I have a question concerning calculating the Hastings ratio in a specific case (multinomial proposal distribution).
I consider a discrete vector $M$ with integer values that sum up to some number $N$. In each step of the Metropolis-Hastings algorithm I need to propose a new value for $N$ and a new value of $M$.
Let's denote $q(x \mid y)$ as the proposal distribution, i.e., the conditional probability of proposing a new state $x$ given the previous value was equal to $y$.
We propose a new value $N^*$ according to:
$$ q_1\left(N^*\mid N^{\textrm{prev}} \right) \stackrel{d}{=} \lfloor T \rfloor, \textrm{ where } T \sim tNorm(N^{\textrm{prev}}, \sigma). $$ For proposing a new value of $M$, we introduce: $$ q_2\left( M^* \mid N^{*}, M^{\textrm{prev}} \right) \sim Multinomial \left( N^*,\frac{ M^{\textrm{prev}} + s}{ \sum \limits_{i} M^{\textrm{prev},\;i} + s} \right) $$ where $s$ is a realisation of a random variable $$ S \sim N(0, \sigma) $$ and introduces an additional noise.
To perform a Metropolis-Hastings accept-reject step, we need to calculate the acceptance probability: $$ r = \min \left(1, \frac{f\left( x^* \right) q\left( x^{\textrm{prev}} \mid x^* \right)}{f\left( x^{\textrm{prev}}\right) q\left( x^* \mid x^{\textrm{prev}}\right)} \right) $$
where $x = [M,N]$.
$$ \frac{q\left( x^{\textrm{prev}} \mid x^* \right)}{q\left( x^* \mid x^{\textrm{prev}}\right)} = \frac{q_1( N^{\textrm{prev}}\mid N^{*} ) \;\; q_2\left( M^{\textrm{prev}}\mid N^{\textrm{prev}}, \frac{ M^{*} + s}{ \sum \limits_{i} (M^{*,i} + s)} \right)}{ q_1( N^{*}\mid N^{\textrm{prev}} ) \;\; q_2 \left( M^{*}\mid N^{*}, \frac{ M^{\textrm{prev}} + s}{ \sum \limits_{i} (M^{\textrm{prev},i} + s)} \right) } = $$ $$ = \frac{\text{dabsNorm}(N^{\textrm{prev}} \mid N^*, \sigma) }{\text{dabsNorm}(N^* \mid N^{\textrm{prev}}, \sigma)} \frac{\text{dmultinom} \left( M^{\textrm{prev}}\mid N^{\textrm{prev}}, \frac{ M^{*} + s}{ \sum \limits_{i} (M^{*,t} + s)} \right)} {\text{dmultinom} \left( M^* \mid N^*, \frac{ M^{\textrm{prev}} + s}{ \sum \limits_i (M^{\textrm{prev},i} + s)}\right)}, $$ where $\text{dabsNorm}$ denotes density of absolute value of the floor of a random variable following the normal distribution and $\text{dmultinom}$ density of a random variable following the multinomial distribution.
Is the above calculation correct?~
Edit 1: (based on the first comment) I was missing randomness coming from $S$. Multinomial distribution should be integrated with respect to s, or at least $s^*$ and $s^\textrm{prev}$ should appear:
$$ \frac{q\left( x^{\textrm{prev}} \mid x^* \right)}{q\left( x^* \mid x^{\textrm{prev}}\right)} = \frac{q_1( N^{\textrm{prev}}\mid N^{*} ) \;\; q_2\left( M^{\textrm{prev}}\mid N^{\textrm{prev}}, \frac{ M^{*} + s^*}{ \sum \limits_{i} (M^{*,i} + s^*)} \right)}{ q_1( N^{*}\mid N^{\textrm{prev}} ) \;\; q_2 \left( M^{*}\mid N^{*}, \frac{ M^{\textrm{prev}} + s^{\textrm{prev}}}{ \sum \limits_{i} (M^{\textrm{prev},i} + s^{\textrm{prev}})} \right) } = $$ $$ = \frac{\text{dabsNorm}(N^{\textrm{prev}} \mid N^*, \sigma) }{\text{dabsNorm}(N^* \mid N^{\textrm{prev}}, \sigma)} \frac{\text{dmultinom} \left( M^{\textrm{prev}}\mid N^{\textrm{prev}}, \frac{ M^{*} + s^*}{ \sum \limits_{i} (M^{*,t} + s^*)} \right)} {\text{dmultinom} \left( M^* \mid N^*, \frac{ M^{\textrm{prev}} + s^{\textrm{prev}}}{ \sum \limits_i (M^{\textrm{prev},i} + s^{\textrm{prev}})}\right)}. $$
