[continued] An alternative way would be to apply stochastic optimal control as I replied to another question of yours. The situation there becomes trivial since the value function will not depend on time.

Since you deal with the finite-time horizon, the price is just a martingale even if the increments would be dependent: you just need that the conditional expectation of the increment (given previous history) is zero. As @Uwe suggested, OST would tell you that whenever you buy an asset, the expected gain is profit is zero no matter which selling time you choose. Although optimal stopping theory often considers just a single stopping time, if you take care of details and apply this fact sequentially, you get that the total expected profit for several trades is zero as well.

@Alexander: I extended my answer just to give a flavor of how the method works. Yet again, if you just change some parameters in your model, the procedure stays the same - and you can immediately see whether you'll get a change in the optimal policy.

@TheUser there is also a generalization of the notion of being "universal", see it used e.g. in this answer: math.stackexchange.com/a/248199/5887

Cross-posted

The book by Meyn and Tweedie "Markov Chains and Stochastic Stability" discusses many criteria for the convergence of $\lim_n\|\mu K^n - \pi\|$. Since this difference can be regarded as a size of $K^n$ on functions orthogonal to $\pi$, I guess shall help you as well.

@Douglas: completely agree, but since OP asked how to find a moment using MGT, I thought it's worth reminding how to do this.

In case you know moment-generating function of $\tau$, you can simply use the fact that $$ \mathsf E[\tau^2] = m''(0) $$ where $m(t) = \mathsf E[\mathrm e^{\tau t}]$ is MGT of $\tau$

Another website: math.stackexchange.com is perhaps more suitable for such questions