2
$\begingroup$

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?

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

$\endgroup$
3
  • 1
    $\begingroup$ 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 $V_1, \ldots, V_n$?? $\endgroup$ Commented Nov 7, 2015 at 9:52
  • 1
    $\begingroup$ 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? $\endgroup$ Commented Nov 8, 2015 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
    Commented Nov 8, 2015 at 11:03

1 Answer 1

1
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ Oh sorry that was supposed to be subseteq not equal. Hehe. Anyway I'll read this later. Thanks $\endgroup$
    – BCLC
    Commented Nov 8, 2015 at 15:19
  • $\begingroup$ I approved self-study, but this is not homework $\endgroup$
    – BCLC
    Commented Nov 8, 2015 at 19:44
  • $\begingroup$ Thanks P.Windridge . So saz is wrong? $\endgroup$
    – BCLC
    Commented Nov 29, 2015 at 22:08
  • 1
    $\begingroup$ 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 :) $\endgroup$ Commented Nov 30, 2015 at 12:23
  • $\begingroup$ lol anyway saz admitted he's wrong. thanks ^-^ $\endgroup$
    – BCLC
    Commented Nov 30, 2015 at 12:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.