poisson process are independents}\\
    &=& \lambda (t-s)-\lambda t+ N_s\hspace{0.2cm}\text( since\hspace{0.1cm} N_t-N_s\rightsquigarrow\mathcal{P}(\lambda(t-s)))\\
    &=& N_s-\lambda s\\
    &=& \overline{N}(s).\end{aligned}$$

2.  $$\begin{aligned}
    \mathbb{E}(\overline{N}(t+s)-\overline{N}(t))&=& \mathbb{E}(N(t+s)-N(t)-\lambda s)\\
    &=& \lambda (t+s-t)-\lambda s= 0.\end{aligned}$$

3.  $$\begin{aligned}
    [\overline{N}(t+s)-\overline{N}(t)]^2&=&[N(t+s)-N(t)-\lambda s]^2\\
    &=&[N(t+s)-N(t)]^2+\lambda^2s^2-2\lambda s(E(t+s)-N(t)).\end{aligned}$$

    Since $N(t)\rightsquigarrow \mathcal{P}(\lambda t)$, using the relation $\mathbb{E}(N_t)=var(N_t)= \lambda t$, it follows that : $$\begin{aligned}
    \mathbb{E}[\overline{N}(t+s)-\overline{N}(s)]^2&=&\lambda(t+s-t)+\lambda^2(t+s-t)^2+\lambda^2s^2-2\lambda s(\lambda s)=\lambda s. \end{aligned}$$

4.  $$\begin{aligned}
     \mathbb{E}[\overline{N}^2_t-\lambda t/\mathcal{F}_s]&=& \mathbb{E}[\overline{N}^2_t/\mathcal{F}_s]-\lambda t\\
    &=&\mathbb{E}[\overline{N}^2_t+\overline{N}^2_s-2\overline{N}_t\overline{N}_s+2\overline{N}_t\overline{N}_s-
    \overline{N}_s^2/\mathcal{F}_s]-\lambda t\\
    &=&\mathbb{E}[(\overline{N}_t-\overline{N}_s)^2/\mathcal{F}_s]+\mathbb{E}[\overline{N}_s(\overline{N}_t-\overline{N}_s)/\mathcal{F}_s]+\mathbb{E}[\overline{N}_t\overline{N}_s/\mathcal{F}_s].\end{aligned}$$

    Using the fact that $\overline{N}_t$ have independent increments and using the first part of the theorem, it follows that $$\begin{aligned}
     \mathbb{E}[\overline{N}^2_t-\lambda t/\mathcal{F}_s]&=&\mathbb{E}[(\overline{N}_t-\overline{N}_s)^2]+\overline{N}_s\mathbb{E}(\overline{N}_t-\overline{N}_s)+\mathbb{E}[\overline{N}_t-\overline{N}_s+\overline{N}_s/\mathcal{F}_s]-\lambda t\\
     &=&\lambda (t-s)+0+\overline{N}_s\mathbb{E}[\overline{N}_t-\overline{N}_s+\overline{N}_s/\mathcal{F}_s]-\lambda t\\
     &=&\lambda t-\lambda s+0+0+\overline{N}^2_s-\lambda t\\
     &=&\overline{N}^2_s-\lambda s\end{aligned}$$ This complete the proof.

Stochastic integral
-------------------

Let $\mathbb{M}^p([0,T], \mathbb{R})$ be the subspace of $\mathbb{L}^p([0,T], \mathbb{R})$ such that for any process

$(X_t)\in \mathbb{M}^p([0,T], \mathbb{R})$ we have $$\begin{aligned}
\mathbb{E}\left(\int_0^T|X(t)|^pdt\right)<\infty.\end{aligned}$$

Consider a Brownian motion $W$ and a stochastic process $(X_t)$ both adapted to a given filtration $(\mathcal{F}_t)$. We will define the following expression called stochastic integral $$\begin{aligned}
I_t(X)=\int_0^tX(s)dW(s).\end{aligned}$$ We will also give some of its properties.

Let’s start with the stochastic integral of simple process.

**\[Elementary process or simple process\]**

A process $(X_t)_{t\in\mathbb{R}}\in\mathbb{L}^p([0,T], \mathbb{R})$ is called simple or elementary process if there exist a partition $0=t_0<t_1<...<t_n=T$ such that $$\begin{aligned}
X_s(\omega)=\sum_{j=0}^{n}1_{]t_j, t_{j+1}]}\theta_j(\omega),\end{aligned}$$ where $\theta_j$ is a bounded $\mathcal{F}_{t_j}$-measurable random 