Proving X1X2 ~ Y1Y2: Distribution of X1 and Y1 is the Same as X2 and Y2

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Homework Help Overview

The problem involves proving or finding a counterexample for the statement that if X1 ~ Y1 and X2 ~ Y2, then X1X2 ~ Y1Y2, focusing on the distributions of the random variables involved.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to reason that the statement might be true by providing an example involving uniform and binomial distributions. Other participants seek clarification on the notation used, specifically the meaning of "~" in this context.

Discussion Status

Participants are exploring the validity of the original statement, with some providing examples and counterexamples. A counterexample has been presented, indicating that the initial assumption may not hold true.

Contextual Notes

There is a need for clarity on the definitions and properties of the distributions involved, as well as the implications of the notation used in the problem statement.

shan
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Homework Statement


If X1 ~ Y1 and X2 ~ Y2, then X1X2 ~ Y1Y2, prove or find a counterexample. (the distribution of X1 has the same distribution of Y1 etc)


Homework Equations


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The Attempt at a Solution


I'm guessing the statement is true. For example if X1 and Y1 were both uniform and X2 and Y2 are binomial, then a uniform * binomial is distributed the same as a uniform * binomial.
 
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It would help if you also told us what "~" meant.
 
Ah sorry, it's the stuff I put in the brackets ie.

X1 ~ Y1 : the distribution of X1 has the same distribution of Y1, X1 and Y1 being random variables.

The same with X2 ~ Y2 and X1X2 ~ Y1Y2.
 
If anyone was interested, the answer was no. The counter example was:

Let [tex]P(X_1 = 1) = P(X_1 = -1) = 0.5[/tex]. Let [tex]X_2 = -X_1[/tex] and [tex]Y_1 = Y_2 = X_1. X_1 X_2 = -X_1^2 = -1[/tex] almost surely, but [tex]Y_1 Y_2 = X_1^2 = 1[/tex]
 

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