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How can we construct a probability space $(\Omega, \mathcal{F}, P)$ such that there exists indep. r.v's $X_1,X_2,\ldots$ and $Y$ where $X_i$ are Bernoulli r.v. and $Y$ is a Cauchy r.v..

My thought is let $\Omega =\{(w_1,w_w,\ldots,):w_i=1 \text{ or } 0 \} $. $\mathcal{F}=2^\Omega$. Then let $X_i(\omega)=1$ if $\omega_i=1$ and $X_i(\omega)=0$ if $\omega_i=0$. Define also $P(\{w:w_i=1\})=p \in (0,1)$, $P(\emptyset)=0$. Then, we get $X_1,X_2,\ldots $ independent Bernoulli$(p)$ r.v's.

By using Central limit theorem, we can get $Z_1$,$Z_2$ independent standard normal. Finally, $Y:=Z_1/Z_2$ is Cauchy.

Is my argument valid or rigorously enough?

Do I need to show they are independent?

Or check $P$ is really a probability measure?

$Y$ as a limit of $F$-measurable function should be $F$-measurable as well. So it is on the same probability space?

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  • $\begingroup$ You did not explain how you constructed TWO independent standard normal random variables from the sequence $(X_n)$. Until you do, it is difficult to say what is your argument. $\endgroup$ – Did Oct 19 '13 at 17:35
  • $\begingroup$ Another problem being the sigma-algebra you chose... $\endgroup$ – Did Oct 19 '13 at 17:36
  • $\begingroup$ so any suggestions?? $\endgroup$ – poormaths Oct 20 '13 at 6:48
  • $\begingroup$ Sure: that you explain how you constructed two independent standard normal random variables from the sequence (Xn). $\endgroup$ – Did Oct 20 '13 at 7:43

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