Dgdrxrt

We have the following setting

We have a 2-dimensional Brownian motion $(X,Y)$, and we define the process $M_t$ as $$M_t=e^X_tcos(Y_t)$$

The problem is to show that the process $M_t$ is a local martingale, and that we have that $$langle Mrangle_t=int^t_0e^X_sds$$

I know that $M_t$ would be a local martingale if it is a continuous, adapted process and if there exists a sequence of stopping times $(tau_n)_ngeq0$, such that $tau_nuparrowinfty$ almost surely, and that $(M_tau_nwedge t-M_0)_tgeq0$ is a uniformly integrable martingale for all $n$.

However, I am not aware of any straight forward way to proving that something is a local martingale, and I do not see an easy approach to proving it directly from the definition. Are there ''standard'' routes to follow when trying to prove that a process is/isn't a local martingale?

Moreover, when trying to show that the quadratic variation of $M$ can be written in such a way, I cannot figure out why the cosine term would drop out.

Any help is appreciated!

If we apply Ito's lemma to the function $f(x,y) = e^x cos(y)$ you'll find that
$$dM_t = M_t dX_t - e^X_t sin(Y_t) dY_t$$
since $langle X rangle_t = langle Y rangle_t = t$ and $langle X,Y rangle_t = 0$. Since the stochastic integral against a continuous local martingale is again a local martingale, this shows that $M_t$ is a continuous local martingale. We also deduce from this that
$$d langle M rangle_t = M_t^2 d langle X rangle_t + e^2X_t sin(Y_t)^2 dlangle X rangle_t = e^2X_t (cos^2(Y_t) + sin^2(Y_t))dt = e^2X_t dt$$
(where we have again used that $langle X,Y rangle_t = 0$) and so
$$langle M rangle_t = int_0^t e^2X_s ds$$