In Sinai's book on stochastic processes, the definition for discrete time stochastic processes is "a sequence of random variables $\{X_{n}\}_{n\in{}T}$ defined on a common probability space $(\Omega,\mathcal{B},P)$".

While I think it makes sense that the sequence of random variables may share a commen state space $(\Omega,\mathcal{B})$, I'm wondering why they also share the same probability measure? Aren't the random variables supposed to have different distributions as they move along time?

Can someone help explain where I am mistaken? Thank you!


Careful. The distributions of the $X_n$ might be different or not. The distribution of $X_n$ is the probability measure $P\circ X_n^{-1}$. Now this is notationally elegant, but what it says is that for a Borel set $B\subseteq\mathbb{R}$, the distribution tells me how likely the value of $X_n$ lies in $B$. And that is $$P\big(\underbrace{\{\omega\in\Omega:X_n(\omega)\in B\}}_{X_n^{-1}}\big)=P\big(X_n^{-1}(B)\big).$$ So the distributions can be different even though the underlying probability space is the same. If you want to discuss things like whether $X_n$ and $X_{n+1}$ are independent, you have to go further and use their joint distribution.


Michael gave a great answer, and I know this question was asked 3 years ago. However, I spent a long time trying to understand this, and for anybody else confused, I thought I would elaborate a little on Michael's answer:

The main reference is these notes by Evans. For other helpful references, see Kozdron's notes, Zitkovic's notes, Gallager's notes, and the book by Serfozo. An important note is that many of these explanations, definitions, and examples are taken directly from the above references. I do not claim to have come up with these myself; I merely assembled the material below from the references above!

In a given $\textit{probability space}$ $(\Omega, \mathcal{F}, \mathcal{P})$, a $\textit{random variable}$ maps sample points $\omega \in \Omega$ to $\mathbb{R}^n$. There's also another condition on the measurability of the function, but it's usually satisfied, so we don't worry about it here.

It's easiest to think about one dimension $n = 1$. Some examples include: For a coin, $\Omega = \{\mbox{Heads, Tails}\}$. If we were betting and with a payoff structure of $\$ \, 1$ for a head, and $\$ \, 0$ for a tail, the associated random variable would be: \begin{equation} X(\omega) = \begin{cases} 1 \qquad &\mbox{for} \quad \omega = \mbox{heads}, \\ 0 &\mbox{for} \quad \omega = \mbox{tails}. \end{cases} \end{equation}

To take another example, consider the sum of the numbers when rolling two die. The sample space is $\Omega = \{36 \, \mbox{pairs possible from} \{1, 2, 3, 4, 5, 6 \} \}$, and the random variable is: \begin{equation} X((n_1, n_2)) = n_1 + n_2, \end{equation} where $n_1, n_2$ are the values of the first, second dice rolled, respectively.

Additionally, associated with every random variable is a \textit{probability distribution function}, whch gives the probability that a discrete random variable is equal to some value. In particular, the probability distribution function $f_X : \mathbb{R}^n \to [0, 1]$ of a random variable $X$ is defined by: \begin{equation} f_X(x) = \mathcal{P} \{\omega \in \Omega \mid X(\omega) = x \}. \end{equation} For probability distribution functions associated with discrete random variables, the name \textit{probability mass function} is used; for continuous random variables, the name \textit{probability density function} is used.

For example, the probability mass function for the coin example is: \begin{equation} f_X(x) = \begin{cases} \frac{1}{2} \qquad &\mbox{if} \quad y = 1, \\ \frac{1}{2} &\mbox{if} \quad y = 0. \end{cases} \end{equation}

The probability mass function for the die example is: \begin{equation} f_X(S) = \frac{\min (S - 1, 13 - S)}{36}, \end{equation} where $S \in \{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \}$.

For spinner (such as a compass needle) that spins in a random direction, the random variable $X$ takes values in $[0, 360)$. The probability density of $X$ is $\frac{1}{360}$. While the probability of getting any chosen number is $0$, the probability of getting a number in any open interval is nonzero. This can be calculated by multiplying the (Lebesgue) measure of the set by $\frac{1}{360}$. For instance, choosing the set to be $[0, 90]$, we find the probability to be $(90 - 0) \times \frac{1}{360} = \frac{1}{4}$.

