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$$X_k=\{A(k)X_1\}, \mbox{ with } A(k)=b_2A(k-1)+b_1A(k-1)$$

$$X_k=\{A(k)X_1\}, \mbox{ with } A(k)=b_2A(k-1)+b_1A(k-2)$$

Update on 9/25/2020

Based purely on pattern recognition techniques, I've found this:

$$X_k=\{A(k)X_1\}, \mbox{ with } A(k)=b_2A(k-1)+b_1A(k-1)$$

with $A(0)=\frac{X_0}{X_1}$ and $A(1)=1$. I don't have a proof, but this looks like something very easy to prove. In addition, it helps prove whether or not the stochastic uniform/independence solution is correct or not. More about this next week.

regardless of $m$ and $0\leq \alpha_0,\cdots,\alpha_m\leq 1$, when more and more terms (that is more and more $k$'s) are used to estimate these probabilities. I thus assume (maybe erroneously) that it must also be true for the theoretical properties. This is illustrated further in the Appendix (last section).

regardless of $m$ and $0\leq \alpha_0,\cdots,\alpha_m\leq 1$, when more and more terms (that is more and more $k$'s) are used to estimate these probabilities. I thus assume (maybe erroneously) that it must also be true for the theoretical probabilities. This is illustrated further in the Appendix (last section).