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I want to show that the stochastic process $$ S_t^i = S_0^i \exp\left( \int_0^t \left(\mu_s^i - \frac{1}{2} \sum_{j=1}^m (\sigma_s)^{ij} \right)^2 d s + \sum_{j=1}^m \sigma_t^{ij} S_t^i dW_t^j \right) $$ solves the stochastic differential equation $$ dS_t^i = \mu_t^i S_t^i dt + \sum_{j=1}^m \sigma_t^{ij} S_t^i dW_t^j. $$

I just want to plug in this expression and thereby verifying that it solves the SDE, just as in the non-stochastic case. For this I first recognize that the SDE is a shorthand for $$ S_t^i = S_0^i+ \int_0^t \mu_s^i S_s^i ds + \int_0^t \sum_{j=1}^m \sigma_s^{ij} S_s^i ~ dW_s^j. $$ Now first evaluating the integral $$ \int_0^t \sum_{j=1}^m \sigma_s^{ij} S_s^i ~ dW_s^j $$ Here I have no idea how to start, I know I must apply Ito's formula in some clever way, but when I look it up it is always written with some mysterios $d X_i \cdot d X_j$ for the second order terms, and remarks like "these are evaluated according to the rules $dB_i dB_j = \delta_{ij} dt, dt dB_i = dB_i dt = 0$" which I cannot make sense of, so I have no idea how to apply Ito's formula here to solve the stochastic integral? Any help?

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  • $\begingroup$ There is a typo in the definition of $S_t^i$: It should read $\sum_{j=1}^m \sigma_t^{ij} dW_t^j$ instead of $\sum_{j=1}^m \sigma_{t}^{ij} S_t^i dW_t^j$, right? $\endgroup$ – saz Jul 2 '15 at 18:11
  • $\begingroup$ You are right, it was a typo! $\endgroup$ – StefanH Jul 3 '15 at 9:53
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To keep notation simple, we write $S_t$ instead of $S_t^i$, $\sigma_t^j$ instead of $\sigma_{t}^{ij}$ and so on. That's okay, because $i$ is a fixed number throughout this calculation.

Suppose $(X_t)_{t \geq 0}$ is an Itô process of the form

$$dX_t = b(t) \, dt + \eta(t) \, dW_t$$

where $\eta = (\eta_1,\ldots,\eta_m)$ and $(W_t)_{t \geq 0}$ is an $m$-dimensional Brownian motion. Then Itô's formula states that

$$\begin{align*} f(X_t)-f(X_0) &= \int_0^t f'(X_s) \, dX_s + \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s \end{align*}$$

for $f \in C^2$ where

$$\begin{align*} \int_0^t f'(X_s) \, dX_s &:= \int_0^t f'(X_s) \eta_s \, dW_s + \int_0^t f'(X_s) b(s) \, ds \\ d\langle X \rangle_s &:= \sum_{j=1}^m \eta_j^2(s) \, ds. \end{align*}$$

In your setting, we have $$b(t) = \left(\mu_t - \frac{1}{2} \sum_{j=1}^m \sigma_t^j \right)^2 \qquad \eta(t) := (\sigma_t^1,\ldots,\sigma_t^m).$$

Applying Itô's formula for $f(x) := \exp(x)$ shows that $(S_t)_{t \geq 0}$ is a solution to the given SDE.

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  • $\begingroup$ I am unsure about the $S_0$, strictly we have $S_t = S_0 f(X_t)$, not $S_t = f(X_t)$, so when I apply the Differential could I just write $dS_t = S_0 df(X_t)$ and then apply Ito's Formula to $df(X_t)$ (then of course everything would work out), or should it be $dS_t = dS_0 f(X_t) + S_0 df(X_t)$ in analogy to the product rule from ordinary calculus (but what then would $dS_0$ be...)? $\endgroup$ – StefanH Jul 3 '15 at 9:52
  • $\begingroup$ @Stefan Just treat it as a constant, i.e. use Itô's formula for $c f(X_t)$ and then replace $c$ by $S_0$. $\endgroup$ – saz Jul 3 '15 at 10:36
  • $\begingroup$ But $S_0$ is not constant, it depends on $\Omega$, i.e. $S_0 : \Omega \to \mathbb R$ is a random variable... $\endgroup$ – StefanH Jul 3 '15 at 14:21
  • $\begingroup$ @Stefan I know; that's why I wrote "treat it as a constant". $\endgroup$ – saz Jul 3 '15 at 16:24
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    $\begingroup$ @Stefan The reason why it works: First apply Itô's formula with $S_0=1$. Then we get that $$\exp \left( \int_0^t b(s) \, ds + \int_0^t \eta(s) \, dW_s \right) = \int_0^t \mu_s \tilde{S}_s \, ds + \sum_j \int_0^t \sigma_s^j \tilde{S}_s \, dW_s^j.$$ (Here $\tilde{S}_t$ denotes the process defined as $S_t$ but with $\tilde{S}_0=1$). Now multiply both sides with $S_0$ and you are done. (You have to assume that $S_0$ is measurable wrt $\mathcal{F}_0$.) $\endgroup$ – saz Jul 3 '15 at 16:43

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