The problem is from this lecture video. Fish are caught in a Poisson process with parameter $\lambda$. You fish for 2 hours, if in that time you catch at least one fish, then you stop at 2 hours time. Otherwise, you continue to fish until you catch one fish. The question is, what is the expected number of fish $E[N]$. Where $N$ is the random variable for the number of fish.
In the lecture, the professor states that the answer is simply:
$$E[N]=\lambda\tau + P_P(0,2)(1)$$
Here, $P_P$ is the probability distribution for the number of arrivals $k$ in some interval $\tau$:
$$P_P(k, \tau) = \frac{(\lambda\tau)^ke^{-\lambda\tau}}{k!}$$
I don't get why the first term does not have a factor of $1-P_P(0,2)$ which is based on the total expectation theorem:
$$E[N]=P(F)E[N|F] + P(F^c)E[N|F^c]$$
Here, let $F$ be the event that you fish for 2 hours and thus $F^c$ will be the event that you fish for more than 2 hours since you have not caught any fish in the first 2 hours.
Best Answer
Rather unfortunately, the notation that is used is not ideal. A better way to conceptualize the problem is to write the model as follows. As stated in the lecture, let $N_\tau$ be the number of fish caught in $\tau$ hours, if you were to simply fish for that amount of time, irrespective of the number of fish caught. Then $$N_\tau \sim \operatorname{Poisson}(\lambda \tau).$$ Now let $S$ be the random number of fish caught under the conditions of the problem, in which if $N_2 = 0$, then you continue to fish until one more is caught. The reason I use a different variable here is because $S$ is not Poisson and there is no definite stopping time specified; thus I do not want to use the name $N$ as I don't want to confuse it with $N_\tau$. Now we can write $$S = \begin{cases}N_2, & N_2 \ge 1 \\ 1, & N_2 = 0. \end{cases}$$ So $$\operatorname{E}[S] = \operatorname{E}[N_2 \mid N_2 \ge 1]\Pr[N_2 \ge 1] + \operatorname{E}[1 \mid N_2 = 0]\Pr[N_2 = 0].$$ I think this is what you had in mind when applying the law of total expectation. Rather than introducing a new event $F$, one can simply use the existing notation $N_2 \ge 1$ to represent the event that one fishes for only $2$ hours. All that is left is to execute the calculation:
$$\Pr[N_2 \ge 1] = 1 - e^{-2\lambda}, \\ \operatorname{E}[N_2 \mid N_2 \ge 1] = \sum_{n=1}^\infty n \frac{e^{-2 \lambda}}{1 - e^{-2 \lambda}} \frac{(2 \lambda)^n}{n!} = \frac{2\lambda}{1-e^{-2\lambda}},$$ hence $$\operatorname{E}[S] = 2\lambda + e^{-2\lambda}$$ as claimed. But why does this simplify so nicely? Well, let's look at it another way now. We write $$S = N_2 + \mathbb 1(N_2 = 0),$$ where $\mathbb 1$ is the indicator function that equals $1$ if $N_2 = 0$ and $0$ otherwise. Equivalently, $\mathbb 1(N_2 = 0)$ is a Bernoulli random variable in which the parameter $p = \Pr[N_2 = 0]$. Then by linearity of expectation, $$\operatorname{E}[S] = \operatorname{E}[N_2] + \operatorname{E}[\mathbb 1(N_2 = 0)] = 2\lambda + \Pr[N_2 = 0] = 2\lambda + e^{-2\lambda}.$$ This is what the solution in the lecture did. Note we did not use the law of total expectation--we did not condition on the outcome of $N_2$--but we did use the linearity of expectation. The reason this works is because we reframed the model for $S$ in a way that is not directly conditioned on $N_2$.