1
$\begingroup$

I now have two random sums $S_{N_1}\equiv\sum_{i=1}^{N_1}X_i$ and $S_{N_2}\equiv\sum_{i=1}^{N_2}Y_j$ where $N_1$ is dependent on $X_i$'s and $N_2$ on $Y_j$'s. In particular, $X_i$'s and $Y_j$'s are i.i.d. exponential random variable given the corresponding states (from a continuous time Markov) are known. And, yes, $X_i$'s and $Y_j$'s are inter-arrival times from two independent Markov-modulated Poisson processes.

I want to know the probability that $$ \mathbb{P}\{S_{N_1} < S_{N_2}\} $$ What am I supposed to begin with?

A rough idea now I am having goes as follows. I first condition on $N_1$ and $N_2$ to get $$ \mathbb{P}\{S_{N_1} < S_{N_2}\} = \sum_{N_1, N_2}\mathbb{P}\{S_{n_1} < S_{n_2}\}\mathbb{P}\{N_1\}\mathbb{P}\{N_2\} $$ then, since $S_{n_1}$ and $S_{n_2}$ will be some distributions (not gamma since the rates in $X_i$'s or $Y_j$'s are distinct) given the CTMC states, I condition again on the states $$ \mathbb{P}\{S_{N_1} < S_{N_2}\} = \sum_{N_1, N_2}\sum_{\mathcal{S}_{1},\mathcal{S}_{2}}\mathbb{P}\{S_{n_1} < S_{n_2}\}K\mathbb{P}_{N_1}\mathbb{P}_{N_2} $$ where $K$ is some term resulting from the conditioning on the CTMC states. Up here, I have no idea how to proceed with the $\mathbb{P}\{S_{n_1} < S_{n_2}\}$ with $S_{n_1}$ and $S_{n_2}$ are sums of exponentials with distinct rates.

Any thoughts will be greatly appreciated

$\endgroup$
  • $\begingroup$ Will you give an example of how the N’s might depend on the X’s and Y’s? $\endgroup$ – Matt F. Oct 30 '17 at 1:06
  • $\begingroup$ Actually, for instance $N_1$ is defined as $N_1 = \inf\{n : X_1 < C, X_2 < C....X_n < C and X_{n+1} > C\} wherein $C$ is a given constant. And $N_2$ has the same definition with $X_i$'s replaced by $Y_j$'s. $\endgroup$ – Liäm Oct 30 '17 at 1:41
  • $\begingroup$ Thanks, Matt. In your case, $S_{N_1}$ should be $\infty$ w.p.1 because whatever X is, we cannot find an N such that $X_{n+1} > 7$. Can you be more specific? $\endgroup$ – Liäm Oct 30 '17 at 5:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.