My problem goes as follows : Consider $X_n$, where $n\geq1$, as the sequence of independent poisson r.v's with mean 5. Further, we define
$N$ = min{$n\geq3$ : $X_{n-2}=2$, $X_{n-1}=1$, $X_n$=0}
and our goal is the find $E(N)$ (i.e, the expectation of the number of trials it takes to get the ordered sequence {2,1,0}. Here is my approach:
I first started by defining two complementing events, $M$ and $L$, where (written informally):
$M$ = first time that {2,1} appears in the sequence, and $L$ = first time that {2} appears in the sequence,
This follows that
$E(N)$ = $E(E(N|M))$ = $E[P(X=0)(E(N|\text{next in seq. is 0}) + P(X\neq0)(E(N|\text{next in seq. not 0}))]$
$=E[(e^{-5}(E(M))+1) + (1-e^{-5})(E(M)+1+E(N))]$
$=E(M)e^{5}+e^{5}$
(first thing I'm unsure of is the case where $X\neq0$. Does the $E(N)$ encompass the +1 already in the case where the counter restarts?)
We can then create an equation for $E(M)$:
$E(M)$ = $E[P(X=1)(E(M|\text{next in seq. is 1}) + P(X\neq1)(E(M|\text{next in seq. not 1}))]$
=$E[5e^{-5}(E(L)+1) + (1-5e^{-5})(E(L)+1+E(M))]$
=$e^{5}/5 +(e^{5}/5)E(L)$
and since now we only need to look for $X=2$, $L$ follows a geometric with $p = P(X=2)$, thus we can solve for $E(M)$ and then $E(N)$.
Now the answer that this gives is not the answer provided in the textbook : $\frac{1}{e^{-5}5e^{-5}5^{2}e^{-5}/2!}$.
I'm not sure where my logic is flawed or if my approach is overall faulty. Additionally, notation for the double expectation is also irking me. I just end up dropping the second expectation without really knowing why (some clarification would be much appreciated).
The usual approach of solving a system of linear equations to determine the mean absorption time in a Markov chain can be applied here. Let $\tau_i$ be the expected number of trials remaining to observe $\{2,1,0\}$ given that the past $i$ trials are of this sequence, for $i=0,1,2$, then from $\tau_i = 1+\sum_{j\in\{0,1,2\}}P_{ij}\tau_j $ we have the equations \begin{align} \tau_0&=1+\left(1-\frac{1}{2} e^{-\lambda } \lambda ^2\right) \tau_0+ \frac{1}{2} e^{-\lambda } \lambda ^2 \tau_1\\ \tau _1&=1+ \left(-\frac{1}{2} e^{-\lambda } \lambda ^2-e^{-\lambda } \lambda +1\right) \tau _0+\frac{1}{2} e^{-\lambda } \lambda ^2 \tau _1+e^{-\lambda } \lambda \tau _2\\ \tau_2&=1+\left(1-e^{-\lambda }\right) \tau_0. \end{align} Solving yields $\tau_0=e^\lambda+2\lambda^{-3}e^{3\lambda}$, and for $\lambda=5$, this is approximately $52452.69112$.