There is a whole literature on stochastic differential equations of geometrically compounded normal distributions because it is such a heavy part of quantitative finance. And, magically, everything works out great: the solution to the standard SDE $$dX_t = \mu X_t dt + \sigma X_t dW_t$$ has the very desirable outcome that its expectation is $e^{\mu t}$ due to the normality - i.e., it is the same expectation as a deterministic geometric growth at rate $\mu$.
I am trying to find even a rough, asymptotic, or even a qualitative answer to what happens if there is a binomial distribution that grows geometrically. For example, assume (for the discrete case) that $X_0 = 100$ and in each period $X$ either goes up by $z$% or down by $z$%, each with probability 0.5.
I doubt that there exists a proper closed form solution to this case (either discrete or in a limiting continuous process), though I would be thrilled if one did. But does anyone know of even a qualitative approach that helps describe the outcome?
Intuitively it seems that the expectation at any time $t$ for this process, $E_t$, is greater than $X_0$ since the result can grow without bound, but can never reach zero on the downside. For example, after ten periods (in the discrete case) the odds of ten upticks in a row is the same as ten downticks (about 1 in 1024), and the ten upticks give more absolute growth than the ten downticks give in negative growth. But that is hardly a formal argument or even helpfully descriptive in a way that lets one bound results or make useful claims analytically.
After much thought, I think I have got the basic answer...but if anyone has more insight I would appreciate it.
Much like standard geometric Brownian motion, it probably makes sense to first transform the process by taking logarithms. And, instead of $X_0$ being 100 (as I said), let's call it 1.
Now each 'step' in a discrete model is an increase or decrease of a fixed amount, since the log transform makes that: now each step is up or down log(z) units. Further, $log(X_0)=0$ since we have re-based to 1.0 to start.
Now, in the long run (either meaning many iterations or a shrinking $\delta t$), the CLT means the distribution becomes normal with mean zero and a knowable variance. Thus it looks just like 'normal' Browninan motion.
As always in finance, the last step is to take that solution and undo the transform by exponentiating it.
Net result, it seems, is that asymptotically one ends up with the usual log-normal distribution...thanks, as always, to the CLT in the transformed problem.