1
$\begingroup$

I have some probability exercise.

Suppose two players play some game countably many times, where result can be that person $A$ wins or person $B$ wins. So set $\Omega = \{x_A, x_B \}^{\mathbb{N}}$, where $x_A$ means result that $A$ wins and the same for $B$. $\sigma $-algebra $F$ is a product of power set of set $\{x_A,x_B\}$ finitely many times. Person $A$ wins with probability $p$ and person $B$ with probability $(1-p)$.

Lets have shift $T : \Omega \to \Omega $ which is $T(x_1, x_2,... )=(x_2, x_3 ,... )$.

How to show that $T$ is measurable function?

Thanks for help.

  • 0
    For which $\sigma$-algebra? – 2017-02-14
  • 0
    I have edited question. Hope it is okay now. @flytothesurface – 2017-02-14

1 Answers 1

1

Let $B$ be a cylinder set, and note that the collection of all cylinder sets generates our sigma-algebra. So $B = \prod_{n \in \mathbb{N}}A_n$, where $A_n \subset \{x_A, x_B \}$ and, for all but finitely many $n$, $A_n = \{x_A, x_B \}$. Then,

$$T^{-1}(B) = \{x_A, x_B \} \times B$$

is a cylinder set too, so measurable. We are done because we only have to check measurable sets that generate the sigma-algebra in question.

  • 0
    You have introduced new symbols $\chi_A$ and $\chi_B$. Was that intentional? – 2017-02-14
  • 0
    Sorry, that was an accident. – 2017-02-14
  • 0
    Thanks for the answer. It is all good, except for the thing that I don't really see how this $B$ generates our $\sigma$- algebra. Does $A_n$ needs to be = $\{x_A, x_B \}$ for all $n$ in natural numbers or only for finitely many? @aduh – 2017-02-17
  • 1
    Only for finitely many, in general. If $A_n = \{x_A, x_B \}$ for all $n$, then $A_n = \Omega$. My claim is not that $B$ alone generates our sigma-algebra. $B$ is a particular cylinder set, and the collection of _all_ cylinder sets generates our sigma-algebra. I will edit to make that clearer. – 2017-02-17
  • 0
    Great. I do understand everything now. Next question in this exercise is that I have given 2 events. First: $W_A=\{(x_1,x_2,...)∈Ω;x_1=x_A\}$ , so that in the first game person $A$ wins and the second is that $W_B=\{(x_1,x_2,...)∈Ω;x_1=x_B\}$. I need to show that for every event $C$ in our σ-algebra $F$ it holds: $P(T^{1}(C)|W_A)=P(T^{-1}(C)|W_B)=P(C)$, where $P$ is probability measure and it is a product of distributions $\begin{pmatrix} x_A & x_B\\ p & (1-p) \end{pmatrix}$ Would you maybe have some ideas how to prove that? Thank you very much for any help. I would really appreciate it. – 2017-02-18
  • 0
    I'd just use the definition of conditional probability. It looks like it'll be a straightforward calculation. If you get stuck, then you should ask a separate question. – 2017-02-18