# Markov Process-Variance of time until jump

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?

• Your main question is answered on several threads on this site, linked to by a search on Markov generator, so we should consider that part already asked and answered. Thus respondents should focus on the second question about the variance. – whuber Apr 16 '15 at 14:44
• Are you sure that's your generator matrix? Look at the first row. – Cristián Antuña Apr 17 '15 at 15:39
• Corrected the generator – kullapparos Apr 17 '15 at 17:01

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.