Results 1 to 5 of 5

Math Help - What is the variance of the division of 2 random variables?

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    3

    What is the variance of the division of 2 random variables?

    More specifically, if X and Y are Poisson-distributed, what is the variance of X/Y (assuming that X and Y are independent)?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by floats View Post
    More specifically, if X and Y are Poisson-distributed, what is the variance of X/Y (assuming that X and Y are independent)?

    Thanks!
    As far as I know, Poisson random variables can be 0 (with probability e^{-\lambda}), so why would the ratio \frac{X}{Y} make sense?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    3
    I see, that is a good point.

    What if we consider X/(Y+1) instead? What would the variance of that be?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by floats View Post
    What if we consider X/(Y+1) instead? What would the variance of that be?
    To get the variance, you need the expectation, and the expectation of the square. Since the variables are independent, you need E[X]E\left[\frac{1}{Y+1}\right] and E[X^2]E\left[\frac{1}{(1+Y)^2}\right]. I guess you know how to get E[X] and E[X^2] ( \lambda and \lambda+\lambda^2 since I know the variance is \lambda). So let's consider the other ones.

    E\left[\frac{1}{Y+1}\right]=e^{-\lambda}\sum_{n=0}^\infty \frac{1}{n+1}\frac{\lambda^n}{n!}=e^{-\lambda}\sum_{n=0}^\infty \frac{\lambda^n}{(n+1)!} =\frac{1}{\lambda}e^{-\lambda}\sum_{n=0}^\infty \frac{\lambda^{n+1}}{(n+1)!} ; then let m=n+1 in the sum to see that it equals e^{\lambda}-1.

    E\left[\frac{1}{(Y+1)^2}\right]=e^{-\lambda}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\frac{\l  ambda^n}{n!} is more tricky. I'd let f(\lambda)=\sum_{n=0}^\infty \frac{1}{(n+1)^2}\frac{\lambda^{n+1}}{n!} and compute f'(\lambda)=\sum_{n=0}^\infty \frac{1}{n+1}\frac{\lambda^n}{n!}, which we have computed above ; now integrate to get f(\lambda), and multiply by e^{-\lambda}\frac 1\lambda to get E[\frac{1}{(Y+1)^2}].
    Last edited by mr fantastic; March 23rd 2010 at 07:25 PM. Reason: Made the square brackets around fractions bigger (such a lovely reply deserved it).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2010
    Posts
    3
    I got the E[\frac{1}{Y+1}] part but was struggling with the E[\frac{1}{(Y+1)^2}] part.

    Thanks for the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Variance, Means and Random Variables.
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 31st 2011, 10:58 AM
  2. Independent Random Variables - Variance of 2 Die
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 29th 2010, 09:54 PM
  3. Marginal variance with two random variables
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 20th 2010, 05:43 AM
  4. Covariance/variance with two random variables
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 2nd 2009, 01:12 PM
  5. random variables and variance
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: November 26th 2008, 01:25 AM

Search Tags


/mathhelpforum @mathhelpforum