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Why random walk is considered a stochastic process? Definition of stochastic processes assumes that each random variable is based on the same probability space.

If we consider random walk: $$X_{n}=V_{1}+V_{2}+...+V_{n}$$ where $V$ are bernoulli random variables then probability space of $X_{1}$ is different than probability space of $X_{10}$, $X_{10}$ can take different values than $X_{1}$

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  • $\begingroup$ You consider all the 'values' that $X_n $ takes for some fixed $n $ normall (eg positiom after $10$ steps). So you are right. $\endgroup$
    – AnyAD
    Apr 26 '17 at 13:16
  • $\begingroup$ So what is definition of random walk? Sequence of $X_{n}$ defined as above? And such thing is not a stochastic process? $\endgroup$
    – mokebe
    Apr 26 '17 at 13:20
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    $\begingroup$ Whose definition of stochastic processes assumes that each random variable is based on the same probability space? See for example math.stackexchange.com/a/17617/6460 In any case, all your $X_n$ take values in the integers (though some with probability $0$) $\endgroup$
    – Henry
    Apr 26 '17 at 13:22
  • $\begingroup$ Yes, the sum of the $V_i $'s. $\endgroup$
    – AnyAD
    Apr 26 '17 at 13:23
  • $\begingroup$ I am only talking about a random walk which is a very basic type of stochastic process. $\endgroup$
    – AnyAD
    Apr 26 '17 at 13:25
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The random walk is considered a stochastic process because it is ;-)

Usually you use the "infinite coin toss" for modelling a random walk and then the problem doesn't occur so take

$\Omega = \left\{\omega = (\omega_1, \omega_2, \ldots), \omega_i \in \{0,1\}, i\in\Bbb N \right\} = \{0,1\}^\Bbb N$ and $V_i(\omega) = \omega_i$

with $V_i$s are bernoulli random variables.

Then $X_n(\omega) = V_1(\omega) + \ldots + V_n(\omega)$ and all r.v. are defined on the same probability space…

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