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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.

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For simplicity, I would suggest you look at a translation of the whole problem by $-20$ dollars per second; so you make $0$ dollars per second, and you have a certain probability of making $30$ dollars per second. –  TMM Nov 1 '12 at 11:06
yes but that certain probability is time dependable.. how i calculate the average profit per second with the probability change over time and when it gives money it resets back to the basic value of 5% –  Excadrix Nov 1 '12 at 11:59
@Excadrix - If you like an answer, you can upvote as well as accepting it. –  nbubis Nov 1 '12 at 12:42

1 Answer 1

up vote 2 down vote accepted

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 $A=24.72^-$.

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thank you a lot! i have a question thought. why p=2% and not 5%? –  Excadrix Nov 1 '12 at 12:20
Because p(1)=5% is p+q, not p. –  Did Nov 1 '12 at 12:56
yes but if we had bigger step per example q = 10% then how the starting 5% could be p+q ? –  Excadrix Nov 1 '12 at 12:59
Then we would use the correct family of coefficients (p(k)), whatever their values are. –  Did Nov 1 '12 at 13:03
ok i think i got it ;) thank you very much! –  Excadrix Nov 1 '12 at 13:04

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