Here's a short vade mecum about these generator matrices.

You can understand the process as a discrete Markov chain with exponential holding times before two consecutive jumps. The parameters of the holding times are given by the diagonal of the matrix : . And the transitions are given by the proportions between holding times: (i.e. you normalize the lines (except the diagonal) of the matrix ). Thus, if is the "discrete skeleton" of (meaning that , is the state of after its first jump, etc.), then is a Markov chain of transition matrix given above (of course, due to the definition of ), and the process is described as follows: it stays at the initial position during a time which is exponential with parameter , then it jumps to , where it spends a time following an exponential distribution with parameter , etc.

It is equivalent to be given or both and .

The matrix itself can be used to describe the process, in the following way. Imagine that each "edge" between different states , comes with a "clock" that ticks after an exponential time of parameter . When is at site , it waits until the first of the clocks on the neighbouring edges ticks, and jumps across it; then all the clock are set back to 0 to decide the next jump, and so on.

This gives the following "infinitesimal" description: for , as .

One last thing: if is a stationary distribution for the discrete-time Markov chain (you know how to find that), then defines a stationary measure for (which you may need to normalize in order to get a probability distribution). Notice that is the expectation of the holding time at site (expected value of an exponential r.v.), so that going from (holding time = 1) to (exponential holding times) consists in giving more weight to sites where the Markov chain stays longer. That makes sense.

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So, you should:

- write down what is for each site (this is given by the diagonal, or by the sum of the off-diagonal terms); this gives you the parameters of the sojourn times at different sites.

- write down the matrix obtained by normalizing the off-diagonal entries of (i.e. dividing them by to get lines summing up to 1) and putting zeroes on the diagonal

- compute the stationary distribution of from (solving ...)

- deduce the stationary distribution of using the above formula ( and normalization)