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Consider the following two random walks:

The first random walk $\{S_n\}$ has i.i.d. step size $$ X_i\sim\mathcal{N}(1,1) $$ The second random walk $\{S'_n\}$ has i.i.d. step size $$ Y_i\sim\mathcal{N}(4,4) $$ And let $t_1,t_2$ be the minimizer of the two random walks: $$ t_1=\arg\min_n S_n $$ $$ t_2=\arg\min_n S'_n $$

I'm now interested in the distribution of $t_1$ and $t_2$, are they kind of a rescale of each other? It makes some sense since a step in random walk $2$ is the same as four steps in random walk $2$.

However, I don't know how to prove or disapprove this, any help would be appreciated :-)

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  • $\begingroup$ @AnthonyQuas, Thanks for your reply! But why does $S_n'/2$ have the same distribution as $S_n$? Just let $n=1$, then the former is a Normal(2,1) and the latter is a Normal(1,1) $\endgroup$
    – Oliver
    Commented May 16, 2017 at 22:58
  • $\begingroup$ Sorry. My mistake. $\endgroup$
    – Anthony Quas
    Commented May 16, 2017 at 23:24
  • $\begingroup$ @AnthonyQuas, Haha thanks anyway. Let me know if you have any thoughts :-) $\endgroup$
    – Oliver
    Commented May 18, 2017 at 0:22

1 Answer 1

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The topic belongs to the fluctuation theory of random walks. See e.g. Feller II, 2nd ed. (1970), chapter XII.

Here is some useful (condensed) information:

(1) (for each $n$) the position of the (first) maximal (resp. minimal) term in $S_0=0,S_1,\ldots,S_n$ and the number of strictly positive (resp. strictly negative) terms in $(S_1,\ldots,S_n)$ have the same distribution. In the case of positive drift and continuous step distribution the minimizer and the total number of strictly negative terms therefore have the same distribution.

(2) (as a consequence of Sparre Andersen's theorem) the generating function $g(z)$ of the number of strictly negative terms can be expressed in the form

$$g(z) =\exp\left\{-\sum_{k=1}^\infty (1-z^k)\frac{\mathbb{P}(S_k<0)}{k}\right\}$$

(3) For the normal random walk with $N(\delta,\sigma^2)$-distributed steps the rhs can be rewritten in a way that shows that it depends only on the sign of $\delta$ and the quotient $\frac{\delta^2}{\sigma^2}$

In this sense the two distributions can be considered as scaled versions of each other.

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  • $\begingroup$ Thanks a lot for the information! I checked the related chapters in the book, however I do have some questioned about the answer you wrote above. In $(2)$, how did you get the generating function $g(z)$? It seems that this was a little different than what the book had. Also, in $(3)$, we know that $g(z)$ is a function only of $\frac{\delta}{\sigma}$, but from that how can we tell that the two distributions are scaled versions of each other? $\endgroup$
    – Oliver
    Commented May 22, 2017 at 21:47
  • $\begingroup$ (i) the formula in (2) is another consequence of the Sparre Andersen transformation, see e.g. Theorem 4.4 here (ii) "kind of rescale" leaves room for interpretation. I just meant the fact that exchanging the "scale parameter" $\frac{\delta^2}{\sigma^2}$ transforms one distribution into the other. $\endgroup$
    – esg
    Commented May 23, 2017 at 16:11
  • $\begingroup$ (1)It's a little weird since on the book, XII.7 Theorem 1 states that $\log \frac {1}{1-\tau(s)} = \sum_{i=1}^\infty \frac {s^n}{n}\mathbb{P}\{S_n>0\}$. And those two doesn't seem to match. Anyway, I'll check that paper also. (2) So I guess what you meant was $\delta/\sigma$ is the only parameter? Not the 'scale parameter' in the statistical 'scale family' sense? $\endgroup$
    – Oliver
    Commented May 25, 2017 at 19:23
  • $\begingroup$ (1) Yes, they do match. The book treats ascending ladder epochs/no. of positive terms, the formula above rewrites that for descending ladder epochs/no. of negative terms. (2) Yes, that's what I meant. $\endgroup$
    – esg
    Commented May 26, 2017 at 18:11
  • $\begingroup$ Right, I think I get it now :-) Thanks! $\endgroup$
    – Oliver
    Commented May 29, 2017 at 17:10

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