Let's say you have a two-state markovian source whose transition matrix is $P=\begin{pmatrix}1-\sigma & \sigma\\ \tau & 1-\tau\end{pmatrix}$, for the state 0 the data rate is 0 and for the state 1 the data rate is R.
Then at any time (discrete) the amount of incoming data $B_k$ is $B_k = \left\{\begin{matrix} 0, p^0_k \\ R,p^1_k \end{matrix}\right.$, where the $p_k$ can be computed as such:
Let's say that you have a 0.5 probability of starting with each state. Then the probability that at the time $k$ you are in the state $i, i \in \begin{Bmatrix}0,1\end{Bmatrix}$ is the $i^{th}$ value of the vector $\begin{pmatrix}\frac 1 2\\ \frac 1 2\end{pmatrix}\cdot P^k$ I manage to get an expression of $P^k$ by using an eigendecomposition. You can have the result here (wolfram)
Then I need to get the log moment generating function of such a source. This is defined by: $$ \Gamma(\theta) = \lim_{n\to\infty} ({ \frac 1 n \times ln(E(e^{\theta \times A_n}))}) $$ With $E$ being the expectation of a random variable, and $A_n$ being the the total arrival, so $\sum_{k=1}^{n} B_k$
That's where I have some trouble.
Can I transform the expectation the product into a product of expectation?
Any advice?
Thank you
Here is the result, if can help:
$\Gamma(\theta)=ln(\frac 1 2 *[(1-\tau)e^{\theta R}+(1-\sigma)+\sqrt{(1-\tau)e^{\theta R}+(1-\sigma)^2-4e^{\theta R}(1-\tau-\sigma)}])$