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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.

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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.

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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 $$\lim_{n\rightarrow \infty}P(X_n>a)=1 \iff \frac{\ln(2/3)-p\ln(2r/3)}{\sqrt{p(1-p)}\ln(2r/3)}>0 \iff r>\Big(\frac{2}{3}\Big)^{\frac{1-p}{p}} $$ $$\lim_{n\rightarrow \infty}P(X_n<a)=1 \iff \frac{\ln(2/3)-p\ln(2r/3)}{\sqrt{p(1-p)}\ln(2r/3)}<0 \iff r<\Big(\frac{2}{3}\Big)^{\frac{1-p}{p}}$$ 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.

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