2
$\begingroup$

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

  1. Why $Z$ is continuous semi-martingale and $Y$ is continuous semi-martingale with bounded variation? What is required to prove them so?
  2. How to show exactly that $\left[ZY\right] = 0$

Your pointer will be highly helpful

$\endgroup$
4
$\begingroup$

For the first one, we have:

$$ dS_t = \mu_t S_t dt + \sigma_tS_t dW_t $$ and note that

$$ (dS_t)^2 = \sigma_t ^2 S_t^{2} dt. $$

We apply Ito formula to

$$ f(S_t) = \ln S_t. $$

As $f'(x) = x^{-1}$ and $f^{''}(x)= -x^{-2}$, we get:

$$ d \ln S_t = S_t^{-1} dS_t - 0.5 S_t^{-2} (dS_t)^2 $$

which is equivalent to

$$ d \ln S_t = S_t^{-1} dS_t - 0.5 \sigma_t^2 dt. $$

Integration from $0$ to $T$ gives:

$$ \ln S_T - \ln S_0 = \int_0^T S_t^{-1} dS_t - 0.5 \int_0^T \sigma_t^2 dt. $$

For the second one:

A process has bounded variation if almost all its paths are functions with bounded variation.

This needs to be proven for $Y_t$.

For the rest of the questions, you can take as facts (proven in many books on stochastic calculus) that:

  1. Ito processes are continuous semimartingales (with the two pieces, continuous local martingale and continuous finite variation process, visible).

  2. Finite variation processes (bounded variation on every finite time interval, with probability $1$) have $0$ quadratic variation.

  3. The quadratic covariation between a finite variation process and a continuous semimartingale is $0$.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.