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Given a stochastic process $Z = (Z_n)_{n \geq 0}$ on a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F_n}\}, \mathbb{P})$ where $\mathscr{F_n} = \mathscr{F_n}^{Z} \doteq \sigma(Z_0, Z_1, ..., Z_n)$, where

$$P(Z_{n+1} = 2Z_n) = 1/2 = P(Z_{n+1} = 0)$$ and $$Z_0 = 1$$

we can define the independent random variables $V_0 = 1$,

$V_1, V_2, ...$ ~ $P(V_i = 0) = P(V_i = 2) = 1/2$

$\to Z_n = \prod_{i=0}^{n} V_i$.

How do you prove rigorously that $$\sigma(Z_0, Z_1, ..., Z_n) = \sigma(V_0, V_1, ..., V_n)$$? I get it intuitively: we know up to vn iff we know up to zn


What I tried:

$$Z_n = \prod_{i=0}^{n} V_i \to \sigma(Z_n) \subseteq \sigma(V_0, ..., V_n)$$

$$Z_{n-1} = \prod_{i=0}^{n-1} V_i \to \sigma(Z_{n-1}) \subseteq \sigma(V_0, ..., V_{n-1}) \subseteq \sigma(V_0, ..., V_n)$$

$$.$$

$$.$$

$$.$$

$$Z_{0} = \prod_{i=0}^{0} V_i \to \sigma(Z_0) \subseteq \sigma(V_0) \subseteq \sigma(V_0, ..., V_n)$$

So $$\sigma(Z_0), \sigma(Z_1), \dots, \sigma(Z_n) \subseteq \sigma(V_0, ..., V_n)$$

$$\to (\sigma(Z_0) \cup \sigma(Z_1) \cup \dots \cup \sigma(Z_n)) \subseteq \sigma(V_0, ..., V_n)$$

Does this mean

$$\sigma(\sigma(Z_0) \cup \sigma(Z_1) \cup \cdots \cup \sigma(Z_n)) = \sigma(Z_0, Z_1, ..., Z_n) \subseteq \sigma(V_0, ..., V_n)$$ ?

How about the other direction?

$$\sigma(Z_0, Z_1, ..., Z_n) \supseteq \sigma(V_0, ..., V_n)$$ ?

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    $\begingroup$ Think about what you're asking in an intuitive sense. The sigma algebra generated by some RVs is like the "information" the RVs carry. Does information/knowledge about $Z_0, \ldots, Z_n$ always allow you to reconstruct $V_1, \ldots, V_n$?? $\endgroup$ – P.Windridge Nov 7 '15 at 9:52
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    $\begingroup$ no need to be sorry, I was just suggesting a strategy for tackling the problem. Suppose you observe the sequence $Z_0 = 1, Z_1 = 2, Z_2 = 4, Z_3 = 0, Z_4 = 0, \ldots, Z_{10} = 0$. This tells you that $V_1 = 2$, $V_2 = 2$, $V_3 = 0$. But what about $V_4, \ldots, V_{10}$? Do you have all the information you need to determine these values? $\endgroup$ – P.Windridge Nov 8 '15 at 9:24
  • $\begingroup$ @P.Windridge I mean I'm sorry for giving off the impression that I wanted an intuitive understanding. Ah, I guess we can determine the $V_i = Z_{i}/Z_{i-1}$ for $i \ge 0$ So we say that $\sigma(Z_{i-1}, Z_i) = \sigma(V_i)$ ? $\endgroup$ – BCLC Nov 8 '15 at 11:03
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The $\sigma$-algebras are different.

Consider the $n = 2$ case.

Assume the $Z_i$ ($i = 1,2$) are constructed from the $V_i$ as you wrote in the question. (I assume the probabilities you wrote for the $Z_i$ are conditional probabilities for $Z_{i+1}$ given $Z_i$, as otherwise your question is inconsistent).

Examine the event $E = \{ V_1 = 0, V_2 = 0 \}$.

(Note on this event we have $Z_0 = 1, Z_1 = 0$.)

By definition of the generated $\sigma$-algebra we have $E \in \sigma(V_1,V_2)$.

The three possible observations for $(Z_1,Z_2)$ are $(0,0)$, $(2,0)$ or $(2,4)$.

The event $(Z_1,Z_2) = (0,0)$ is $F_1 = \{ V_1= 0, V_2 = 0\} \bigcup \{ V_1= 0, V_2 = 2 \}$. Note the union here. The event $(Z_1,Z_2) = (2,0)$ is $F_2 = \{ V_1 = 2, V_2 = 0\}$. The event $(Z_1,Z_2) = (2,4)$ is $F_2 = \{ V_1 = 2, V_2 = 2\}$.

The smallest $\sigma$ algebra containing these three disjoint events $F_1,F_2,F_3$ can be written down explicitly; it is the power set $$ \{ \emptyset, F_1, F_2, F_3, F_1\cup F_2, F_1\cup F_3, F_2\cup F_3, F_1\cup F_2 \cup F_3\}. $$

The event $E$ is not there!

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  • $\begingroup$ Oh sorry that was supposed to be subseteq not equal. Hehe. Anyway I'll read this later. Thanks $\endgroup$ – BCLC Nov 8 '15 at 15:19
  • $\begingroup$ I approved self-study, but this is not homework $\endgroup$ – BCLC Nov 8 '15 at 19:44
  • $\begingroup$ Thanks P.Windridge . So saz is wrong? $\endgroup$ – BCLC Nov 29 '15 at 22:08
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    $\begingroup$ Hi BCLC I don't have time to look at the other answer. However, I don't think I made a mistake, so if the other answer contradicts mine then yes I think there's a mistake :) $\endgroup$ – P.Windridge Nov 30 '15 at 12:23
  • $\begingroup$ lol anyway saz admitted he's wrong. thanks ^-^ $\endgroup$ – BCLC Nov 30 '15 at 12:36

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