I've read the answer of George Lowther on this question and I don't understand his proof. I know just a few things on Markov Process. If I define $\left(X_n\right)_{n \in \mathbb{N}}$ with $X_0=1$ and $$ X_{n+1}=\begin{cases} 2X_n,&\text{with probability }1/2,\\ 3X_n,&\text{with probability }1/3,\\ 6X_n,&\text{with probability }1/6. \end{cases} $$ How can write $$ P\left(X_n=k\right)=1_{\{2\mid k\}}P\left(X_n=k/2\right)/2+1_{\{3\mid k\}}P\left(X_n=k/3\right)/3+1_{\{6\mid k\}}P\left(X_n=k/6\right)/6. $$ I'm not even sure to understand this indicatrice function $1_{\{2\mid k\}}$. Cuuld you explain me how to obtain such results ? And why not the whole proof he made ?

  • $\begingroup$ I have no idea what that notation means. I observe there is an $n$ on the left-hand-side (and no $t$), while there is no $n$ on the right-hand-side (but a $t$). I don't know what $p_t(k/2)$ means in relation to the notation $P[X_n=k/2]$, or perhaps $P[X_t=k/2]$, and I certainly don't know what those indicator-like-things are. $\endgroup$ – Michael May 14 '18 at 19:58
  • $\begingroup$ @Michael Even from the original answer ? $\endgroup$ – Atmos May 14 '18 at 20:00
  • $\begingroup$ You seem to have fixed the $t$ versus $k$ issue. I suspect the indicator function $1_{2|k}$ is for the event "2 divides $k$" (i.e., $k$ is even). $\endgroup$ – Michael May 14 '18 at 20:02

$$p_{t+1}(k):=P(X_{t+1}=k)= \sum_{x}P(X_{t+1}=k|X_{t})P(X_{t-1}=x)=\sum_{x}P(X_{t+1}=k|X_{t-1}=x)p_t(x).$$

To compute $P(X_{t+1}=k|X_{t}=x)$, notice that the only nonzero probabily values that $k$ can take are $2x,3x,6x$, or alternatively we need at least one of $k/2,k/3,k/6$ to be integers. Thus $$P(X_{t+1}=k|X_{t}=x)=1_{x=k/2}/2+1_{x=k/3}/3+1_{x=k/6}/6.$$

Here's an example. Suppose $k=12$. Then $x=6,4,2$ will contribute to $P(X_{t+1}=12)$. Otherwise if say $k=27$, then $x=9$ contributes. Finally if $k=7$, no $x$'s contribute.

So the sum for $p_{t+1}(k)$ reduces to $x\in\{k/2,k/3,k/6\}$ , and the result follows.


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