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Independent trials, each of which is a success with probability $p$, are performed until there are $k$ consecutive successes. Let $N_k$ denote the number of necessary trials to obtain $k$ consecutive
successes, and show that:

$$mathbbE(N_k|N_k−1) = N_k−1 + 1 + (1 − p)mathbbE(N_k).$$

You are conditioning on $N_k-1$, which means you are conditioning on the fact that you have $k-1$ consecutive successes. Let $X sim textBern(p)$ be the outcome of the next trial and consider the cases:

• If $X=1$ then you now have $k$ consecutive successes;


• If $X=0$ then you now have $0$ consecutive successes and you have to start all over again.

Hence, by application of the law-of-total probability you have:

$$beginequation beginaligned
mathbbE(N_k|N_k-1)
&= mathbbP(X=1|N_k-1) mathbbE(N_k|N_k-1,X=1) + mathbbP(X=0|N_k-1) mathbbE(N_k|N_k-1,X=0) \[6pt]
&= p mathbbE(N_k|N_k-1,X=1) + (1-p) mathbbE(N_k|N_k-1,X=0) \[6pt]
&= p(N_k-1+1) + (1-p) (N_k-1+1 + mathbbE(N_k)) \[6pt]
&= p(N_k-1+1) + (1-p) (N_k-1+1) + (1-p) mathbbE(N_k) \[6pt]
&= N_k-1+1 + (1-p) mathbbE(N_k). \[6pt]
endaligned endequation$$

I'll give you the basic reasoning here, but you can write it out formally yourself.

Let $N_k$ be the number of trials necessary to obtain $k$ consecutive successes.

We want to show

$$E[ N_k | N_k-1]= N_k-1 + 1 + (1-p)E[N_k]$$

Firstly, given $N_k-1$, consider the possible values of $N_k-1+1$. If the trial immediately following the $N_k-1$'th trial is a success, then clearly $N_k = N_k-1+1$ with probability $p$.

Next, suppose the trial following $N_k-1$ is a failure. So that we have $N_k-1+1$ total trails with the last one being a failure. Well, since the last one is a failure, then by the independence of each trial, under this situation we would have the conditional expected number of trials until $k$ consecutive successes as

$$N_k-1 + 1 + E[N_k]$$

Since we already have $N_k-1 + 1$ trials, but the last one being a failure "resets" our expectation back to $E[N_k]$.

Hence you can express the original expectation as

$$E[ N_k | N_k-1]= p(N_k-1 + 1) + (1-p)(N_k-1 + 1 + E[N_k])$$

$$=N_k-1 + 1 + (1-p)E[N_k]$$

This answer uses the law of total expectation: $E[ E[X|Y] ] = E[X]$. Here we take $Y$ as the result of the $N_k-1+1$ trial, and $X$ as $N_k|N_k-1$.