0
$\begingroup$

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!

$\endgroup$
  • 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
1
$\begingroup$

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 $X_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.