how to show that definition for stochastic process in continuous time applies to stock prices

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I know that the formal definition of a stochastic process is: {$X(t,\omega)\,\,t\ge0$} is a stochastic process if:

  1. For any fixed $t\ge0$, $X(t,\omega)$ is a random variable

  2. For any fixed $\omega$ being an element of $\Omega$, $X(t,\omega)$ is a function of $t$

But how do I apply this to {$S(t),\,\,t\ge0$}? Can I formally say that:

{$S(t)\,\, t\ge0$} is defined as a stochastic process if I let {$S(t,\omega)\,\, t\ge0$} be a function from $[0,\infty)\times\Omega$ to $\mathbb{R}$

  1. For any fixed $t\ge0$, $S(t,\omega)$ is a random variable

  2. For any fixed $\omega$ being an element of $\Omega$, $S(t,\omega)$ is a function of $t$

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This really is just a matter of notation. Usually, when considering random variables (or indeed, stochastic processes), you don't specify the dependency on $\omega\in\Omega$. That is, if $X$ is a real-valued random variable, then $X$ is a function $\Omega\to\mathbb{R}$, and for $\omega\in\Omega$, $X(\omega)\in\mathbb{R}$; but commonly we just write $X\in\mathbb{R}$. That is, we don't specify the variable $\omega$, since this is inherently unknown. Note however, that this is actually abuse of notation.

For your process $\{S(t); t\geq 0\}$, for each $t\geq 0$ you can say that $S(t)$ is a function $\Omega\to\mathbb{R}$. If you want the dependency on $\omega\in\Omega$ to be explicit, you can use $S(t)(\omega)$, $S(t;\omega)$ or even $S(t,\omega)$ as you suggest, though I would prefer any of the former two. As I said; it's a matter of notation.

That being said, it doesn't make sense to say that $S(t;\omega)$ is a random variable, since here $\omega$ is explicity given, and hence $S(t; \omega)$ is in fact just a real number. A random variable is not a number, but a function. You could however say that $S(t; \cdot):\Omega\to\mathbb{R}$ (or, just $S(t)$) is a random variable.