DES - Demonstration on How to Calculate the Transition Rate Matrix Q

Arguments:

DES - Definition of ‘Transition Function Matrix for CTHMC’

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Summary: Declaration of $H(t)$ and $Q$ in respect to each other

Complete Demonstration Starting from:

$$H(t+a)=H(t)\cdot H(a)$$

We want to arrive at:

$$H(t)=e^{Qt}$$

(Obtained from a solution to a Cauchy Problem) Where:

$$Q\triangleq \underset{dt\to 0}{\lim }\frac{H(dt)-I}{dt}$$

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Summary: Check that the limit is actually equal to $Q$

Complete Demonstration Starting from:

$$\begin{array}{rl}& \underset{dt\to 0}{\lim }\frac{H(dt)-I}{dt}\\ & H(0)=I\\ & H(t)=e^{Qt}\end{array}$$

We want to arrive at:

$$\underset{dt\to 0}{\lim }\frac{H(dt)-I}{dt}=Q$$

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Summary: Brake the Recursion of $H(t)$ and $Q$

Complete Demonstration Starting from:

$$\begin{array}{rl}& Q=\left[\begin{array}{ccc}{q}_{1,1}& q_{1,2}& \dots \\ q_{2,1}& q_{2,2}& \dots \\ ⋮& ⋮& \ddots \end{array}\right]\\ & q_{i,i}=-\sum _{j\ne i}{q}_{i,j}\\ & H(dt)=I+Q\cdot dt+o(dt)\\ & H(t)=e^{Qt}\end{array}$$

We want to arrive at:

$$F_i(t)=1-e^{-q_{i,i}t}$$

Useful to know:

$$\begin{array}{rl}& F_i(t)=P(V(i)\le t)\\ & P(V(i)\le t+dt\mid V(i)>t)=\sum _{j\ne i}{p}_{i,j}(dt)\end{array}$$

Note that the matrix $H(t)$ describes the probability transitions from state $i$ to state $j$, so the probability of changing state in time $dt$ is equal to the sum of all rows of $p_{i,j}$ excluding $j=i$ because that is the probability of remaining in state $i$.

Where:

• $F_i(t)$ cdf of state holding time of state $i$
• $V(i)$ State Holding Time of state $i$

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Summary: Compute $H(t)$ from $Q$

Complete Demonstration Starting from:

$$\begin{array}{rl}& p_{i,j}(t)\triangleq P(X(s+t)=j\mid X(s)=i)\\ & p_{i,j}(dt)=q_{i,j}\cdot dt\text{for i=j}\end{array}$$

We want to arrive at:

$$p_{i,j}=-\frac{q_{i,j}}{q_{i,i}}$$

Useful to know:

$$\begin{array}{rl}{p}_{i,j}& =P\left(\bigcup _{t\ge 0}\left\{X(t+dt)=j\mid X(s)=i\forall s\in [0,t]\right\}\right)\\ & ={\int }_0^{\infty }\left\{X(t+dt)=j\mid X(s)=i\forall s\in [0,t]\right\}\end{array}$$

NOTE: in this equations $p_{i,j}$ does NOT depend on time, when we write $p_{i,j}$ without the dependency on time, we are talking about the probability that the next event (whenever it will happen) will change the state from $i$ to $j$

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Declaration of $H(t)$ and $Q$ in respect to each other

Arguments:

$$\begin{array}{rl}& H(t+a)=H(t)\cdot H(a)\\ & H(0)=I\end{array}$$

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Check that the limit is actually equal to $Q$

Arguments:

$$\begin{array}{rl}& \underset{dt\to 0}{\lim }\frac{H(dt)-I}{dt}\\ & H(0)=I\\ & H(t)=e^{Qt}\end{array}$$

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Brake the Recursion of $H(t)$ and $Q$

Arguments:

$$\begin{array}{rl}& Q=\left[\begin{array}{ccc}{q}_{1,1}& q_{1,2}& \dots \\ q_{2,1}& q_{2,2}& \dots \\ ⋮& ⋮& \ddots \end{array}\right]\\ & q_{i,i}=-\sum _{j\ne i}{q}_{i,j}\\ & H(dt)=I+Q\cdot dt+o(dt)\\ & H(t)=e^{Qt}\end{array}$$

We know that:

$$H(dt)=I+Q\cdot dt+o(dt)$$

So we can say that:

$$\begin{array}{rl}& p_{i,j}(dt)=0+q_{i,j}\cdot dt\text{for i=j}\\ & p_{i,j}(dt)=1+q_{i,j}\cdot dt\text{for i=j}\end{array}$$

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Compute $H(t)$ from $Q$

Arguments:

$$p_{i,j}(t)\triangleq P(X(s+t)=j\mid X(s)=i)$$