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I have the following question:

I am a student and my teacher defined therandom sampleas 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????

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# The definition of random sample mean and convergence

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