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I know a simple random walk is defined as $X_t=X_{t-1}+w_t$, but how can I modify this equation is show it is a Markov process?

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    $\begingroup$ If the Markov process follows the Markov property, all you need to show is that the probability of moving to the next state depends only on the present state and not on the previous states, i.e., $P(X_t \mid X_{t-1}, \ldots X_{1}) = P (X_t \mid X_{t-1})$. $\endgroup$
    – Maxtron
    Jan 21, 2020 at 4:15

1 Answer 1

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Showing it is a Markov process means showing that $$P(X_t = x_t | X_1 = x_1,\dots,X_{t-1} = x_{t-1}) = P(X_t=x_t | X_{t-1} = x_{t-1})$$

In your case, you have \begin{align*} P(X_t = x_t | X_1 = x_1,\dots,X_{t-1} = x_{t-1}) & = P(X_{t-1} + w_t = x_t|X_1 = x_1,\dots,X_{t-1} = x_{t-1})\\ & = P(x_{t-1} + w_t = x_t|X_1 = x_1,\dots,X_{t-1} = x_{t-1})\\ & = P(w_t = x_t - x_{t-1}|X_1 = x_1,\dots,X_{t-1} = x_{t-1})\\ \end{align*}

and since $w_t$ is independent of $X_k$ for $k=1,\dots,t-1$ we have \begin{align*} P(X_t = x_t | X_1 = x_1,\dots,X_{t-1} = x_{t-1}) & = P(X_{t-1} + w_t = x_t|X_{t-1} = x_{t-1})\\ & = P(w_t = x_t - x_{t-1})\\ \end{align*}

and we are done.

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