Let $\mathcal{T} \subset [0, \infty)$. A $\textbf{stochastic process}$ is a family of random variables $\{ X_t \}_{t \in \mathcal{T}}$, indexed by $\mathcal{T}$, defined on the probability space $(\Omega, \mathcal{F}, \mathcal{P})$.

If $\mathcal{T} = \mathbb{N}$ (or $\mathcal{T} = \mathbb{N}_0$), we call $\{ X_t \}_{t \in \mathcal{T}}$ a $\textbf{discrete-time process}$. For $\mathcal{T} = [0, \infty)$, we call $\{ X_t \}_{t \in \mathcal{T}}$ a $\textbf{continuous-time process}$.

For example, if $\mathcal{T}$ is a singleton set, like $\mathcal{T} = \{ 1 \}$, then $\{ X_t \}_{t \in \mathcal{T}} = X_1$ is just a single random variable.

We are now ready to tackle the distinction between the probability measure $\mathcal{P}$ and the probability distribution $f_X$ of a random variable $X$. The key is that $\mathcal{P} : \mathcal{F} \to [0, 1]$, and does not vary in time. However, $X : \Omega \to \mathbb{R}^n$ (in other words, $\Omega$ maps sample points to $\mathbb{R}^n$, while $\mathcal{P}$ maps measurable sets to $[0, 1]$) and thus $f_X : \mathbb{R}^n \to [0, 1]$.

Keeping track of the domain and ranges of these functions helps one figure out what $\mathcal{P}, f_X$ are.

We now consider an example.

Recognizing that although we usually identify $t$ as time, mathematically it serves as a placeholder, so we can associate it with very different random variables. In the simplest case, we take $\mathcal{T}$ to be a singleton set $\{ 1 \}$, so $\{ X_t \}_{t \in \mathcal{T}} = X_1$ is just a random variable.

Using the example from Wikipedia, we consider this discrete-time stochastic process and consider a group of people $G$. Let the random variable $X$ map each person to his or her height (i.e. $\in \mathbb{R}$). The probability density function (notice that although it is a discrete-time process, the random variable, height, is continuous) $f_X$ needs to be integrated to find the probability: \begin{equation} \mathcal{P}[a \leq X \leq b] = \int_a^b \, f_X(x) \, \mathrm{d} x. \end{equation}

Here, the key is that $\mathcal{P}$, the probability measure, gives the probability of any (measurable) set, but the distribution $f_X$, when graphed, plots the values of the random variable; namely, the heights (x-axis) and the probability density of people in the group $G$ having that height (y-axis).

To see this for a continuous-time stochastic process, we use the example from Kozdron's notes:

Consider a probability space $(\Omega, \mathcal{F}, \mathcal{P})$, and let $Z$ be a random variable with ($\mathcal{P}(Z = 1)$ means $\mathcal{P}\{\omega \in \Omega \mid Z(\omega) = 1\}$) $\mathcal{P} (Z = 1) = \mathcal{P} (Z = -1) = \frac{1}{2}$. Define the continuous time stochastic process $X = \{ X_t, t \geq 0 \}$ by: \begin{equation} X_t(\omega) = Z(\omega) \sin t \qquad \mbox{for all} \quad t \geq 0. \end{equation}

If we fix $t$, $X(t, \cdot) : \Omega \to \mathbb{R}$ is the random variable $X_t$ which has the following distribution: \begin{equation} \mathcal{P}(X_t = \sin t) = \mathcal{P}(X_t = - \sin t) = \frac{1}{2}. \end{equation}

If we fix $\omega$, $X(\cdot, \omega) : [0, \infty) \to \mathbb{R}$ describes a sample path of $X$ at $\omega$. In this case, there are two possible trajectories, each occurring with probability $\frac{1}{2}$:

Case 1: $Z(\omega) = 1$ The trajectory is $t \to \sin t$.

Case 2: $Z(\omega) = -1$ The trajectory is $t \to - \sin t$.

Notice how $\mathcal{P}$ is fixed, since we are still in the probability space $(\Omega, \mathcal{F}, \mathcal{P})$.

Namely, we see that random variables have different probability functions (also called probability distributions) as time progresses if and only if the random variables that characterize the stochastic process are time-dependent.

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    $\begingroup$ THANK YOU! Your explanation here is exactly what I was looking for. The examples are especially helpful. $\endgroup$ – Idempotent Sep 27 '19 at 14:27

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