From Justin's email:
a quick work-around might involve the following.
- Allow for two time (t) specific random draws: x1(t) and x2(t).
- At the beginning of life, the random draw used to identify labour transitions is set xl(t) = x1(t)
- In each subsequent year, x2(t) is compared against an exogenous probability p;
a. If x2(t) < p, then xl(t) = x1(t)
b. Otherwise, xl(t) = xl(t-1).
Attempt as a first-stage fix of #163