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

  1. Sep 22, 2008 #1
    I know Var(aX + bY)...
    I know Var(XY)...
    I don't know of a solution to Var(X/Y)... is there one?
  2. jcsd
  3. Sep 22, 2008 #2


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    Not a quick one. In fact, I'm not sure what "solution" to [tex] Var(XY) [/tex] you mean, unless the two variables are (stochastically) independent.

    The problem is with the denominator - it is entirely possible for a random variable [tex] X [/tex] to have finite mean, variance (even higher-order moments) but for the moments of [tex] 1/X [/tex] fail to exist. Without more information there is no other answer to provide.
  4. Sep 22, 2008 #3


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

    - Warren
  5. Sep 22, 2008 #4


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    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 probabilty 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.)
    Last edited by a moderator: Sep 22, 2008
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