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

[mqwit]

I know Var(aX + bY)...
I know Var(XY)...
I don't know of a solution to Var(X/Y)... is there one?

 

[statdad]

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.

 

[chroot]

You don't mean X divided by Y do you? You mean X given Y?

- Warren

 

[HallsofIvy]

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.)