# Sample space of infinite coin tosses experiment

In one of the online statistics/probability web sites http://www.statlect.com/asymptotic-theory/mean-square-convergencee, the following definition is given for the mean - square convergence of the sequence of random variables:

Let $\{X_n\}$ be a sequence of square integrable random variables defined on a sample space $\Omega$. We say that $\{X_n\}$ is mean-square convergent (or convergent in mean-square) if and only if there exists a square integrable random variable X such that $\{X_n\}$ converges to X, according to the metric $d(X_n,X)=E[ (X_n-X)^2]$ , that is,$$\lim_{n \to \infty} E [(X_n-X)^2]=0$$

And then an example follows where it is shown that a sequence of random variables $\{\overline X_n\}$ where $$\overline X_n=\frac{\sum_{i=1}^nX_i}{n}$$ and $X_i$ are iid with the same variance and expactation (say $\mu$), converges in mean square to $X=\mu$. For example $X_i\in\{1,0\}$, is a result of the i-th coin toss. That's all fine.

So just to check, lets say that we have an infinite sequence of coin tosses as above. Then, the sample space $\Omega$ is $\Omega=\{\omega : \omega=(\omega_1, \omega_2...)\}$ where $\omega_i\in\{H,T\}$. That is, one "outcome of the experiment" consists of an "infinite - sequence" of heads/tails, because random variables $\overline X_n$, according to the definition above, all have to be defined on the same sample space. Because if I restrict sample space of $\overline X_n$ to $\Omega_n=\{\omega : \omega=(\omega_1\ldots,\omega_n)\}$, then $\overline X_n$ and $\overline X_{n+1}$ will not have the same sample space anymore. Am I correct about this?

Thanks.

• Yes, $\Omega=\{0,1\}^\mathbb N$ works, as would tons of other probability spaces, say $\Omega=[0,1]$. – Did Apr 13 '16 at 6:18