E[(X^2+Y^2)/XY] for Geometric(p) R.V.s

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Discussion Overview

The discussion revolves around calculating the expected value E[(X^2+Y^2)/XY] for two independent Geometric(p) random variables, focusing on the application of the Law of the Unconscious Statistician (LOTUS) and related expectations.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in applying the LOTUS equation to geometric variables and seeks assistance.
  • Another participant suggests that it might be simpler to first find E[(X^2)/(XY)] as an intermediate step.
  • A different participant raises a concern that if either X or Y can take the value of 0, the expectation could be infinite.
  • One participant confirms that the distribution starts at 1 based on the formula provided and agrees that the suggested intermediate step is useful.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the best approach to solve the problem, and there are differing views on the implications of the random variables potentially taking the value of 0.

Contextual Notes

There are unresolved assumptions regarding the behavior of the geometric random variables, particularly concerning their support and the implications for the expectation calculation.

Jonobro
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The Question
Let X and Y be two independent Geometric(p) random variables. Find E[(X^2+Y^2)/XY].

Formulas
Px(k) = py(k) = pq^(k-1)
E(x) = Σx(p(x))

My attempt at a solution
I am really struggling with this question because I want to apply the LOTUS equation but am unsure how to do it for geometric variables. Any help would be appreciated.
 
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I didn't test it but it might be easier to find E[(X^2)/(XY)] as intermediate step.
 
If X or Y has a non-zero probability of being 0, that expectation is infinite.
 
Based on the formula in post 1, the distribution starts at 1.

And I checked it, the suggested intermediate step is useful.
 

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