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The definition of random sample mean and convergence

  1. May 7, 2006 #1
    I have the following question:
    I am a student and my teacher defined the random sample as following:

    Def:A simple and repeated random sample of size n from the population (omega,K)
    for the characteristic X:eek:mega->R is a random vector
    V=(X_1,X_2,...,X_n) defined on (omega)^n and with values in R^n as follows:

    for O_n=(o_1,o_2,...,o_n) in omega^n we define X_i(O_n)=X(o_i)
    Then he proves that X_1,...,X_n are stochastic independent and have the same repartition as the characteristic for study,X,and he calls them sample random variables
    Things are well till now.
    Then he defines a new "sample random variable", called "the sample mean" like this:

    m(X,O_n):eek:mega^n ->R


    Now(and here is the problem) he states that the
    sequence of random variables m(X,O_n) for n converges almost sure to M(X) the mean of the characteristic.
    What i don't understand is the follwing:
    how is this a sequence of random variables?
    I mean,when we define a sequence of functions don't they have the same domain?
    Here, m(X,O_n) has the domain omega^n, which changes with n!!!!!!
    Isn't this a problem??
    I looked on the definition of almost sure convergence and it says that all X_n are defined on the same set!!!

    And another problem is the proof of this teacher for the almost sure convergence of the sequence m(X;O_n)-
    he applies the law of large numbers to the sequence X_n!!!!
    But how the hell can he do that,since

    m(X;O_n) we have X_1,....,X_n defined on omega^n and in
    m(X;O_n+1) the random variables X_1,...X_n,X_n+1 are defined on omega^(n+1)
    so X_1,...,X_n in m(X;O_n+1) are different from X_1,...,X_n in omega^n.
    Is this teacher right????
  2. jcsd
  3. Nov 3, 2006 #2


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    Science Advisor
    Homework Helper

    Good points; why don't you raise them with your teacher?
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