2]-(E[X])^2
Cov=(X,Y)=E[(X-E[X])(Y-E[Y])]
=E[XY-YE[X]-XE[Y]+E[X]E[Y]]
=E[XY]-E[Y]E[X]-E[X]E[Y]+E[X]E[Y]
=E[XY]-E[X]E[Y]

E[X]=E[X|Y=-2]= 1/9(-1)+1/12(0)+0(1)=-1/9
E[X|Y=-1]=1/18(-1)+1/9(0)+1/8(1)=5/72
E[X|Y=0]=1/12(-1)+0(0)+1/4(1)=1/6
E[X|Y=1]=0(-1) + 1/8(0) +1/18(1)=1/18
E[X]=-1/9+5/72+1/6+1/18=13/72

now the variance Var(X)
(-1/9)^2+(5/72)^2+(1/6)^2+(1/18)^2=1/81+25/5184+1/36+1/324
=83/1728
Var=83/1728-169/5184=5/324

E[Y]= [Y|X=-1]=1/9(-2)+1/18(-1)+1/12(0)+0(1)=-5/18
[Y|X=0]=1/12(-2)+1/9(-1)+0(0)+1/8(1)=-11/72
[Y|X=1]=0(-2)+1/8(-1)+1/4(0)+1/18(1)=-5/72
E[Y]=-5/18 -11/72-5/72=-41/72

Cov=(83/1728x5/324)-(13/72x-41/72)=37/5000

Could you check part A ...I really need help with part b and c

 

[tnich]

Your numerical answer is correct for for $E\{X\}$, but if I were grading your homework, I might mark off a point for your notation.
$$E\{X|Y=-2\} = \frac {\sum_{X = -1}^1 X p(X, -2)}{\sum_{X = -1}^1 p(X, -2)}$$
so you shouldn't use the notation $E\{X|Y=-2\}$ to mean something else. What you can do is write
$$E\{X\} = {\sum_{X = -1}^1 \sum_{Y = -2}^2X p(X, Y)}$$
and that is what you ended up doing when you calculated the numerical result.