I have a trouble understanding the random walk, where $/xi_1,...,/xi_n$ is iid integer valued rv with the probability mass function $f(x)$.

I want to get the expression $p(x,y) = f(y-x)$.

$p(x,y)= P(X_1 = y | X_0 = x) = P(X_0 + \xi_1 = y | X_0 = 1) = P(\xi_1 = y-X_0 | X_0=1) = f(y-x)$.

I could understand up to the last equality. Using the definition of conditional probability, $P(\xi_1=y-X_0 | X_0=1) = P(\xi=y-X_0 and X_0=1)/P(X_0=1)$ but I cannot connect to the final result mathematically rigorously.

I came into a Markov chain class without much exposure to probability theory so I think I lack some understanding that was assumed.

Any help would be appreciated thanks!

  • 1
    $\begingroup$ So many typos in there... Try rather $$P(X_0 + \xi_1 = y \mid X_0 =x)=P(X_0 + \xi_1 = y,X_0=x \mid X_0 =x)=P(x + \xi_1 = y,X_0=x \mid X_0 =x)=P(x + \xi_1 = y \mid X_0 =x)=P(x + \xi_1 = y).$$ $\endgroup$ – Did Jan 16 '16 at 19:12
  • $\begingroup$ Yes, I somehow typed 1 where it should have been x. I think I lacked the understanding of the conditional probability--the first and last equality. Thank you. I think I understand better. $\endgroup$ – Henri L Jan 17 '16 at 3:25

Your equations are not correct. They should be as follows: $$ \begin{align} p(x,y)&\triangleq \mathbb{P}(X_1=y|X_0=x)\\ &= \mathbb{P}(X_0+\Xi_1=y|X_0=x)\\ &=\mathbb{P}(\Xi_1=y-x|X_0=x)\\ &\overset{(a)}{=}\mathbb{P}(\Xi_1=y-x)\\ &=f_\Xi(y-x), \end{align} $$ where $(a)$ follows since $\Xi_1$ is independent of $X_0$.


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