Stochastic Calculus

1. Using Ito’s formula with the $\displaystyle C^2$-function $\displaystyle f(x)=x^2$ and the process $\displaystyle Z_t:=\int_0^t f(s)W_s$, on obtient : $\displaystyle d(Z_t^2) = 2Z_t dZ_t + (dZ_t)^2 = 2Z_t f(t)dW_t + f(t)^2dt.$ Then $\displaystyle dX_t = 2Z_t f(t)dW_t$.
   Moreover from Itô’s isometry and Fubini,
   $$\begin{aligned} E \int_0^T Z_t^2 f(t)^2 dt &=\int_0^T E [Z_t^2] f(t)^2 dt \\ &\le\int_0^T \int_0^T f(s)^2 f(t)^2 \ dsdt= \left( \int_0^T f(t)^2 dt \right)^2<\infty.\end{aligned}$$ Thus $\displaystyle X_t$ is the stochastic integral of a $\displaystyle ({\mathcal F}_t)$-adapted process in $\displaystyle L^2(\Omega\times[0,T])$, et then $\displaystyle (X_t)$ is a $\displaystyle ({\mathcal F}_t)-$martingale (not only a local-martingale).

2. Using Ito’s formula with the $\displaystyle C^2$-function $\displaystyle f(x)=e^x$ and the process $\displaystyle Z_t:=\int_0^t f(s)dW_s - \frac12 \int_0^t f(s)^2ds$, we obtain:
   $$dY_t = Y_t dZ_t + \frac{1}{2} Y_t (dZ_t)^2 = Y_t f(t)dW_t.$$ Noticing that $\displaystyle \int_0^t f(s)dW_s \sim \mathcal N\left(0,\int_0^t f(s)^2ds\right),$ on get that $\displaystyle e^{Z_t} \in L^2(\Omega)$ and then $\displaystyle t\mapsto E [e^{Z_t}]$ is continuous. Thus $\displaystyle E \int_0^T Y_t^2 dt<\infty$, and we conclude in the same way as $\displaystyle 1)$.

Let $\displaystyle S$ be a positive semi-martingale verifying the SDE. The map $\displaystyle t\mapsto S_t$ is not differentiable a.s. since the Brownian Motion is not. If a function $\displaystyle R$ is differentiable, then $\displaystyle \frac{dR_t}{R_t}=d\log(R_t)$. So we feel like computing $\displaystyle d\log(S_t)$. Hence we apply Itô’s formula with:

1. the function $\displaystyle f:x\mapsto \log(x)$ sur $\displaystyle \R_+^*$ ; we have $\displaystyle f'(x)=1/x$ et $\displaystyle f''(x)=-1/x^2$ ;
2. the process $\displaystyle S_t$ ; we have $\displaystyle dS_t=S_t(\mu dt+\sigma dW_t)$ et $\displaystyle (dS_t)^2=S_t^2\sigma^2 dt$.

This yields $$d\log(S_t)=\frac{dS_t}{S_t}-\frac12\frac{(dS_t)^2}{S_t^2}=\mu dt+\sigma dW_t-\frac{\sigma^2}{2}dt.$$ We integrate between $\displaystyle 0$ et $\displaystyle t$ both sides: $$\log(S_t)-\log(S_0)=\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t,$$ thus $\displaystyle S_t=S_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t}$. We verify that $\displaystyle S_t>0$. Conversely, this last process is a solution of the SDE.

Remark — In the literature, the infinitesimal quadratic variation is often denoted by $\displaystyle d\langle S\rangle_t$, but we must remember that “morally” it is a squarred increment of $\displaystyle S_t$, explaining the notation $\displaystyle (dS_t)^2$ we used, which is quite convenient!