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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 $\sigma(X)$.

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  • $\begingroup$ Problem 1: What is $Y$? (Is $Y=T$?) And what does it mean for a process to determine a random variable? $\endgroup$ – Florian May 25 '11 at 12:41
  • $\begingroup$ @ Florian Yeah, it should be $T$. $\endgroup$ – Hawii May 25 '11 at 18:13
  • $\begingroup$ 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. $\endgroup$ – user940 Jun 8 '11 at 18:38
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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 $t$.

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    $\begingroup$ It's a good solution, but the converse in part 1 seems overly complicated. You could just use $\{T\leq t\}=\{X_t=1\}\subseteq \sigma(X).$ $\endgroup$ – user940 Jun 8 '11 at 11:54
  • $\begingroup$ @Byron Yes, of course. Thanks. I have edited the answer to reflect this improvement. $\endgroup$ – Jason Swanson Jun 8 '11 at 16:42

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