Results 1 to 2 of 2

Math Help - Mean square convergance

  1. #1
    Newbie
    Joined
    Mar 2013
    From
    Poland
    Posts
    1

    Mean square convergance

    Hey, can someone could help me with this proof?
    If (X_{n})_{n in N}, (Y_{n})_{n in N} and X,Y are independent random variables, then
    l.i.m (X_{n}\cdot Y_{n})=(l.i.m X_{n}) \cdot (l.i.m. Y_{n})
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie ButterflyM's Avatar
    Joined
    Mar 2013
    From
    Cambridge
    Posts
    8

    Re: Mean square convergance

    Yes, I saw once a proof similar to that. Not quite sure I can get it formally together, but remember independence implies that the covariance is zero. This further, in general, implies mean independence, that is, E[XY]=E[X]E[Y]

    Now, without going into probability limits, when you just multiply the series by 1/N and take the normal limit, you have by the law of large numbers lim_{N\rightarrow \infty}XY/N=E[XY].

    For independence, as shown, it follows that E[X]E[Y] or again expressed in limits  lim_{N\rightarrow \infty}X/N lim_{N\rightarrow \infty}Y/N
    Last edited by ButterflyM; March 23rd 2013 at 03:22 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergance of SF(x)
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: October 5th 2010, 02:33 PM
  2. Replies: 2
    Last Post: June 27th 2010, 02:26 PM
  3. Interval of Convergance
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 1st 2010, 03:08 PM
  4. Convergance
    Posted in the Differential Geometry Forum
    Replies: 8
    Last Post: November 15th 2009, 11:21 AM
  5. Series convergance?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 22nd 2009, 07:42 PM

Search Tags


/mathhelpforum @mathhelpforum