Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following are some problems I encountered when self-learning GTM 261 "Probability and Stochastics".

Definition (determinability)

If $X$ and $Y$ are random variables taking values in $(E,\mathcal{E})$ and $(D,\mathcal{D})$, then we say that $X$ determines $Y$ if $Y=f\circ X$ for some $f:E\rightarrow D$ measurable with respect to $\mathcal{E}$ and $\mathcal{D}$.

Problem 1

Let $T$ be a positive random variable and define a stochastic process $X=(X_t)_{t\in\mathbb{R}_+}$ by setting, for each $\omega$ $$ X_t(\omega) = \begin{cases} 0 & \text{if } t < T(\omega) \\ 1 & \text{if } t \geq T(\omega) \end{cases} $$ Show that $X$ and $T$ determine each other. If $T$ represents the time of failure for a device, then $X$ is the process that indicates whether the device has failed or not. That $X$ and $T$ determine each other is intuitively obvious, but the measurability issues cannot be ignored altogether.

In particular, I do not know how to show the measurability part.

Problem 2

A slight change in the preceding exercise shows that one might guard against raw intuition. Let $T$ have a distribution that is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}_+$; in fact, all we need is that $\mathbb{P}\{T = t\} = 0$ for every $t\in\mathbb{R}_+$. Define $$ X_t(\omega) = \begin{cases} 1 & \text{if } t = T(\omega)\\ 0 & \text{otherwise} \end{cases} $$ Show that, for each $t\in\mathbb{R}_+$, the random variable $X_t$ is determined by $T$. But, contrary to raw intuition, $T$ is not determined by $X=(X_t)_{t\in\mathbb{R}_+}$. Show this by the following steps below:

a. For each $t$, we have $X_t = 0$ almost surely. Therefore, for every sequence $(t_n)$ in $\mathbb{R}_+$, $X_{t_1} = X_{t_2} = \ldots = 0$ almost surely.

b. If $V\in \sigma(X)$, then $V = c$ almost surely for some constant $c$. It follows that $T$ is not in $\sigma(X)$.

share|improve this question
    
Problem 1: What is $Y$? (Is $Y=T$?) And what does it mean for a process to determine a random variable? –  Florian May 25 '11 at 12:41
    
@ Florian Yeah, it should be $T$. –  Hawii May 25 '11 at 18:13
    
The underlying question in problem 2 is purely about measurability; the introduction of the probability measure $\mathbb{P}$ is not necessary. In general, $T$ is determined by the process $(X_t)$ if and only if $T$ only takes on countably many values. –  Byron Schmuland Jun 8 '11 at 18:38
add comment

1 Answer 1

up vote 5 down vote accepted

Problem 1: Fix $t\in\mathbb{R}_+$. Then $X_t=1_{\{T\le t\}}$. Since $\{T\le t\}\in\sigma T$, it follows that $X_t$ is $\sigma T$-measurable. Therefore, $\sigma X_t\subset\sigma T$, and this implies $$ \sigma X = \bigvee_{t\in\mathbb{R}_+}\sigma X_t \subset\sigma T. $$ By Theorem 4.4, $X$ is a measurable function of $T$, and so $T$ determines $X$.

For the converse, fix $t\in\mathbb{R}_+$. Then $$ \{T\le t\} = \{X_t = 1\} \in \sigma X_t \subset \sigma X. $$ Since $t$ was arbitrary, this gives $\sigma T\subset\sigma X$, and so $X$ determines $T$.

Problem 2(a): For each $t$, we have $P(X_t \ne 0) = P(T = t) = 0$. Hence, $X_t=0$ a.s. Moreover, given a sequence $(t_n)$, $$ P(\exists n\text{ such that }X_{t_n}\ne 0) = P\bigg(\bigcup_{n=1}^\infty\{X_{t_n}\ne 0\}\bigg) \le \sum_{n=1}^\infty P(X_{t_n}\ne 0) = 0. $$ Thus, $P(X_{t_n}=0,\forall n)=1$, that is, $X_{t_1}=X_{t_2}=\cdots=0$ a.s.

Problem 2(b): Suppose $T$ is $\sigma X$ measurable. Then by Proposition 4.6, we have $$ T = f(X_{t_1},X_{t_2},\ldots), $$ for some sequence $(t_n)$ and some Borel-measurable function $f:\mathbb{R}^\infty \to \mathbb{R}_+$. Define $t=f(0,0,\ldots)\in\mathbb{R}_+$. Then, by part (a), we have $$ T = f(X_{t_1},X_{t_2},\ldots) = f(0,0,\ldots) = t\text{ a.s.} $$ But this contradicts the hypothesis that $P(T=t)=0$ for all $t$.

share|improve this answer
1  
It's a good solution, but the converse in part 1 seems overly complicated. You could just use $\{T\leq t\}=\{X_t=1\}\subseteq \sigma(X).$ –  Byron Schmuland Jun 8 '11 at 11:54
    
@Byron Yes, of course. Thanks. I have edited the answer to reflect this improvement. –  Jason Swanson Jun 8 '11 at 16:42
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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