Finding Marginal Probability Mass Functions for Random Variables X and Y

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SUMMARY

The discussion focuses on calculating the marginal probability mass functions (p.m.f) for random variables X and Y based on a provided joint p.m.f table. The correct formulas for marginal p.m.f are established as P_X(x) = ∑ P_{X,Y}(x,y) over all y and P_Y(y) = ∑ P_{X,Y}(x,y) over all x. The marginal p.m.f for X is defined as P_X(0) = 0.45 and P_X(1) = 0.55, while for Y, it is P_Y(0) = 0.35, P_Y(1) = 0.30, and P_Y(2) = 0.35. The notation for values outside the defined range is also discussed.

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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

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:
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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:
thanks for your response :)

So I should define the marginal pmf's as?

<br /> P_X \left( x \right) = \left\{ {\begin{array}{*{20}c}<br /> {0.45\;...\;x = 0} \\<br /> {0.55\;...\;x = 1} \\<br /> \end{array}} \right.<br />

<br /> P_Y \left( y \right) = \left\{ {\begin{array}{*{20}c}<br /> {0.35\;...\;y = 0} \\<br /> {0.3\;...\;y = 1} \\<br /> {0.35\;...\;y = 2} \\<br /> \end{array}} \right.<br />
 
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.
 
Last edited:
Yep sure, that makes sense, thanks for your help!
 

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