DES - Chapman-Kolmogorov Equation

Summary:

The Transition Function Matrix $H(t)$ supports this formula:

$$H(t)=H(u)\cdot H(t-u)0\le u\le t$$

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Chapman-Kolmogorov Equation:

Knowing that, from the definition:

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

We can say, knowing the Total Probability Rule that:

$$p_{i,j}(t)=\sum _{\genfrac{}{}{0ex}{}{}{r\in \mathcal{X}}}{p}_{i,r}(u)\cdot p_{r,j}(t-u)$$

Where:

$$p_{i,r}(u)=P(X(s+u)=r\mid X(s)=i)$$

And,

$$p_{r,j}(t-u)=P(X(s+t)=j\mid X(s+u)=r)$$

With the notion of how the matrix $H(t)$ is defined we can say that:

• $p_{i,j}(t)$ is the entry in position $(i,j)$ of the matrix $H(t)$
• $p_{i,r}(u)$ is the entry in position $(i,r)$ of the matrix $H(u)$
• $p_{r,j}(u)$ is the entry in position $(r,j)$ of the matrix $H(t-u)$

So $p_{i,j}(t)$ is the product of the $i$-th row of $H(u)$ and the $j$-th row of $H(t-u)$

Matrix Form of the Chapman-Kolmogorov Equation

So the Chapman-Kolmogorov Equation can be rewritten in matrix form:

$$H(t)=H(u)\cdot H(t-u)0\le u\le t$$

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Demonstration:

DES - Demonstration of the Chapman-Kolmogorov Equation

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~Example: