Is a GMM-HMM equivalent to a no-mixture HMM enriched with more states? I'm trying to model sequence data that has 5 hidden states. Observation data conditional to each state is gaussian except for one state for which mixture of 2 gaussians seems more appropriate. Unfortunately, the R package that I'm using (depmix) does not seem to support (without extending the package) a GMM as a possible response distribution. So I was considering the possibility of adding a 6th state so I could interpret one of this enriched set of states as a state for which observation distribution is the first gaussian in my above mixture and another one as a state for which observation distribution is the second gaussian.
Am I wrong thinking that the two approaches are equivalent? 
 A: It is not exactly equivalent: the 6-state HMM can model everything the GMM-HMM can, but not the other way around.
Suppose you start with the GMM-HMM, with $s_5$ being the GMM state, and turn it into the 6-state HMM with states $s_6$ and $s_7$ instead of $s_5$.
Let $p_6$ and $p_7$ be the prior probabilities of the two components of the GMM (that are then transformed into states $s_6$ and $s_7$).
For every transition from a state $s_i$ to $s_5$ in the GMM-HMM (with probability $t$), create two transition probabilities in the 6-state HMM:


*

*$s_i$ to $s_6$ with probability $t \cdot p_6$

*$s_i$ to $s_7$ with probability $t \cdot p_7$
For every transition from $s_5$ to a state $s_i$ in the GMM-HMM (with probability $t$), create two transition probabilities, respectively from $s_6$ and $s_7$, going to $s_i$, both with the same probability $t$.
If I am not mistaken, the resulting 6-state HMM is equivalent to the GMM-HMM.
However, the other way around doesn't always work. Imagine you are starting the the 6-state HMM.
Suppose that the transition probabilities for $s_i \rightarrow s_6$ and $s_i \rightarrow s_7$ are not equal do not have the same ratio as $s_j \rightarrow s_6$ and $s_j \rightarrow s_7$ (EDIT). You could not carry this information into the GMM-HMM.
In short, the 6-state HMM should be able to represent everything the GMM-HMM can, and more.
A: No you are not wrong thinking that. 
If $Y \mid X_1 \sim \alpha f_1(y) + (1-\alpha)f_2(y)$, then you can also let $X_2 \sim \text{Bernoulli}(\alpha)$ independently and say 
$$
Y \mid X_1, X_2 = 1 \sim f_1(y)
$$
and
$$
Y \mid X_1, X_2 = 0 \sim f_2(y).
$$
This is because 
$$
f_{Y|X_1}(y \mid x_1) =  \sum_{i=1}^2f_{Y|X_1,X_2}(y \mid x_1, x_2) f(x_2) = \alpha f_1(y) + (1-\alpha)f_2(y).
$$
Keep in mind the sequence through time $\{X_2^t\}_t$ is iid, and so the Markov structure is overkill (but still perfectly fine).
