## Joint Distribution (easy qn)

1. The problem statement, all variables and given/known data
The following table gives the joint probability mass function (p.m.f) of the random variables X and Y.

http://img170.imageshack.us/img170/555/tableph9.jpg

Find the marginal p.m.f's $$P_X \left( x \right)$$ and $$P_Y \left( y \right)$$

2. The attempt at a solution

I think I have just missed the point of this somewhere.
I know that:
$${P_X \left( x \right) = \sum\limits_{all\;y} {P_{X,Y} \left( {x,y} \right)} }$$
and
$${P_Y \left( y \right) = \sum\limits_{all\;x} {P_{X,Y} \left( {x,y} \right)} }$$

I just don't know how to apply this to the question properly.

For $$P_X \left( x \right)$$ it's the sum of $${P_{X,Y} \left( {x,y} \right)}$$ over all y (y=0,1,2). So do we just take the first row?
i.e. 0.15+0.20+0.10 = 0.45?

Following this, would
$$P_Y \left( y \right)$$ be 0.35?

Any help would be greatly appreciated.
Cheers
 Hello, The possible X values are x=0 and x=1, so if you compute $$P_x \left( 0 \right) = p(0,0)+p(0,1)+p(0,2)=X1$$ Find x1 $$P_x \left( 1 \right) = p(1,0)+p(1,1)+p(1,2)=X2$$ Find x2 *You basically do this for how many possible X values you have. Then the marginal pmf is then $$P_x \left( x \right) = \left\{ x1forx= 0; x2forx=1;0,otherwise}$$ Then compute the marginal pmf of Y obtained from the column totals. Hope that makes sense.
 thanks for your response :) So I should define the marginal pmf's as? $$P_X \left( x \right) = \left\{ {\begin{array}{*{20}c} {0.45\;...\;x = 0} \\ {0.55\;...\;x = 1} \\ \end{array}} \right.$$ $$P_Y \left( y \right) = \left\{ {\begin{array}{*{20}c} {0.35\;...\;y = 0} \\ {0.3\;...\;y = 1} \\ {0.35\;...\;y = 2} \\ \end{array}} \right.$$

## Joint Distribution (easy qn)

Yes, that's correct. From what I've been taught, you also have to put {0 otherwise} but depending on how the notation that you've been taught in class/book, then it's fine.

Also, for the marginal pmf of Y you can also put for {.35 y = 0,2 . Again, a notational way to write it.
 Yep sure, that makes sense, thanks for your help!