+ Reply to Thread
Results 1 to 3 of 3

Thread: Quirky question about F-distribution.

  1. #1
    Points: 949, Level: 16
    Level completed: 49%, Points required for next Level: 51

    Posts
    24
    Thanks
    3
    Thanked 0 Times in 0 Posts

    Quirky question about F-distribution.




    Hi

    We can create an F-distribution if we have two independent chi-squared random variables, and we divide them by their own degress of freedom, and divide the result by each other.

    F = \frac{\chi^{2}_{1}/\nu_{1}}{\chi^{2}_{2}/\nu_{2}}.

    We have that:
    E(\chi^{2}_{1}/\nu_{1})=1
    E(\chi^{2}_{2}/\nu_{2})=1

    Now intuitively I would then think that E(F)=1, but we have
    E(F) = \frac{\nu_{2}}{\nu_{2}-2}


    This can offcourse be proven mathematically, but is there an intuitive way of seeing that the expected value can not be 1, even though each part of the fraction has expected value 1? And is it an intuitive way to see that it has to be greater than 1, not \le 1?

  2. #2
    Devorador de queso
    Points: 95,940, Level: 100
    Level completed: 0%, Points required for next Level: 0
    Awards:
    Posting AwardCommunity AwardDiscussion EnderFrequent Poster
    Dason's Avatar
    Location
    Tampa, FL
    Posts
    12,937
    Thanks
    307
    Thanked 2,630 Times in 2,246 Posts

    Re: Quirky question about F-distribution.

    Break it up into (\chi_1^2 / \nu_2 ) * \frac{1}{\chi_2^2 / \nu_2}. Since these two chi-squares are independent we see that

    E\left[(\chi_1^2 / \nu_2 ) * \frac{1}{\chi_2^2 / \nu_2}\right] = E\left[(\chi_1^2 / \nu_2 )\right] * E\left[\frac{1}{\chi_2^2 / \nu_2}\right]

    We know that the first expectation is 1 so your question really boils down to:

    Why is E\left[\frac{1}{\chi_2^2 / \nu_2}\right] \neq 1 even though E\left[\chi_2^2 / \nu_2\right] = 1

    The answer is Jensen's inequality. And I'm actually going to stop here. There are some nice intuitive ways to understand Jensen's inequality but I'll let you do some research on it since that (sort of) answers your question.
    I don't have emotions and sometimes that makes me very sad.

  3. The Following User Says Thank You to Dason For This Useful Post:

    Tarly (01-20-2014)

  4. #3
    Points: 949, Level: 16
    Level completed: 49%, Points required for next Level: 51

    Posts
    24
    Thanks
    3
    Thanked 0 Times in 0 Posts

    Re: Quirky question about F-distribution.


    Cool, from that inequality on wikipedia what it seems it seems that it is always that case that if E(X) = \mu, then E(1/x) \ge1/E(X), not only for chi-squared variables, if we look at positive values, since here f(x)=1/x is convex. And since 1/x is strictly convex, we do not have equality and get strictly inequality, which then says that that E(F) is strictly greater than 1. I thought that maybe the explanation(for E(1/x)>1/E(x)) was because we have chi-squared, and since the chi-squared is skewed this could possibly provide an answer. But the answer is not in the distribution of the random variable?, it is because of the function 1/x?, and we would get the result no matter skew or symmetry?

    Thanks, this was a very interesting inequality!

    I guess if one think really hard, then maybe one can convince oneself intuitively that if we only have positive values, then E(1/x) must be larger than 1/E(x), but for now I do not see how to do this.
    Last edited by Tarly; 01-20-2014 at 02:28 PM.

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats