# The definition of random sample mean and convergence

1. May 7, 2006

### bgd2007

Hi.
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 Xmega->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)mega^n ->R

m(X,O_n)=[X_1(O_n)+X_2(O_n)+...+X_n(O_n)]/n.
Good.

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
in

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. Nov 3, 2006

### EnumaElish

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