# Joint Distribution (easy qn)

## Homework Statement

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 [Broken]

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

Last edited by a moderator:

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.

Last edited:
$$P_X \left( x \right) = \left\{ {\begin{array}{*{20}c} {0.45\;...\;x = 0} \\ {0.55\;...\;x = 1} \\ \end{array}} \right.$$
$$P_Y \left( y \right) = \left\{ {\begin{array}{*{20}c} {0.35\;...\;y = 0} \\ {0.3\;...\;y = 1} \\ {0.35\;...\;y = 2} \\ \end{array}} \right.$$