Probability between 2 values that change over time. Average value per second? Lets say I make 20 dollars per second and there is a certain starting probability that instead of 20 I make 50 dollars. let's say that this starting probability is 5%.
Every second that passes and I haven't made 50 dollars, the probability raises by 3%. When I make 50 dollars then the probability for that resets back to 5%.
So my question is, how can I calculate the average value per second of this process?
Thank you.
 A: At each step, one makes either $\$a$ or $\$b$ with $a=20$ and $b=50$. The $b$-steps are separated by i.i.d. numbers of steps distributed as the following positive integer random variable $N$. 
For every $k\geqslant1$, conditionally on $[N\geqslant k]$, the probability of the event $[N\geqslant k+1]$ is $1-p(k)$, where it seems that $p(k)=2\%+k3\%$ (but what happens at step $33$ is not very clear since $2\%+32\cdot3\%\lt1$ but $2\%+33\cdot3\%\gt1$.) Thus, for every $k\geqslant1$,
$$
\mathbb P(N\geqslant k)=\prod_{i=1}^{k-1}(1-p(i)).
$$
The number of steps $S_n$ needed to achieve $n$ $b$-steps is $S_n=\sum\limits_{k=1}^nN_k$, during which one collected $\$M_n$, where $M_n=a\sum\limits_{k=1}^nN_k+(b-a)n$. 
When $n\to\infty$, $S_n\approx n\mathbb E(N)$ and $M_n\approx n(a\mathbb E(N)+b-a)$ hence $\$A$ the average amount collected by step is such that
$$
A=\frac{a\mathbb E(N)+b-a}{\mathbb E(N)}=a+\frac{b-a}{\mathbb E(N)}.
$$
One has
$$
\mathbb E(N)=\sum_{k\geqslant1}\mathbb P(N\geqslant k)=\sum_{k\geqslant0}\prod_{i=1}^{k}(1-p(i)),
$$
that is, in the present case,
$$
\mathbb E(N)=\sum_{k=0}^{32}\prod_{i=1}^{k}\frac{98-3i}{100}.
$$
Numerically, $\mathbb E(N)=6.358^-$ and $A=24.72^-$.
