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

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SUMMARY

The statement "If X1 ~ Y1 and X2 ~ Y2, then X1X2 ~ Y1Y2" is proven false through a counterexample. Specifically, if X1 is a random variable with P(X1 = 1) = P(X1 = -1) = 0.5, and X2 is defined as -X1 while Y1 and Y2 are both equal to X1, then X1X2 results in -1 almost surely, whereas Y1Y2 results in 1. This demonstrates that the distributions of the products do not necessarily align even when the individual distributions are the same.

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

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