People come to a website every month and some of them subscribe. If they subscribe, they have to pay 1 dollar a month. Every month, we get one new subscriber with probability $p=0.1$. We never get more than one new subscriber. At the beginning of every month, all current subscribers (who joined last month or earlier) toss a coin with probability $q=0.5$ of heads independently. If they see heads, they unsubscribe. In the steady state, how much money will the website be making on average every month?
Is it true that if $p$ is very high and $q$ is very low, the money the website makes every month will just grow unbounded? That is trivially true if $p>0$ and $q=0$. But what are the conditions on $p$ and $q$ when the money per month will stay finite as time progresses?
If this were a finite state Markov chain, we would just get the steady state probabilities of the states and get the average of that distribution. But this chain has an infinite number of states.
Let $S_n$ denote the number of subscribers in month $n$, which is also the total payments received in month $n$. Now, out of $S_n$ subscribers, the number of subscribers next month is a Binomial r.v. with parameter $1-q$ and consequently, the expected number of subscribers next month given $S_n$ is $$E[S_{n+1}|S_n] = (1-q)S_n + p $$ where $p$ appears because independent of $S_n$, the expected number of additional subscriber is $p$.
Using the law of iterated expectations, $$E[S_{n+1}] = E[E[S_{n+1}|S_n]] = E[(1-q)S_n + p] = (1-q)E[S_n] + p $$ So this gives us a recursion. Since $E[S_1] = p$, we find \begin{align*} E[S_2]&= (1-q)p+p = [1+(1-q)]p\\ E[S_3]&= (1-q)[[1+(1-q)]p]+p = [1+(1-q)+(1-q)^2]p\\ \vdots&\\ E[S_n]&= [1+(1-q)+(1-q)^2+\cdots+(1-q)^{n-1}]p = \frac{1-(1-q)^n}{q}p \end{align*}
Therefore, in the long run, as $n\to \infty$, the expected number of subscribers (thus expected revenue) will be
$$E[S_{\infty}] = \frac{1}{1-(1-q)}p = \frac{p}{q} $$ And as you claimed, this number increases unboundedly as $q\to 0$.