Simple Bayes? Probability of a state at time t in hidden markov model Suppose we have a HMM with $2$ states -- $A$ and $B$, with $P(A) = 0.4$ and $P(B) = 0.6$. $A$ has a probability of $0.9$ of outputting "hot," and $B$ has a probability of $0.1$ of outputting "hot." Given that at time step $1$, we output hot, what are the probabilities of being in $A$ and $B$? My reasoning is that $P(A|\text{hot}) = P(A$ and hot$)/P($hot$)$. $P($hot$) = 1$ as it's given, and $P(A $ and hot$) = 0.4\times 0.9$. If we calculate $P(B|$hot$)$ similarly, $P(A|$hot$)+P(B|$hot$)$ doesn't equal $1$, even though we must be in either state $A$ or $B$. What am I missing here?

$P(\text{hot})$ isn't one. All you know is that the event "hot" occurred; this doesn't imply that it has probability 1. You need to calculate using Bayes' rule. Using $H$ to denote the event "hot", we see: $$P(A\mid H)={P(H\mid A)P(A)\over P(H)}={P(H\mid A)P(A)\over P(H\mid A)P(A) + P(H|B)P(B)}.$$ Now plug in $P(A)=0.4$, $P(B)=0.6$, $P(H\mid A)=0.9$, $P(H\mid B)=0.1$.

Finally, calculate $P(B\mid H) = 1-P(A\mid H)$.