I am really stuck in this question:

Let $\{S_t\}$ and $\{S'_t\}$ be two stochastic processes, satisfying \begin{equation} dS_t = S_t ( \sigma_t \,dB_t + r_t \,dt), \quad dS'_t = S'_t (\sigma'_t \,dB_t + r_t \,dt), \end{equation} where $\sigma$, $\sigma'$ and $r$ are adapted processes, and $B$ is a common standard Brownian motion.

$1.$ Assuming that the processes $\sigma$ and $\sigma'$ are both constant, prove that for each $t>0$, $S_t$ and $S'_t$ are perfectly correlated ( correlation $=1$) if and only if $\sigma = \sigma'$.

$2. $ Assuming that $r=0$ and that $\sigma$ and $\sigma'$ are bounded, prove that if $S_0 = S'_0>0$, and $S_1$ and $S'_1$ are perfectly correlated, then $\int_{0}^{1} (\sigma_t - \sigma'_t )^2 \,dt =0$ a.s..

$3.$ Without imposing any condition on $\sigma$ and $\sigma'$ and suppose that they are different, i.e. $\mathbb{P} ( \int_0^t ( \sigma_s - \sigma'_s )^2 \,ds =0) <1$. Can $S_t$ and $S'_t$ be perfectly correlated for each $t>0$ ?

(I only know that for any two random variables $X$ and $Y$, they are perfectly correlated if and only if $X= aY$, for some $a>0$. However, I don't know how to use this fact.)

  1. Suppose that $S_t$ and $S_t'$ are perfectly correlated for each $t>0$ with correlation $1$. Then it follows from the very definition that $S_t = S_t'$ for all $t>0$. Applying Itô's formula (with $f(x)=x^2$ yields $$\begin{align*} \underbrace{(S_T-S_T')^2}_{0} &= 2 \int_0^T \underbrace{(S_t-S_t')}_0 d(S_t-S_t') + \int_0^T \underbrace{\langle S-S' \rangle_t}_{(S_t \sigma-S_t' \sigma')^2 \, dt}. \end{align*}$$ Using again $S_t = S_t'$, we get $$\int_0^T S_t^2 (\sigma-\sigma')^2 \, dt = 0.$$ Hence, $\sigma=\sigma'$. On the other hand, if $\sigma=\sigma'$ it follows from the uniqueness of solutions of SDEs that $(S_t)_t=(S_t')_t$. In particular, $S_t = S_t'$.
  2. Since $r=0$ and $\sigma_t,\sigma_t'$ are bounded, both processes $(S_t)_t$ and $(S_t')_t$ are martingales. The assumption $S_0 = S_0'>0$ therefore yields $\mathbb{E}(S_1) = \mathbb{E}(S_1')>0$. Consequently, since $S_1$ and $S_1'$ are perfectly correlated, this yields $S_1 = S_1'$. We apply again Itô's formula: $$\begin{align*} (S_T-S_T')^2 &= 2 \int_0^T (S_t-S_t') \, d(S-S')_t + \int_0^T \langle S-S' \rangle_t \\ &= 2 \int_0^t (S_t-S_t') (S_t \sigma_t - S_t' \sigma_t') \, dB_t + \int_0^t (S_t \sigma_t-S_t' \sigma_t')^2 \, dt. \ \end{align*}$$ Taking the expecation on both sides gives for $T=1$ $$\mathbb{E} \left( \int_0^1 (S_t \sigma_t - S_t' \sigma_t')^2 \, dt \right)=0. \tag{1}$$ This, in turn, implies that $(S_t)_{t \geq 0}$ is a solution to the SDEs $$S_t = S_t \sigma_t \, dB_t = S_t' \sigma_t' \, dB_t, \qquad 0 \leq t \leq 1.$$ Again the uniqueness of solutions yields $(S_t)_{t \leq 1} = (S_t')_{t \leq 1}$. Plugging this into $(1)$ finally proves the claim.
  3. No, in general we cannot expect this; have a look at 2.
  • $\begingroup$ Thanks for your detailed solution. I don't understand how $(1)$ implies that $(S_t)$ is a solution to the two SDEs. $\endgroup$ – ashburn Dec 11 '14 at 20:39
  • $\begingroup$ Also, you used the fact that $\mathbb{E} \int_0^T (S_t - S^{'}_t ) ( S_t \sigma_t - S^{'}_t \sigma^{'}_t ) \,dB_t = 0.$ I thought we need to check that $\mathbb{E} \big\{ \int_0^T (S_t - S^{'}_t ) ( S_t \sigma_t - S^{'}_t \sigma^{'}_t) \big\}^2 \,dt $ is finite in order to make this conclusion. $\endgroup$ – ashburn Dec 11 '14 at 22:30
  • $\begingroup$ @ashburn $(1)$ shows $S_t \sigma_t = S_t' \sigma_t'$ (with respect to the given norm). Therefore, the process is a solution to both SDEs. Concerning integrability: Yes; use the fact that $\sigma$, $\sigma'$ are bounded in order to check this. $\endgroup$ – saz Dec 12 '14 at 5:22
  • $\begingroup$ Does the process $(\sigma_t)$ contain continuous paths? If not, then we can only assert that $S_t \sigma_t = S^{'}_t \sigma^{'}_t$ a.s., for t-a.e. on $[0,1]$. But then, it should still be true (if we use the fact that two processes that differ only at times of measure zero have the same stochastic integral), since $ S_t = \int_{0}^{t} S_u \sigma_u \,dB_u = \int_{0}^{t} S^{'}_u \sigma^{'}_u \,dB_u = S^{'}_t$. Is my deduction correct? $\endgroup$ – ashburn Dec 12 '14 at 13:12
  • $\begingroup$ @ashburn No. Note that $\mathbb{E} \left( \int_0^t |f(s)-f'(s)|^2 \, ds \right)=0$ implies $\mathbb{E}\left( \left| \int_0^t f(s) \, dW_s - \int_0^t f'(s) \, dW_s \right|^2 \right)=0$ (by Itô's isometry). Now apply this for $t=1$, $f(s) = \sigma_s S_s$ and $f'(s) = \sigma'_s S_s'$. $\endgroup$ – saz Dec 12 '14 at 15:57

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