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:

$$\mathbb{E}(N_k|N_{k−1}) = N_{k−1} + 1 + (1 − p)\mathbb{E}(N_k).$$


You are conditioning on $N_{k-1}$, which means you are conditioning on the fact that you have $k-1$ consecutive successes. Let $X \sim \text{Bern}(p)$ 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:

$$\begin{equation} \begin{aligned} \mathbb{E}(N_k|N_{k-1}) &= \mathbb{P}(X=1|N_{k-1}) \mathbb{E}(N_k|N_{k-1},X=1) + \mathbb{P}(X=0|N_{k-1}) \mathbb{E}(N_k|N_{k-1},X=0) \\[6pt] &= p \mathbb{E}(N_k|N_{k-1},X=1) + (1-p) \mathbb{E}(N_k|N_{k-1},X=0) \\[6pt] &= p(N_{k-1}+1) + (1-p) (N_{k-1}+1 + \mathbb{E}(N_k)) \\[6pt] &= p(N_{k-1}+1) + (1-p) (N_{k-1}+1) + (1-p) \mathbb{E}(N_k) \\[6pt] &= N_{k-1}+1 + (1-p) \mathbb{E}(N_k). \\[6pt] \end{aligned} \end{equation}$$


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 $N_{k}|N_{k-1}$.


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