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2

For a time interval $[0,T]$, Girsanov theorem states that given a process $\lambda$ such that process $U$, defined by $$dU_t = -\lambda_tU_tdW_t, \; U_0=1,$$ is a $P$-martingale, then one can define a new measure $Q$ equivalent to $P$ by $$\frac{dQ}{dP} = U_T,$$ and a standard Brownian motion under $Q$, $W^\star$, by $$dW^\star_t = dW_t + \lambda_tdt.$$ In ...

1

@Olaf gave a clear answer. Another way to see this is as system of 2 SDEs: \begin{cases} dS_u &= \mu(S_u,u) du + \sigma(S_u,u) dW_u \\ dy_u &= S_u du + S_u dW_u. \end{cases} E.g. if we want to simulate this system using Euler discretuzation then we perform \begin{cases} S_u + \Delta S_u &= S_u + \mu(S_u,u) \Delta t + \sigma(S_u,u) \epsilon ...

7

A stochastic differential equation is nothing more than a short-hand notation for a corresponding integral equation. So the initial SDE you provided actually means $$\int_0^t d S_u = \int_0^t \mu(S_u, u) du + \int_0^t\sigma(S_u, u) dW_u$$ This is how the SDE is defined (see e.g. here). The reason is that you cannot differentiate a Brownian motion. It does ...

1

The first part has already been answer by @Uditg_ucla, so I am only providing answer of your 2nd part. Rewriting your SDE in more sophisticated way: $$dS=k(b-S)dt+\sigma S dz$$ You want SDE for $S^2$. Using Taylor series, it can be written as: $$df(S)=f'(S)dS + \frac{1}{2!}f''(S)(dS)^2+\cdots$$ $$df(S)=2SdS+(dS)^2$$ df(S)=2S[k(b-S)dt+\sigma S ...

0

the answer is simple: look at key differences between these two models. GBM is diffusion, OU is mean-reversion

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Given efficient markets, asset prices should be unpredictable in the sense that any upcoming returns are uncorrelated with current or past returns. Hence for traded assets the price should follow something more similar to a GBM than an O-U process. However, many financial metrics are not prices; for example interest rates or volatility. O-U processes may ...

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The two processes are not pathwise equal. Here is a simulation (sample path) of the two processes $(t-\tau)dW_{\tau}$ and $W_\tau d\tau$: Note that both processes have the same value at the final time.

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If $X_t$ is square integrable, then the integral \begin{align*} \int_0^t X_{\tau} dW_{\tau} \end{align*} is a martingale. Here, the integrand $X_{\tau}$ does not depend on the integral limit $t$. However, in your case, the integrand, $t-\tau$, depends on $t$, then the condition for the martingality of the integral fails.

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