What is the variance of X/Y? Does it exist?

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Homework Help Overview

The discussion revolves around the variance of the ratio of two random variables, X and Y, specifically Var(X/Y). Participants are exploring the conditions under which such a variance might exist and the implications of different interpretations of the expression.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the existence of Var(X/Y) and discussing the implications of the distributions of X and Y. There is uncertainty regarding whether X/Y refers to the division of X by Y or the conditional expectation of X given Y. Some participants highlight the need for specific probability distributions to address the problem accurately.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided insights into the challenges of defining the variance without additional information about the distributions involved. There is no explicit consensus yet, but the conversation is delving into important considerations regarding the definitions and assumptions at play.

Contextual Notes

Participants note that the existence of moments for 1/X can complicate the analysis, and the need for clarity on the probability distributions of X and Y is emphasized. The distinction between X/Y and X given Y is also a critical point of discussion.

mqwit
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I know Var(aX + bY)...
I know Var(XY)...
I don't know of a solution to Var(X/Y)... is there one?
 
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Not a quick one. In fact, I'm not sure what "solution" to Var(XY) you mean, unless the two variables are (stochastically) independent.

The problem is with the denominator - it is entirely possible for a random variable X to have finite mean, variance (even higher-order moments) but for the moments of 1/X fail to exist. Without more information there is no other answer to provide.
 
You don't mean X divided by Y do you? You mean X given Y?

- Warren
 
I don't see how you can answer any of those questions without specifying the probability distributions of X and Y.

If, for example, X is 1 with probability 1/2 and 2 with probability 1/2, Y has the same distribution, then X/Y is 1/2 with probabililty 1/4, 1 with probability 1/2, 2 with probability 1/4 so the mean is (1/4)(1/2)+ (1/2)(1)+ (1/4)(2)= 9/8 and the variance is (1/4)(1/4- 9/8)2+ (1/2)(1- 9/8)2+ (1/4)(2- 9/8)2= (1/4)(-7/8)2+ (1/2)(-1/8)2+ (1/4)(7/8)= (1/4)(49/64)+ (1/2)(1/64)+ (1/4)(49/64)= 100/256= 25/64.

Now, what probability distribution are you talking about?

(And if "X/Y" means "X given Y" rather than "X divided by Y" you will need to specify the joint distribution as well.)
 
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