Markov Process-Variance of time until jump

Question

A Markov process on E = {1, 2} is constructed according to holding time parameters λ1 = 2 and λ2 = 4; the defining Markov chain has transition probabilities

p11 = p12 = 0.5

and

p21 = 1.

How do I calculate the generator matrix for this process?

Also what is the variance of the time (starting from state 1) until the process jumps to state 2?

EDIT USING: \begin{align} \nonumber g_{ij} = \left\{ \begin{array}{l l} \lambda_i p_{ij} & \quad \textrm{ if }i \neq j \\ & \quad \\ -\lambda_i(1-p_{ii}) & \quad \textrm{ if }i = j \end{array} \right. \end{align} I have calculated the generator matrix to be:

-1  1
4  -4

However I still have no clue how to solve the second part of the question,

Any help would be appreciated

what is the variance of the time (starting from state 1) until the process jumps to state 2?

Answer 1

Distribution of time out of state '1' is Exponentially distributed with parameter=1, and possibility of jump out of state '1' is only to '2'.

Hence, Distribution of time between jumps to '1' and '2' is$$T\sim exp(-t)$$So, variance of the time (starting from state '1') until the process jumps to state '2' is 1.