Thread:

1. 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!

For the given Joint Probability Mass Function $f(x,y)=\frac{1}{36}(x+y)$ over the nine points with $x=1,2,3$ and $y=1,2,3$, Determine $E(X)$ and $V(X)$
So for $E(X)$ we need to find $f_{X}(x)$.

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

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

Now onto the variance;

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

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

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

$V(X)=1.471$

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

Kasper

2. Originally Posted by Kasper

$f_{X}(x)=\int_Y f(x,y) dy = \int_{1}^{3} \frac{1}{36}(x+y) dy = \frac{1}{18}x+\frac{1}{9}$
You seem to be confused about discrete ad continuous distributions.

For a discrete distribution with joint pmf f(x,y), the marginal pmf is found by

$f_{X}(x) = \sum_{y} f(x,y)$

and

$f_{Y}(y) = \sum_{x} f(x,y)$

Integration is done in the case of continuous distributions.

3. Riiight. Can't just go integrating for discrete cases.

So in my case then,

$f_{X}(x)=\Sigma_Y f_{XY}(x,y)=\frac{1}{36}(x+1)+\frac{1}{36}(x+2)+\f rac{1}{36}(x+3)=\frac{1}{12}x+\frac{1}{6}$?

Thanks for the help, good to know I'm on the right track, short of integrating across integers.

4. That looks right.

5. Thanks man, much appreciated
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