To my knowledge, a Stochastic Process is a set of Random Variables indexed by time.

Let's take the Bernoulli process as an example, which is a sequence of independent and identically distributed (iid) random variables.

But aren't those random variables in the Bernoulli process the same? Since every random variable is just a function that maps an outcome from the sample space to a real number, $F: S \to \mathbb{R}$.

If they are just a set of random variables that measure a coin-flip $\{Head, Tail\}$ to $\{T, F\}$, what makes these random variables different so that I need to index them by time ($X_1, X_2, ...$)?

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    $\begingroup$ If two random variables are independent, they can't be the same variable. Otherwise there's no way to distinguish, say, $E[X_1^2]$ and $E[X_1 X_2]$. $\endgroup$ Mar 2 at 2:58
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    $\begingroup$ Intuitively, a random variable is the outcome of an experiment. If we toss a fair coin $10$ times, we have $10$ i.i.d. Bernoulli r.v's, but obviously the outcome of the tosses may be different. $\endgroup$
    – saulspatz
    Mar 2 at 3:15
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    $\begingroup$ You may also find it worth examining the various types of equivalence for random variables detailed here on the Wikipedia page. The r.v.s in a stochastic process are equal in distribution, but that's weaker than being equal (or even almost-equal). To put a finer point on it: If two random variables are truly equal, then the probability of them being equal is 1. But the probability of two independent coin flips being equal is 1/2. $\endgroup$ Mar 2 at 4:26
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    $\begingroup$ Yes, technically a random variable is a measurable function on a probability space, but I was talking about the intuitive idea they attempt to capture. There's some discussion of this here $\endgroup$
    – saulspatz
    Mar 2 at 4:39
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    $\begingroup$ @OODWaterball Try this: $X_{i}$ is the random variable that measure the $i$-th coin-flip $\left\{Head_{i},Tail_{i}\right\}$ to $\left\{T,F\right\}$, this explains how the independence works in your example and why these random variables are different. $\endgroup$
    – Q9y5
    Mar 3 at 12:42

Maybe an example will help. Let's take the sample space to be the unit interval $[0,1]$, and the probability measure $P$ thereon to be Lebesgue measure. If $\omega\in[0,1]$ let $$ \sum_{n=1}^\infty {\omega_n\over 2^2} $$ be the base 2 expansion of $\omega$. Now define $X_n(\omega) :=\omega_n$ to be the $n$th binary digit of $\omega$. [In the ambiguous case of an $\omega$ with a terminating expansion, I choose the one than terminates in all $0$s.] The functions $X_n:[0,1]\to\{0,1\}$ are all different functions, but they all have the same distribution: $$ P[X_n=0] = P[X_n=1] = {1\over 2},\qquad n=1,2,\ldots. $$ In addition, the $X_n$s are mutually independent, so they constitute an instance of a "Bernoulli process".


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