Apa artinya mengatakan bahwa suatu peristiwa “terjadi pada akhirnya”?

Pertimbangkan jalan acak 1 dimensi pada bilangan bulat dengan keadaan awal : $\mathbb{Z}$ $x\in\mathbb{Z}$

S_{n} = x + \sum_{i = 1}^{n} ξ_{i}

$\begin{equation} S_n=x+\sum^n_{i=1}\xi_i \end{equation}$

di mana kenaikan adalah IID sehingga $\xi_i$ . $P\{\xi_i=1\}=P\{\xi_i=-1\}=\frac{1}{2}$

Seseorang dapat membuktikan bahwa (1)

P^{x} {S_{n} reaches +1 eventually} = 1

$\begin{equation} P^x{\{S_n \text{ reaches +1 eventually}\}} = 1 \end{equation}$

di mana subscript menunjukkan posisi awal.

Biarkan menjadi waktu bagian pertama yang negara . Dengan kata lain, . Orang juga dapat membuktikan bahwa (2) $\tau$ $+1$ $\tau:=\tau(1):=\min\{n\geq0:S_n=1\}$

E τ = + \infty

$\begin{equation} E\tau = +\infty \end{equation}$

Kedua bukti tersebut dapat ditemukan di http://galton.uchicago.edu/~lalley/Courses/312/RW.pdf . Melalui membaca artikel, saya mengerti kedua bukti.

Namun, pertanyaan saya adalah apa arti "akhirnya" dalam pernyataan pertama dan juga secara umum. Jika sesuatu terjadi "pada akhirnya", itu tidak harus terjadi dalam waktu yang terbatas, bukan? Jika demikian, apa sebenarnya perbedaan antara sesuatu yang tidak terjadi dan sesuatu yang tidak terjadi "pada akhirnya"? Pernyataan (1) dan (2) dalam beberapa hal bertentangan dengan saya. Apakah ada contoh lain seperti ini?

EDIT

Hanya ingin menambahkan motivasi untuk pertanyaan, yaitu, contoh langsung dari sesuatu yang terjadi "akhirnya", tetapi dengan waktu tunggu yang terbatas diharapkan.

\begin{aligned} P {walker eventually moves left} & = 1 - P {walker never moves left} \\ = 1 - lim_{n \to \infty} \frac{1}{2^{n}} \\ = 1 \end{aligned}

$\begin{split} P\{\text{walker eventually moves left}\} & = 1 - P\{\text{walker never moves left}\} \\ & = 1-\lim_{n\to\infty} \frac{1}{2^n} \\ & = 1 \end{split}$

Oleh karena itu kita tahu bahwa walker akan "akhirnya" bergerak ke kiri, dan waktu tunggu yang diharapkan sebelum melakukannya (yaitu, bergerak kiri) adalah . $1/(1/2)=2$

Melihat sesuatu yang terjadi "akhirnya" tetapi dengan "waktu tunggu" yang tak terbatas diharapkan cukup imajinasi saya. Bagian kedua dari respons @ whuber adalah contoh hebat lainnya.

probability terminology stochastic-processes randomness random-walk Ye Tian
sumber

pada akhirnya tidak berarti dalam waktu yang terbatas. Itulah tepatnya yang dikontraskan: P terbatas, sementara harapan tau tidak terbatas

seanv507

Yah ada contoh kanonik distribusi Cauchy en.wikipedia.org/wiki/Cauchy_distribution .

seanv507

@ seanv507 - Ya, meskipun rata-rata distribusi Cauchy tidak terdefinisi daripada tidak terbatas (sampel rata-rata dari Cauchy dbn akan melompat-lompat ketika

mendekati tak terhingga daripada terus-menerus bertemu + Infinity). Saya berpikir tentang distribusi Pareto ( en.wikipedia.org/wiki/Pareto_distribution ), yang memiliki mean = Infinity ketika parameter bentuknya

dan memiliki fungsi distribusi probabilitas yang terdefinisi dengan baik.

n

$n$

α <= 1

$\alpha <= 1$

RobertF

@ RobertTF terima kasih - Saya seharusnya mengatakan Pareto

seanv507

Ada beberapa kenyamanan dalam semua ini: jika

, maka

, tetapi tidak sebaliknya.

P (τ = \infty) > 0

$P(\tau=\infty)>0$

E [τ] = \infty

$E[\tau]=\infty$

Alex R.

Jawaban:

How would you demonstrate an event "eventually happens"? You would conduct a thought experiment with a hypothetical opponent. Your opponent may challenge you with any positive number $p$ . If you can find an $n$ (which most likely depends on $p$ ) for which the chance of the event happening by time $n$ is at least $1-p$ , then you win.

