0
$\begingroup$

Suppose two players play some game countably many times, where result can be that person $A$ wins or person $B$ wins. 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 $T(x_1, x_2,... )=(x_2, x_3 ,... )$.

Now i have two 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 $W_B=\{(x_1,x_2,...)∈Ω;x_1=x_B\}$, so that in the first game person $B$ wins.

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} $.

Thanks for help.

  • 0
    Have you tried anything, like just unpacking the definitions? – 2017-02-18
  • 0
    Of course i had. I know I have to do it with conditional expectations, but when I rewrite it with definition of conditional expectation, I am not sure if event $C$ is independet of $(W_A)$. – 2017-02-18
  • 0
    You're using a product measure, right? It would be helpful if you wrote what you've tried so we can see where you're stuck. – 2017-02-18
  • 0
    Well, as you can see, I have a lot of problems with this exercise, so I don't really know if I am doing anything correct. $P$ is product measure, but I have tried it with $P(T^{-1}(A)| W_A)= \frac{P(T^{-1}(A) \cap W_A)}{P(W_A} = \frac{P(w_p,A) }{W_A} $ and don't know how to continue. – 2017-02-18
  • 0
    I'm not sure I follow your notation in the last step. Remember that $T^{-1}(C) = \{ x_A, x_B\} \times C$. And then $ \{ x_A, x_B\} \times C \cap W_A = \{ x_A\} \times C$. Now just calculate $$\frac{P(\{ x_A\} \times C)}{P(W_A)}$$ using the fact that $P$ is product measure. – 2017-02-18
  • 0
    Thanks. Yes, I meant that $ {\x_A\} \times C $ with the last step. I only know that $P(\{x_A\}) =p $ but I don't know what is the $P(W_A)$, I guess it should be $p$ also, because we only care what happend in first game. – 2017-02-18

0 Answers 0