The deterministic model is \[ \frac{dX(t)}{dt} = (\lambda -\mu) X(t) \] which can be solved to give \(X(t) = x_0 \exp[ (\lambda-\mu) t]\).
\[ \frac{dp_x}{dt} = \lambda (x-1)p_{x-1} + \mu (x+1)p_{x+1} - (\lambda + \mu) x p_x \]
\[ \frac{dE[X(t)]}{dt} = (\lambda - \mu)E[X(t)] \]
\[ 2X_1 \longrightarrow X_2 \quad \mbox{and} \quad X_2 \longrightarrow 2 X_1 \] with propensities \(h_1(X_1, X_2) = 0.5 k_1 X_1(X_1-1)\) and \(h_2(X_2) = k_2 X_2\)
We formulate the dimer model in terms of moment equations \[ \begin{align*} \frac{dE[X_1]}{dt} &= 0.5 k_1 (E[X_1^2] - E[X_1]) -k_2 E[X_1] \\ \frac{dE[X_1^2]}{dt} &= k_1 (E[X_1^2X_2] - E[X_1 X_2]) + 0.5 k_1(E[X_1^2] - E[X_1]) \\ &\quad+k_2 (E[X_1] - 2 E[X_1^2]) \end{align*} \]
where \(E[X_1]\) is the mean of \(X_1\) and \(E[X_1^2] - E[X_1]^2\) is the variance
The \(i^{{\small\text{th}}}\) moment equation depends on the \((i+1)^{{\small \text{th}}}\) equation
Rewriting \[ \frac{dE[X_1]}{dt} = 0.5 k_1 ( E[X_1^2] - E[X_1]) -k_2 E[X_1] \] in terms of its variance, i.e. \(E[X_1^2] = Var[X_1] + E[X_1]^2\), we get \[ \frac{dE[X_1]}{dt} = 0.5 k_1 E[X_1](E[X_1]-1) + 0.5 k_1 Var[X_1]-k_2 E[X_1] \]
| Distribution | Moments | Cumulants |
|---|---|---|
| Gaussian | \(\mu_3 = 3 \mu_2 \mu_1 - 2 \mu_1^3\) | \(\kappa_3=0\) |
| Poisson | \(\mu_3 = \mu_1+3\mu_1^2 + \mu_1^3\) | \(\kappa_3=\kappa_2=\kappa_1\) |
| Log-Normal | \(\mu_3 = (\mu_2/\mu_1)^3\) | \(\kappa_3=(\kappa_2/\kappa_1)^3 + 3\kappa_2^2/\kappa_1\) |
\[ X \rightarrow \emptyset \] with propensity function \[ h(X) = \frac{k_1 X}{k_2 + X} \] The associated forward Kolmogorov equation is \[ \frac{dp_x(t)}{dt} = p_{x+1}(t)\frac{k_1(x+1)}{k_2 +x +1} - p_x(t)\frac{k_1x}{k_2 +x}, \] where \(p_x(t)\) is the probability of observing \(x\) individuals at time \(t\)
To find the associated moment equations of we multiply by \((k_2+x+1)(k_2+x)\) to give \[ \frac{dp_x(t)}{dt}(k_2+n+1)(k_2+n) = p_{x+1}(t)k_1(n+1)(k_2+n) - p_x(t) k_1 x (k+x+1). \] which yields the equation for the (stochastic) mean \[ \frac{dE(X)}{dt}=-\frac{k_1 E(X)}{k_2+1/2+E(X)}-\frac{dVar(X)}{dt} \]
If we consider a slightly more complicated propensity function \[ h(X) = \frac{k_1 X^2}{k_2 + X^2} \] we get \[ \frac{dE(X)}{ dt} =\frac{1}{2k_2}\left[-4k_1 E(X^3) - (2k_2+1)\frac{d E(X^2)}{dt} - 2\frac{d E(X^3)}{dt} - \frac{d E(X^4)}{dt}\right]. \]
Suppose we have \(N\) chemical species \(\{X_1, \ldots, X_N\}\) and \(L\) reactions \(\{R_1, \ldots, R_L\}\). Reaction \(R_l\) corresponds to \[ \underline s_{l1} X_1 + \ldots + \underline s_{lN} X_N \overset{k_l}{\longrightarrow} \overline s_{l1} X_1 + \ldots + \overline s_{lN} X_n, \] where \(\mathbf{ \underline{s}_{l}}\) and \(\mathbf{ \overline s_{l}}\) are the number of reactants and the products in each species involved in reaction \(l\). The associated moment pde is \[ \frac{\partial M(\theta; t)}{\partial t} = \sum_{{\bf n}={\bf 0}}^{\infty} {\bf \theta}^{{\bf n}} \sum_{l=1}^L \sum_{{\bf i}} a_{l, {\bf i}} \sum_{{\bf k}={\bf 0}}^{{\bf n}} {\bf s_l}^{\bf k} {\bf n \choose \bf k} \mu_{{\bf n} - {\bf k} + {\bf i}}(t) \] where \(k_1+\ldots + k_N \ne 0\), \[ {\bf s_l}^{\bf k} {\bf n \choose \bf k} = s_{l1}^{k_1} {n_1 \choose k_1}\times \ldots \times s_{lN}^{k_N} {n_N \choose k_N} \] and \[ \mu_{{\bf n} - {\bf k} + {\bf j}}(t) = \mu_{n_1 - k_1 +j_1, \ldots, n_N- k_N+j_N}(t) \]
In particular, when there is only a single species in the population, we get \[ \frac{\partial M(t)}{\partial t} = \sum_{n=1}^{\infty} \theta_1^{n} \sum_{l=1}^L \sum_{i=0}^{\infty} a_{l,i} \sum_{k=1}^{n} s_{l,1}^{k} {n \choose k} \mu_{n-k+i}(t) \;. \] So the equation for the first moment is \[ \frac{d\mu_1(t)}{dt} = \sum_{l=1}^L \sum_{i=0}^{\infty} a_{l,i} \;s_{l,1} \;\mu_{i}(t) \] whilst for the \(n^{\text{th}}\) moment we have \[ \frac{d\mu_{n}(t)}{dt} = \sum_{l=1}^L \sum_{i=0}^{\infty} a_{l, i} \sum_{k=1}^{n} {n \choose k} \;s_{l,i}^{k}\; \mu_{n-k+i}(t) \;. \]
