# Markov Chain Limit Value

Suppose I have the following Markov chain. $$X_0 > 0$$ is a fixed constant and for every $$1 \leq n \in \mathbb{N}$$ we have $$X_n = \begin{cases} 1.5X_{n-1} & \text{with probability } 0.5 \\ rX_{n-1} & \text{with probability } 0.5 \end{cases}$$

I want to find the minimum value of $$r$$ such that the limiting value of the $$X_n$$ is greater than or equal to $$X_0$$, the original amount. I figured that making this a martingale would suffice (i.e. setting $$r = 0.5$$) but when I run a simulation the limiting value is zero every time, so obviously this is wrong.

I'm looking for a value $$r$$ such that the limit is exactly $$X_0$$ since (I'm guessing that) anything larger will lead to an infinite limit almost surely. This problem makes no sense to me though. Please help if you can.

The moral of the story seems to be that your process does not converge. Explanation follows.

We consider the process as you wrote, with $$r = 2/3$$.

Let $$P_i$$ be a random variable that is $$2/3$$ with probability $$1/2$$, and $$3/2$$ with probability $$1/2$$. We are interested in the random variable $$Z_n = \prod_{i = 1}^n P_i$$, and we want to figure out how we can get $$Z_n \to 1$$ (in some sense), since $$X_n = Z_n X_0$$ in your notation.

Consider $$\log(Z_n) = \sum_{i = 1}^n \log(P_i)$$. Then each $$\log(P_i) = \pm ( \log(3) - \log(2))$$. So, after rescaling by dividing by $$\log(3) - \log(2)$$, we can analyze the following process:

$$Y_i = \pm 1$$, with probability $$1/2$$ each, and we want to understand $$S_n = \sum_{i = 0}^n Y_i$$.

Well, $$S_n$$ is a classical random walk, it doesn't converge even in distribution unless you rescale it.

So if $$X_n$$ converged in probability or almost surely, then $$S_n = \frac{1}{\log(3) - \log(2)} \log( X_n)$$ would also, because the function we are applying is continuous and so this theorem applies. In any case, because of infinite recurrence of the simple random walk on $$\mathbb{Z}$$, $$X_n$$ will take on all the possible values infinitely many times.

This point of view also explains why $$2/3$$ is the balancing factor -- any other factor and you either drift towards infinity or negative infinity on the simple random walk side.

We can actually say a lot about the generalized model $$X_n=rX_{n-1}:\text{probability }=p$$ $$X_n=\frac{3}{2}X_{n-1}:\text{probability }=1-p$$Let $$K_{n}\sim\ \text{Binomial}(n,p)$$ and put $$X_{n}:=r^{K_n}\Big(\frac{3}{2}\Big)^{n-K_n}x_{0}$$. Intuitively we may think of $$K_n$$ as counting the number of times we multiplied a successive term by $$r$$ in this random process. The case when $$r>1$$ is obvious, so let's assume that $$r\in (0,1]$$. When $$n$$ is large, we can use the normal approximation to the binomial distribution and deduce for fixed $$a>0$$ that $$P(X_n > a)=P\Bigg(K_n < \frac{\ln(a/x_0)+n\ln(2/3)}{\ln(2r/3)}\Bigg)\approx \phi\Bigg(\frac{\ln(a/x_0)}{\sqrt{np(1-p)}\ln(2r/3)}+\sqrt{n}\cdot \frac{\ln(2/3)-p\ln(2r/3)}{\sqrt{p(1-p)}\ln(2r/3)}\Bigg)$$ where $$\phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt$$. We see $$\lim_{n\rightarrow \infty}P(X_n>a)=1 \iff \frac{\ln(2/3)-p\ln(2r/3)}{\sqrt{p(1-p)}\ln(2r/3)}>0 \iff r>\Big(\frac{2}{3}\Big)^{\frac{1-p}{p}}$$ $$\lim_{n\rightarrow \infty}P(X_n Clearly $$r=\Big(\frac{2}{3}\Big)^{\frac{1-p}{p}}$$ is our threshold which is illustrated in this graph. Taking $$p=1/2$$ yields $$r=2/3$$ as required.