Conditional Expectation of a Poisson Process Given an Overlapping Interval

Question

I'm trying to tackle the following problem:

Let $N_t, t\geq 0$ be a Poisson process with rate $\lambda = 5$ occurrences/day.

Given that there are $3$ occurrences in the first $12$ hours, what is the expected number of occurrences in the first day?

I believe what I am to compute is $\mathbb{E}(N_1 | N_{0.5} = 3$). I'm not sure how to go about this, since the interval $[0, 0.5] \in [0,1]$, although I do have some thoughts:

1. By the memoryless principle, could we argue that the conditional observation, $N_{0.5} = 3$, doesn't quite matter when looking at the entire interval?
2. If my first thought is faulty (which I have a haunch that it is), could I break down $\mathbb{E}(N_1 | N_{0.5} = 3$) into two expectational values, one for each interval of 12 hours? We know that there were 3 occurrences in the first 12 hours, should I be looking for the expectation of the second interval? How would I do that?

I'm not entirely confident where to go from here- any help would be appreciated.

Answer 1

1. By the memoryless principle, $N_{0.5}=3$ doesn't quite matter when looking at right-half $[0.5,1]$ of the entire interval $[0,1]$.
2. Break down $\Bbb{E}[N_1 \mid N_{0.5} = 3]$ into two terms. \begin{align} \Bbb{E}[N_1 \mid N_{0.5} = 3] &= \Bbb{E}[(N_1-N_{0.5})+N_{0.5} \mid N_{0.5} = 3] \\ &= \Bbb{E}[N_1-N_{0.5} \mid N_{0.5} = 3] + \Bbb{E}[N_{0.5} \mid N_{0.5} = 3] \\ &= \Bbb{E}[N_{0.5}] + 3 \\ &= \frac\lambda2 + 3 \\ &= 5.5 \end{align}