Compute $[\mathbb{I}-z\mathbb{P}]^{-1}$

37 Views Asked by At

Unfortunately, I missed some classes and I cannot find information about z-transformation in my book, so I hope somebody can help me! This is an exercise, but I do not have to know the answer, I just want to know how this kind of exercises work and what the z-transform exactly is.

Consider a Markov chain with a two-symbol alphabet and transition matrix $$\mathbb{P}=\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{2}{5} & \frac{3}{5} \end{pmatrix} $$ Prove that $$[\mathbb{I}-z\mathbb{P}]^{-1}= \frac{1}{1-z} \begin{pmatrix} \frac{4}{9} & \frac{5}{9} \\ \frac{4}{9} & \frac{5}{9} \end{pmatrix} + \frac{1}{1-\frac{1}{10}z} \begin{pmatrix} \frac{4}{9} & -\frac{5}{9} \\ -\frac{4}{9} & \frac{5}{9} \end{pmatrix}$$

Again, I do not have to know the complete proof, I just want help with the following:

  • What do they mean with two-symbol alphabet?
  • why do they use $[\mathbb{I}-z\mathbb{P}]^{-1}$ to compute eventually the nth power of $\mathbb{P}$?
  • How do I start to prove this? Can you give me a hint how to begin?

Thanks a lot!