Joint Probability Distribution

Hey, i'm having a little trouble transitioning from single variable probability distributions to bivariate distributions, I'm just wondering if any of you guys have a minute to look over my solution to a question to make sure i'm not going astray on the foundation?

Thanks for your help!

Quote:

For the given Joint Probability Mass Function $\displaystyle f(x,y)=\frac{1}{36}(x+y)$ over the nine points with $\displaystyle x=1,2,3$ and $\displaystyle y=1,2,3$, Determine $\displaystyle E(X)$ and $\displaystyle V(X)$

So for $\displaystyle E(X)$ we need to find $\displaystyle f_{X}(x)$.

$\displaystyle f_{X}(x)=\int_Y f(x,y) dy = \int_{1}^{3} \frac{1}{36}(x+y) dy = \frac{1}{18}x+\frac{1}{9}$

From here,

$\displaystyle E(X)=\Sigma_{x} xf_{X}(x)= (1)(\frac{1}{18}(1)+\frac{1}{9}) + (2)(\frac{1}{18}(2)+\frac{1}{9})+ (3)(\frac{1}{18}(3)+\frac{1}{9}) =\frac{3}{18}+\frac{8}{18}+\frac{15}{18}=1.444$

Now onto the variance;

$\displaystyle V(X)=\Sigma_{X} x^2f_{X}(x) - E(x)^2$

$\displaystyle V(X)= (1)^2(\frac{1}{18}(1)+\frac{1}{9})+(2)^2(\frac{1}{ 18}(2)+\frac{1}{9})+(3)^2(\frac{1}{18}(3)+\frac{1} {9})-1.444^2$

$\displaystyle V(X)=[\frac{3}{18} + \frac{16}{18} + \frac{45}{18}]-2.085 = 3.5555-2.085$

$\displaystyle V(X)=1.471$

Have I gone horribly wrong somewhere? Or does this look like the appropriate approach? Thanks for helping me understand this!

Kasper