So I have a stochastic process which counts a number of clicks in a certain given interval of time $[0,t]$. The number of clicks recorded is given by,
$$ N(t) = \big[\int_{0}^{t} \sum_i \delta(\tau-t_i)\mathrm{d}\tau \big ] $$
And the general $l$-th moment is given by, where $\langle... \rangle$ is a stochastic average.
$$\langle N^l(t) \rangle = \big \langle \big[\int_{0}^{t} \sum_i \delta(\tau-t_i)\mathrm{d}\tau \big ]^l \big \rangle $$
In general, I have been able to show, $$C_m(t) = \langle N(t)[N(t)-1]...[N(t)-m+1] \rangle = \left( \int_0^{t} \prod_{i=1}^{m} \mathrm{d}t_i \rho_m(t_1,t_2..t_m) \right) $$ These are factorial moments for this point process and $\rho_m$ defines the $m$-particle density. One then finds,
$$\rho_2 = \sum_{i\neq j} \langle \delta (\tau_1 - t_i) \delta (\tau_2 - t_j)\rangle $$
Now, the process is stationary, Markovian, such that $$\rho_1(t) = k$$ $$\rho_2(t_1,t_2) = \rho_2(|t_1-t_2|)$$ I need to show that $$\hat{\rho}_2 (s) = k \frac{\hat{f}(s)}{1-\hat{f}(s)}$$
Symbols with a caret indicate a Laplace transform, and $$f = \langle \delta(t_i-t_{i+1}-\tau)\rangle$$
I have also shown,
$$\hat{f}(s) = \int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t = \sum \frac{(-1)^l s^l \langle \tau^l \rangle }{l!}$$ where $\langle \tau^l \rangle = \langle (t_i - t_{i+1})^l\rangle $
Using these results, and the fact that $\rho_1$ is a constant and $\rho_2$ depends only in the difference in time, I need to derive the relation for the Laplace transform of $\rho_2$ The form of the formula is suggestive, that I am summing a geometric series, and it seems to me like I am supposed to go from a convolution to a a sum of product of Laplace transforms (since the Laplace transform of a convolution is a product of Laplace transforms).
$rho_1$ entering the expression for $\rho_2$ is suggestive of reducing $\rho_2$ to a conditional probability.
I have derived that And i have been taking the laplace transform of this expression but can't seem to massage it into the right form.