Consider a random sequence of ones and zeroes (successes and failures) $x_0,x_1,x_2...$ defined as follows. There is another sequence $c_0,c_1,c_2...$ which counts the number of successive failures. If $x_n=0$ (failure), then $c_{n+1}=c_{n}+1$. If $x_n=1$ (success), then $c_{n+1}=0$. $c_0=0$
The value of $x_n$ simply follows a Bernoulli distribution that succeeds with probability $f(c_n)$, where $f$ is some function whose outputs are all between 0 and 1.
Given $f$, is there an easy way to find the expectation of $x_n$ in the long run that doesn't involve simulating the process or using brute force? Please also provide an example.
At any time $n$ for which $c_n$ equals $0$ (meaning that $x_{n-1}=1$, or $n=0$), the probability that the next success happens at time $n+k$ (for any $k\ge0$) is $\big( 1-f(0) \big)\big( 1-f(1) \big)\cdots \big( 1-f(k-1) \big) f(k)$. Therefore the expectation of $x_n$ is asymptotically equal to the reciprocal of the expectation of the number of time steps from one success to the next, or in other words $$ \bigg( \sum_{k=0}^\infty (k+1) \big( 1-f(0) \big)\big( 1-f(1) \big)\cdots \big( 1-f(k-1) \big) f(k) \bigg)^{-1} = \bigg( \sum_{k=0}^\infty (k+1) f(k) \prod_{j=0}^{k-1} \big( 1-f(j) \big) \bigg)^{-1} . $$