Joint Distribution (easy qn)

steven10137

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 [tex]P_X \left( x \right)[/tex] and [tex]P_Y \left( y \right)[/tex]

2. The attempt at a solution

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

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

For [tex]P_X \left( x \right)[/tex] it's the sum of [tex]{P_{X,Y} \left( {x,y} \right)}[/tex] 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
[tex]P_Y \left( y \right)[/tex] be 0.35?

Any help would be greatly appreciated.
Cheers

konthelion

Hello,
The possible X values are x=0 and x=1, so if you compute

[tex]P_x \left( 0 \right) = p(0,0)+p(0,1)+p(0,2)=X1[/tex] Find x1
[tex]P_x \left( 1 \right) = p(1,0)+p(1,1)+p(1,2)=X2[/tex] Find x2
*You basically do this for how many possible X values you have.

Then the marginal pmf is then
[tex]P_x \left( x \right) = \left\{ x1forx= 0; x2forx=1;0,otherwise} [/tex]

Then compute the marginal pmf of Y obtained from the column totals. Hope that makes sense.

steven10137

thanks for your response :)

So I should define the marginal pmf's as?

[tex]
P_X \left( x \right) = \left\{ {\begin{array}{*{20}c}
{0.45\;...\;x = 0} \\
{0.55\;...\;x = 1} \\
\end{array}} \right.
[/tex]

[tex]
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.
[/tex]

konthelion

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.

steven10137

Yep sure, that makes sense, thanks for your help!