*Given the space was occupied for a certain fraction of the previous hour, for what portion of subsequent hours do we expect occupancy?* We approach the problem in three steps. First, we explain the simplest case where boolean occupancy is directly observed. Second, we augment the boolean training with forgetting capability. Finally, we refine the approach to use fractional occupancy in order to make predictions more precise.

### Boolean observed occupancy

Each state of the unrolled Markov chain (Figure \[fig:markovChain\]b) has two outgoing transition paths, analogous to a coin toss where the coin’s bias is unknown. The well-known probability function of a biased coin is $$\psi(\theta,N,N_{H})=\binom{N}{N_{H}}\theta^{N_{H}}(1-\theta)^{N-N_{H}},$$ where $\binom{N}{N_{H}}$ is the number of ways to permute $N_{H}$ heads in a sequence of $N$ tosses, and $\theta$ is the heads bias (with 0.5 being a fair coin). This function can be parameterized on $\theta$, $N$, or $N_{H}$ depending on the purpose. With the bias $\theta=\theta_{0}$ known and the number of tosses $N=N_{0}$ fixed, the probability of obtaining $N_{H}$ heads, $\psi(\theta=\theta_{0},N=N_{0},N_{H})$, is a discrete binomial distribution over $N_{H}$. When $N$ and $N_{H}$ are fixed, $\psi(\theta,N=N_{0},N_{H}=N_{H_{0}})$ is the probability density over the bias $\theta$, with $\int_{0}^{1}\psi(\theta,N=N_{0},N_{H}=N_{H_{0}})\,\D\theta=1$.[^6]

Instead of computing $\psi$ using $N$ and $N_{H}$ 