I was looking at the various examples provided in the discussion Worked examples of applying Ito's lemma
One such example is 9.1 (c). This states that -
if $S_t =\! S_0 + \int\limits_{0}^{t} \mu_u S_u du + \int\limits_{0}^{t} \sigma_u S_u dW_u$ with $\mu=\left(\mu_t \right)_{t\geq0}, \sigma=\left(\sigma_t \right)_{t\geq0}, \int\limits_{0}^{T} |\mu_s| + |\sigma_s|^2 ds < \infty$. Then $\int\limits_{0}^{T} \sigma^2_s ds = -2\log \frac{S_T}{S_0} + \int\limits_{0}^{T} \frac{2}{S_u} dS_u$
Then it says $\frac{S_T}{S_0} = e^{\int\limits_{0}^{T} \sigma_s dW_s - \int\limits_{0}^{T} \left(0.5\sigma_s^2 - \mu_s \right) ds}$, which I understand the derivation.
I then failed to grasp the remaining part which shows that : $\log S_T - \log S_0 = \int\limits_{0}^{T} \frac{1}{S_u} dS_u -0.5 \int\limits_{0}^{T} \sigma_u^2 du$
2nd example goes for 4. This states that -
if $X_t =\! e^{W_t+0.5t} + e^{W_t-0.5t}$, then $dX_t =\! X_t dW_t + e^{W_t+0.5t}dt$.
To prove this, it is taken that $X_t=Z_tY_t, Z_t = e^{W_t-0.5t}, Y_t = e^t + 1$. It sates that the process $Z_t$ is continuous semi-martingale and $Y_t$ is continuous semi-martingale of bounded variation. Therefore it holds that $\left[ ZY \right]=0$. My questions are
- Why $Z$ is continuous semi-martingale and $Y$ is continuous semi-martingale with bounded variation? What is required to prove them so?
- How to show exactly that $\left[ZY\right] = 0$
Your pointer will be highly helpful