Finding Marginal Probability Mass Functions for Random Variables X and Y

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Homework Help Overview

The discussion revolves around finding the marginal probability mass functions (p.m.f) for two random variables, X and Y, based on a provided joint probability mass function table.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of summing probabilities to find marginal p.m.f's for X and Y. Questions arise regarding the correct application of the summation formulas and the interpretation of results from the joint p.m.f table.

Discussion Status

Some participants provide guidance on how to compute the marginal p.m.f's, suggesting specific calculations for the values of X and Y. There is acknowledgment of different notational conventions in expressing the results, but no explicit consensus on a single approach is reached.

Contextual Notes

Participants note potential variations in notation for expressing the marginal p.m.f's, as well as the importance of including values for cases where the random variables take on values outside the defined range.

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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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