3
$\begingroup$

Let $P_i$, $N_i$, and $Z_i$, $i\in\mathbb{N}$ be r.v.'s with the $P_i$, $N_i$, and $Z_i$ being identically distributed with known pdf's $f_P$, $f_N$, and $f_Z$, respectively; and with no dependence between any $P_i$, $N_i$, or $Z_i$. Define the following recurrent process:

$$T_0:=0$$

$$T_i:=\left\{ \begin{array}{ll} Z_i & \quad T_{i-1} = 0 \\ min(T_{i-1}+N_i, 0) & \quad T_{i-1} < 0 \\ max(T_{i-1}+P_i, 0) & \quad T_{i-1} > 0 \end{array} \right.$$

In English: a particle walks randomly on the real number line (starting from the origin) in discrete time steps, where its position cannot change sign from positive to negative or vice-versa in a single step - were it to do so, it is set back at the origin before the next step - and where its evolution depends on the sign of its current position.

What can be said about the pdf's of the $T_i$ in relation to $f_P$, $f_N$, and $f_Z$?

$\endgroup$
2
  • 2
    $\begingroup$ If $P_i, N_i, Z_i$ are all standard normals, then $T_2$ has a pdf of $$\frac{1+\text{erf}(|x|/2)}{4\sqrt{\pi}}e^{-x^2/4}$$ plus a $1/4$ chance of being exactly 0, for a total variance of $3/2+1/\pi$. That complexity does not suggest simple asymptotics. $\endgroup$
    – user44143
    Oct 21, 2018 at 19:55
  • 1
    $\begingroup$ Also: $T_2$ has $1/4$ chance of being exactly 0 whenever $P, N, Z$ have identical continuous distributions with symmetry about the origin. $\endgroup$
    – user44143
    Oct 22, 2018 at 0:31

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.