1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditions on random variable to satisfy limit property

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data
    The problem is to find sufficient and preferably also necessary conditions on random variable X such that its characteristic function g(x) satisfies the limit property:
    [itex]\lim_{t\to0}\frac{1-g(\lambda t)}{1-g(t)}=\lambda^2[/itex]
    I may assume X is symmetric around 0, so the characteristic function is real and even.

    2. Relevant equations
    [itex]g(t)=\int_{-\infty}^{\infty} e^{itx}f_{X}(x) dx[/itex]


    3. The attempt at a solution

    I'm stuck immediately after trying to apply l'Hopital's rule. Any suggestions would be helpful.
    Thank you.
     
  2. jcsd
  3. Sep 4, 2011 #2
    What is the value of [itex]g(0)[/itex]?
     
  4. Sep 4, 2011 #3
    [itex]g(0)=1[/itex] since it is a characteristic function.
     
  5. Sep 4, 2011 #4
    So, what can you say about your limit? Can you evaluate it?
     
  6. Sep 4, 2011 #5
    Using l'Hopital's rule and chain rule, the equivalent condition I need will be
    [itex]\lim_{t\to0}\frac{g'(\lambda t)}{g'(t)} = \lambda [/itex].
     
  7. Sep 4, 2011 #6
    Not really, evaluate these derivatives explicitly:

    [tex]
    \frac{d}{d t} \left(1 - g(\lambda t)\right)
    [/tex]

    [tex]
    \frac{d}{d t} \left(1 - g(t)\right)
    [/tex]
     
  8. Sep 4, 2011 #7
    I must be confused. If I'm not mistaken, the first evaluates to [itex]-\lambda g'(\lambda t)[/itex] and the second [itex]-g'(t)[/itex].
     
  9. Sep 4, 2011 #8
    True, so your limit becomes:

    [tex]
    \lim_{t \rightarrow 0}{\frac{-\lambda g'(\lambda t)}{-g'(t)}} = \lambda \, \lim_{t \rightarrow}{\frac{g'(\lambda t)}{g'(t)}}
    [/tex]

    Before you run in evaluating this limit, you need to consider two cases:

    1) [itex]g'(0) \neq 0[/itex]. Then the limit is simply 1 and you get the result you posted in post #5. However, this is not what you have in the condition of the problem.

    2) [itex]g'(0) = 0[/itex] What is the limit in this case?

    BTW, what does [itex]g'(0)[/itex] mean?
     
  10. Sep 4, 2011 #9
    The condition I posted on #5 is what I desire to have, rather than what will follow from the assumption. Sorry for the confusion.
    By the assumption that X is symmetric around 0 and thus its expected value is 0, it follows g'(0)=0, so it is necessary to use l'Hopital's rule again.
    Thus it seems a sufficient condition is that 0 < var(X) < infinity. Please let me know if I am wrong, but otherwise thank you for your help!
     
  11. Sep 4, 2011 #10
    Yes, you are correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook