Find the pmf of Y, given the pmf of X and Corr(X,Y).

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Homework Help Overview

The discussion revolves around finding the probability mass function (pmf) of a binary random variable Y, given the pmf of another binary random variable X and the correlation between X and Y. Participants explore the relationship between these variables and the implications of their correlation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the joint distribution of X and Y and the relationship between covariance and correlation. They question whether a unique solution exists for the pmf of Y based on the given information.

Discussion Status

Some participants have noted that the initial findings suggest there may not be a unique solution to the problem. Others propose testing with random values to explore the possibility of multiple solutions. The conversation is ongoing, with participants considering different scenarios, such as independence between X and Y.

Contextual Notes

There is an acknowledgment of the complexity introduced by the correlation and the potential for multiple interpretations of the relationship between X and Y. Participants are also aware of the constraints imposed by the definitions of the pmfs and the correlation.

jimholt
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Not really a homework question, but here goes...

Homework Statement



If we have two (binary) random variables X and Y, and we know the probability mass function for X, as well as the correlation between X and Y, can we find the probability mass function for Y?

Homework Equations



Let f(x) be the mass function for X and g(y) be the mass function for Y. Let g(1) = q and g(0) = 1-q. Given f(1) = p and f(0) = 1-p, and that the correlation between X and Y is c, can we find q as a function of p and c (only)?

The Attempt at a Solution



Let h(x,y) be the joint distribution of X and Y. We can show that that Cov(X,Y) = h(1,1)-pq, and so Corr(X,Y) = [h(1,1)-pq]/sqrt[pq(1-p)(1-q)] = c.

Where to go from there? Seems as if we have one equation with two unknowns. (Is there another equation lurking somewhere?)

Thanks for any help folks.
 
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I think what you've shown is that there is not necessarily a unique solution.

To confirm this, i would pick some random values and see if you can find multiple solutions
 
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As another example consider when X & Y are independent then h(x,y) = p(x)p(y).

The co-variance is Cov(X,Y) = h(1,1)-pq = pq-pq = 0, as expected.

Then q can be any value from 0 to 1.
 
Yup, quite right. This brings up another confusing point for me, but let me chew on it a bit before starting a new thread. Thanks for lending me your eyes.
 
now worries, let me know if you want me to have a look at it
 

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