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From here.

The solution given is that

$$E[X_{n+1}\mid X_n] = 1/2\times 2X_n + 1/2\times 0 = X_n$$

Why exactly? In retrospect, I'm not sure I really got it. I'm trying to think about it in terms of indicator functions over events we must define:

Let $A_n = \{ \omega\mid X_{n+1}(\omega) = 2 X_n(\omega) \}$. Note: $A_n \in \mathscr{F}$ and $P(A_n) = 1/2 \ \forall \ n \in \mathbb{N}$

It looks like $X_{n+1} = 1_{A_n} (2X_n) + 1_{A_n^C}(0)$

$\to E[X_{n+1}\mid X_n] = E[1_{A_n} (2X_n) + 1_{A_n^C}(0)\mid X_n]$

$= E[1_{A_n} (2X_n)\mid X_n]$

$= (2X_n) E[1_{A_n}\mid X_n]$

$= (2X_n) E[1_{A_n}]$ (*)

$= (2X_n) P({A_n}) = X_n$ ?

(*) Is $\sigma(A_n)$ independent of $\sigma(X_n)$ ?

What is $\sigma(X_n)$ in terms of $A_n$ anyway? I guess that:

$\sigma(X_0) = \{ \emptyset, \Omega \}$

$\sigma(X_1) = \sigma(A_0) \because X_1 = 2 \times 1_{A_0}$

$\sigma(X_2) \cdots \subseteq \sigma(\sigma(A_0) \cup \sigma(A_1)) = \sigma(A_0, A_1)$

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BCLC
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  • @Did How is this a duplicate? I am asking in terms of indicator functions. The answer in the original question does not seem to be very precise – BCLC Jun 01 '15 at 07:13
  • Quote: "OOOOOOOOOOOOOOOOOHHHHHHHHHHHHHHHHHHHHHHHHH THANKS!!!! HAHAHAHAHAHAHA – BCLC Oct 15 '14 at 13:55" Make up your mind. – Did Jun 01 '15 at 07:38
  • @Did I said "In retrospect, I'm not sure I really got it." – BCLC Jun 01 '15 at 07:46

1 Answers1

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I think you can save yourself a lot of headache by thinking about it in a more heuristic way like this: \begin{align} \mathbb E[X_{n+1}\mid X_n] &= \mathbb E[X_{n+1} 1_{ \lbrace X_{ n+1} = 2 X_n\rbrace} + X_{n+1} 1_{ \lbrace X_{ n+1} = 0\rbrace} \mid X_n]\\ &=\mathbb E[2 X_{n} 1_{ \lbrace X_{ n+1} = 2 X_n\rbrace} \mid X_n]\\ &= 2X_n \mathbb P(X_{ n+1} = 2 X_n \mid X_n) \end{align} that last probability, by definition, is $1/2$

  • Thanks but why is that heuristic? Also, does this mean I'm right except for that last part with the $E[1_{A_n} | X_n] = E[1_{A_n}]$? – BCLC Jun 01 '15 at 07:54
  • @PCLC well it's 'heuristic' i the sense that there are no $\sigma$-algebras, rigorously defined probability spaces and events, etc. You are mostly right, but as you said, $E[1_{A_n}\mid X_n] = E[1_{A_n}]$ might look somewhat dubious at is not fully clear why that is. –  Jun 01 '15 at 11:24
  • Slungpue, actually "no σ-algebras, rigorously defined probability spaces and events, etc." --> "Prove X=(Xn)n≥0 is a martingale w/rt F where X" http://math.stackexchange.com/questions/974969/prove-x-is-a-martingale :P – BCLC Jun 05 '15 at 11:03
  • Actually, why exactly is the last probability 1/2? Sure $P(A_n) = 1/2$, but why is it that $P(A_n | X_n) = 1/2$ ? Also, does it matter if you say $E[X_{n+1} | \mathscr{F}n]$ instead of $E[X{n+1} | X_n]$? – BCLC Jun 05 '15 at 11:04
  • Probability conditioned on a sigma-algebra is a random variable...? http://www.statlect.com/cndprb2.htm – BCLC Jun 05 '15 at 11:16
  • @BCLC well the problem says $X_{n+1} = 2X_n$ with probability $1/2$. That event is defined unconditionally so it doesn't matter what the event is conditioned on. As to $E[X_{n+1}\mid \mathcal F_n]$, we have in this case $E[X_{n+1}\mid \mathcal F_n] = E[X_{n+1} \mid X_n]$ since $X_n$ is a markov chain and the filtration is the filtration generated by $X$ itself. I.e. all the information you have at step $n$ is just the state of $X_n$ at that point. –  Jun 05 '15 at 16:40
  • 1 In general $P(A|B) \neq P(A)$ ? 2 So $\sigma(A_n)$ and $\sigma(X_n)$ are independent? – BCLC Sep 08 '15 at 00:56
  • I think we need $\sigma(A_n) and \sigma(X_n)$ to be independent to say $P(A_n | X_n) = 1/2$? '2 holds iff the events A,B are independent. Choose any two non-independent events to get a counterexample. – Nate Eldredge 44 secs ago' – BCLC Oct 24 '15 at 00:36
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    @BCLC dude... how long have you been working on this problem? –  Oct 24 '15 at 11:37
  • On and off hahaha – BCLC Oct 24 '15 at 17:09