On the probability space ($\Omega, \mathcal{F}, (\mathcal{F}_t), \mathbb{P})$, let $(N_t)$ be a Poisson process with intensity $\lambda$ and $(W_t)$ be a Brownian motion (the two processes being independent from each other.
Applying Itô’s lemma, show that the process $Y_t=e^{\alpha N_t}W_t$, $t\ge0$ is a mixed process.
A mixed process $(X_t)$ is defined as having a representation of the form: $X_t=X_0+\int^t_oh_sds+\int^t_0f_sdW_s+\int^t_0g_sdM_s$, where $M_0$ is a constant, $W$ Brownian motion and $M$ is the compensated martingale associated with a Poisson process $N$ with constant intensity $\lambda$ (i.e. $M_t=N_t-\lambda t)$. $h,f,g$ are adapted processes.
To be honest I don't know how to choose a function directly for $Y$ in order to apply Itô formula, since $Y_t$ should admit the representation $Y_t=Y_0+\int^t_0K_sds+\int^t_0H_sdW_s$, for $K_s,H_s$ adapted stochastic processes (I think it doesn't).
My idea was to divide the process in continuous and pure jump (discontinuous) process as follows: $W_t$ is the continuous process and $V_t:={\alpha N_t}$ is the pure jump process.
In my notes I have the following variant of Itô formula for a pure jump process $V_t$: $f(V_t)-f(V_0)=\sum_{s\le t}\Delta f(V_s)$, where $\Delta f(V_s)=f(V_s)-f(V_{s-}), V_{s-}=\lim_{h\downarrow0}V(t-h)$. Then, following the resolution as in this paper (page 48):
For $f(x)=e^x$, $f(V_t)-f(V_0)=e^{\alpha N_t}-1=\sum^t_{s=0}e^{V_s}-e^{V_{s-}}$. As $\Delta N_s=N_s-N_{s-}=1$, we have $e^{V_s}-e^{V_{s-}}=e^{\alpha N_{s-}}(e^{\alpha}-1)=e^{\alpha N_{s-}}(e^{\alpha}-1)\Delta N_s$
So $e^{\alpha N_{t}}-1=\sum_{s\le t}e^{\alpha N_{s-}}(e^{\alpha}-1)\Delta N_s=(e^{\alpha}-1)\int^t_0e^{\alpha N_{s-}}dN_s$
Regarding the Brownian motion, I have this alternative Itô formula for $f\in C^2$: $f(W_t)-f(W_0)=\int^t_0f'(W_s)dW_s+\frac12\int^t_0f''(W_s)ds$
So take $f\equiv Id$, then I just get $W_t=W_t$
In the end I would just multiply the result I got for both processes, but doesn't sound that right to me.
Since $e^{aN_t}$ and $W_t$ are both semi martingales we have \begin{align} e^{aN_t}W_t&=\int_0^te^{aN_{s-}}\,dW_s+\int_0^tW_s\,d(e^{aN_s})+\big[e^{aN},W\big]_t\,. \end{align} Since $e^{aN_t}$ changes only by jumps of size $$ e^{aN_t}-e^{aN_{t-}}=e^{aN_{t-}+a\Delta N_t}-e^{aN_{t-}}=e^{aN_{t-}}(e^{a\Delta N_t}-1)=e^{aN_{t-}}(e^a-1)\Delta N_t $$ the second integral is $$ \int_0^tW_s\,d(e^{aN_s})=\sum_{s\le t}W_s\,e^{aN_{s-}}(e^a-1)\,\Delta N_s=\int_0^tW_s\,e^{aN_{s-}}(e^a-1)\,dN_s\,. $$ The quadratic variation term is zero because $W$ is continuous and $e^{aN}$ changes only by jumps.
Writing $M_t=N_t-\lambda t$ obviously brings \begin{align} e^{aN_t}W_t&=\int_0^te^{aN_{s-}}\,dW_s+\int_0^tW_s\,e^{aN_{s-}}(e^a-1)\,dN_s\,. \end{align} into the desired form \begin{align} e^{aN_t}W_t&=\int_0^te^{aN_{s-}}\,dW_s+\int_0^tW_s\,e^{aN_{s-}}(e^a-1)\,dM_s+\lambda\int_0^tW_s\,e^{aN_{s-}}(e^a-1)\,ds\,. \end{align}