MHB Success runs in Bernoulli trials

[Mathick]

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?

 

[StoneTemplePython]

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} $