I have been reading a paper on the stochastic epidemic model. Paper deals with the following model.
$dS(t) = (\Lambda– μS(t) – βS(t)I(t)/(1+aI(t)) + r_1I(t) + δR(t)) dt – σS(t)I(t)/(1+aI(t)) dB(t)$,
$dI(t) = (βS(t)I(t)/(1+aI(t)) – (μ + r1 + r2 + α)I(t)) dt + σS(t)I(t)/(1+aI(t)) dB(t),$
$dR(t) = (r_2I(t) – (μ + δ)R(t)) dt.$
I have encountered a place where it is assumed $M_1(t)$ = $\int_{0}^{t}$ σS(τ)/(1+aI(τ)) dB(τ) and $M_1(t)$ is the local continuous martingale with $M_1(0)=0$. I completely stuck how to prove $M_1 (t)$ is a martingale. Please give any hint. Thanks a lot. This article can be found here: https://advancesindifferenceequations.springeropen.com/articles/10.1186/s13662-018-1759-8
user6247850 is right in stating that, in general, this is a continuous local martingale by the construction of the stochastic integral. If you want to prove that this is a true martingale, then it suffices that the following criterion is satisfied: $\mathbb{E} \left[ \langle M_{1} \rangle_t \right] < \infty$ for all $t>0$. The proof of this result is short but non-trivial: it involves reducing the continuous local martingale to obtain $L^2$ bounded martingales, then using Vitali's convergence theorem and strong $L^1$ convergence.
In this case, by properties of quadratic variation and Fubini's theorem:
\begin{equation} \mathbb{E} \left[ \langle M_{1} \rangle_t \right] = \int_{0}^{t} \mathbb{E} \left[ \dfrac{\sigma^2 S_{u}^{2}}{(1 + \alpha I_u)^2} \right] du. \end{equation}
Assuming you are working with a closed population (so that $S+I+R = const.$) and that $\alpha > 0$, the expectation inside the integral is clearly bounded above, so the integral always converges, the criterion holds and this is a true martingale.