1
$\begingroup$

This post gives a definition "of A stochastic process in discrete time"

A stochastic process in discrete time n ∈ $N$ = {0, 1, 2, . . .} is a sequence of random variables (rvs) $X_0, X_1, X_2$, . . . denoted by $X = \{X_n : n ≥ 0\}$.

consider the initial state $i$ be sunny (non-probabilistic), and the sequence is (sunny, sunny, rainy).

In this particular case, is this first sunny (non-probabilistic) a random variable?

$\endgroup$
1
$\begingroup$

Technically a random variable is just a function from the sample space to the real numbers. Thus it's fine to have a random variable in the 'non-probabilistic' sense you describe as it would be just a function that maps anywhere in the sample space to the same number.

$\endgroup$
1
$\begingroup$

EDIT (original answer below):

WHuber's comment is dead on. A stochastic process is a statistical description of the values. A realization is a particular set of values that adhere to this description. You can create an ensemble of randomly generated realizations for a given process. (In physics we talk about ensembles. Do probabilists?)

In the ensemble, ALL values in the process are (or at least can be) random variables. That being said, you could speak of the subset of all realizations such that the first day is sunny. Perhaps this is a new stochastic process based on the original, but conditioned on the first day being sunny. I think this is what I hastily put in my original answer.

I suppose one could define a discrete stochastic process such that every 10th value is equal to $\pi$ or something, in which case these would have a fixed value in each realization. Such a process would not strictly fit the definition in your question. But that seems somewhat pathological.


No. The state of the first day is non-probabilistic. If whether it is sunny on subsequent days depends on the first day, then the probabilities are:

$$P[S, 1] = 1$$

$$ \begin{align} P[S, 2] & = P[S, 2 | S, 1] \cdot P[S, 1] \\ & = P[S, 2 | S, 1] \end{align} $$

$$ \begin{align} P[R, 3] & = P[R, 3 | S, 2; S, 1] \\ & = P[R, 3 | S, 2] \cdot P[S, 2 | S, 1] \cdot P[S, 1] \\ & = P[R, 3 | S, 2] \cdot P[S, 2 | S, 1] \end{align} $$

$\endgroup$
  • $\begingroup$ This contradicts the quoted definition, which characterizes every $X_i$ as a "random variable." Indeed, if your answer were true, it would be impossible for any non-constant process to be stationary! We can partly blame the question, because it misuses the term "random variable" and might also conflate the realization of a process with the process itself. Consider asking the OP for clarification. $\endgroup$ – whuber Aug 13 at 20:00
  • $\begingroup$ @whuber have I adequately addressed your comment? $\endgroup$ – abalter Aug 14 at 6:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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