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Success runs in Bernoulli trials

Mathick

New member
Nov 28, 2014
23
I tried to understand the following problem:

Consider a sequence of Bernoulli trials with success probability $p$. Fix a positive integer $r$ and let $\mathcal{E}$ denote the event that a run of $r$ successes is observed; recall that we do not allow overlapping runs. We use a recurrence relation for $u_n$, the probability that $\mathcal{E}$ occurs on the $n$th trial, to derive the generating function $U(s)$. Consider $n \ge r$ and the event that trials $n, n − 1, . . . , n − r + 1$ all result in success. This event has probability $p^r$.

On the other hand, if this event occurs then event $\mathcal{E}$ must occur on trial $n−k$ ($0 \le k \le r−1$) and then the subsequent $k$ trials must result in success. Thus we derive

$u_n + u_{n−1}p + · · · + u_{n−r+1}p^{r−1} = p^r$, for $n \ge r$.

Everything is fine until 'On the other hand...' How do we know that event $\mathcal{E}$ must occur on trial $n−k$ ($0 \le k \le r−1$)? And also why the LHS of the above equation equals RHS? I mean I understand RHS - the probability of $r$ successes, but I'm struggling to understand LHS. In my opinion, when we take for example $u_{n-1}$, don't we also take into account the successes which happened outside our run in question? I'm clearly wrong but I don't understand why.

Please, could anyone try to explain it to me?
 

steep

Member
Dec 17, 2018
51
This passage comes directly from Feller volume 1. The idea is observe something and calculate it two different but equivalent ways.
- - - - -
So let the coin tossing process run for a little bit (at least $r$ tosses, but consider $2r$ or more to avoid technicalities with boundary conditions / delayed renewal type of arguments)

If you observe a run of $r$ heads, then that occurs with probability $p^r$ -- this is one way of calculating it and it is easy.

Now for the second way of calculating it: if you observe a run of $r$ heads, then we have a 'look back' window over the last $r$ tosses and we know exactly one renewal occurred during this window (why?). Hence we either had a renewal just now, or one toss prior, or two tosses prior, or ..., or $r-2$ tosses, or $r-1$ tosses.

Each renewal ends on a run of heads, so if we had a renewal $r-1$ tosses ago, then the probability of this event is to see $r-1$ in a row subsequent to the renewal (which means $r-1$ 'new' heads and 'reusing' the last heads from the prior run that caused the renewal) times the probability of renewal at that time, given by $u_{n-(r-1)}$. Now for a renewal at $r-2$ the probability is $u_{n-(r-2)}\cdot p^{r-2}$, ..., for renewal at 1 toss ago the event probability is $u_{n-1}\cdot p$ and for renewal 0 tosses ago the probability is $u_n \cdot 1$, These events are mutually exclusive (reference the earlier "why") so their probabilities add. Hence we have the equality

$ p^r = u_n + u_{n−1}p + · · · + u_{n−r+1}p^{r−1} $