Applying the Optional Stopping Theorem to martingale with no upper bound?

66 Views Asked by At

Firstly, this is the first question I've posted, so sorry my formatting isn't quite there yet! I've come across a paper on rumour spreading processes which uses the Optional Stopping Theorem (OST) on a martingale which doesn't appear to have an upper bound, violating the OST condition that the martingale must be bounded.

The martingale is $M(k) = s_k(\frac{n}{n-1})^k$, where $n$ is the total number of nodes in the network, $s_k$ is the number of nodes susceptible to hearing a rumour, and $k$ is the timestep.

The stopping time (T) being used is the time at which no infective nodes are left in the rumour spreading process: $i_T = 0$.

It's clear to see $M(k)$ has a lower bound of $s_0$, but as $M(k)$ is raised to the power of $k$, it has no upper bound?

Nonetheless, the OST is still applied but I don't understand why. Any ideas would be greatly appreciated. Thank you!