In the example, " $S_n$ " is misleading notation because you use it both to refer to one state of a random walk as well as to the entire random walk itself. Let's take care to recognize the distinction. "Reaches $1$ eventually" is meant to refer to a subset $\mathcal{S}$ of the set of all random walks $\Omega$ . Each walk $S\in\Omega$ has infinitely many steps. The value of $S$ at time $n$ is $S_n$ . " $S$ reaches $1$ by time $n$ " refers to the subset of $\Omega$ of walks that have reached the state $1$ by time $n$ . Rigorously, it is the set

Ω_{1, n} = {S \in Ω ∣ S_{1} = 1 or S_{2} = 1 or \dots or S_{n} = 1} .

$\Omega_{1,n} = \{S\in\Omega \mid S_1=1\text{ or }S_2=1\text{ or }\cdots\text{ or }S_n=1\}.$

In your response to the imaginary opponent, you are exhibiting some $\Omega_{1,n}$ with the property that

P_{ξ} (Ω_{1, n}) \geq 1 - p .

$\mathbb{P}_\xi(\Omega_{1,n}) \ge 1-p.$

Because $n$ is arbitrary, you have available all elements of the set

Ω_{1, \infty} = ⋃_{n = 1}^{\infty} Ω_{1, n} .

$\Omega_{1,\infty} = \bigcup_{n=1}^\infty \Omega_{1,n}.$

(Recall that $S \in \bigcup_{n=1}^\infty \Omega_{1,n}$ if and only if there exists a finite $n$ for which $S \in \Omega_{1,n}$ , so there aren't any infinite numbers involved in this union.)

Your ability to win the game shows this union has a probability exceeding all values of the form $1-p$ , no matter how small $p\gt 0$ may be. Consequently, that probability is at least $1$ --and therefore equals $1$ . You will have demonstrated, then, that

P_{ξ} (Ω_{1, \infty}) = 1.

$\mathbb{P}_\xi(\Omega_{1,\infty}) = 1.$

One simple way to appreciate the distinction between "happening eventually" and having an infinite expected first passage time is to contemplate a simpler situation. For $n$ any natural number, let $\omega^{(n)}$ be the sequence

ω^{(n)} = (\underset{n}{\underset{⏟}{0, 0, \dots, 0}}, 1, 1, \dots)

$\omega^{(n)} = (\underbrace{0, 0,\ldots,0}_{n},1,1,\ldots)$

in which $n$ zeros are followed by an endless string of ones. In other words, these are the walks that stay at the origin and at some (finite) time step over to the point $1$ , then stay there forever.

Let $\Omega$ be the set of all these $\omega^{(n)}, n=0, 1, 2, \ldots$ with the discrete sigma algebra. Assign a probability measure via

P (ω^{(n)}) = \frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{(n + 1) (n + 2)} .

$\mathbb{P}(\omega^{(n)}) = \frac{1}{n+1} - \frac{1}{n+2} = \frac{1}{(n+1)(n+2)}.$

This was designed to make the chance of jumping to $1$ by the time $n$ equal to $1-1/(n+1)$ , which obviously approaches arbitrarily closely to $1$ . You will win the game. The jump eventually happens and when it does, it will be at some finite time. However, the expected time when it happens is the sum of the survival function (which gives the chances of not having jumped at time $n$ ),

E (τ) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots,

$\mathbb{E}(\tau) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots,$

which diverges. That is because a relatively large probability is given to waiting a long time before jumping.

whuber
sumber

Am I misunderstanding if I read your first section as boiling down to to an epsilon/delta argument, and thus basically just saying

lim_{n \to \infty} P_{n} = 1

$\lim_{ n \to \infty } P_n = 1$ (where

P_{n}

$P_n$ is the probability of some event after

n

$n$ steps)?

jpmc26

@jpm It doesn't just boil down to it: it is an epsilon-delta argument. In this case "delta" is "

n

$n$ " and "epsilon" is written "

p

$p$ " as a reminder that it is a probability. The emphasis here is on the finiteness of

n

$n$ : limits are defined in terms of finite values and finite operations, not infinite ones.

whuber

I thank an anonymous user for suggesting the use of underbrace in the description of

ω^{(n)}

$\omega^{(n)}$ .

whuber

That something happens eventually means that there is some point in time at which it happens, but there is a connotation that one is not referring to any particular specified time before which it happens. If you say something will happen within three weeks, that is a stronger statement than that it will happen eventually. That it will happen eventually does not specify a time, such as "three weeks" or "thirty-billion years" or "one minute".

Michael Hardy
sumber