Minimizer of two random walks 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 :-)
 A: 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.
