Quadratic variation of $dX_t$ = $\mu \, dt + \sigma \, dB_t$

Question

I am given a Brownian motion $B$ on a probability space$(\Omega,\mathcal{F},\mathcal{Q})$ and a stochastic process X with $dX_t = \mu \space dt + \sigma \space dB_t$ with $\mu \in \mathcal{R} $ and $ \sigma > 0$. I should show what the quadratic variation of X(= $\langle X,X \rangle$) is and that this solution holds for every probability measure $\mathcal{P}$ that is absolutely continuous with respect to $\mathcal{Q}$.

My idea is to use the covariation definition, use the difference between $X_{t_{i+1}}$ and $X_{t_i}$ (which is $\mu \space (t_{i+1} -t_i) + \sigma ( B_{t_{i+1}}-B_{t_i}))?)$ and then make use of a theorem that states that the quadratic variation of a brownian motion is t ($ \langle B \rangle_t = t$ $\forall t \in [0,T]$ $\mathcal{Q}-a.s.$) :

\begin{align*} \langle X,X \rangle_t &= \frac{1}{4}(\langle X+X \rangle_t - \langle X-X \rangle_t) \tag{1} \\ &= \frac{1}{4}(\langle 2X \rangle_t) \tag{2} \\ &= \frac{1}{4} \cdot 4 (\langle X \rangle_t) \tag{3} \\ &= \lim_{n \to \infty} \sum_{t_i \in \Pi^n} (X_{t_{i+1} \land t} - X_{t_i \land t})^2 \tag{5} \\ &= \lim \sum (\mu \space t_{i+1 \land t} + \sigma \space B_{t_{i+1} \land t} + X_0 -(\mu \space t_{i \land t} + \sigma \space B_{t_{i} \land t} + X_0))^2 \tag{6} \\ &= \lim \sum(\sigma(B_{t_{i+1} \land t}-B_{t_{i} \land t})^2 \tag{7} \\ &= \sigma^2 \lim \sum (B_{t_{i+1} \land t}-B_{t_{i} \land t})^2 = \sigma ^2 \langle B\rangle_t \tag{8}\\ &= \sigma^2 t \tag{10} \end{align*}

My question is, is this the right approach, where are faults, what argument can I use in (7) that allows me to set the difference in the t's to $0$ and where do I (have to) use the argument that this holds for every absolutely continuous probability measure $P$ (w.r.t $Q$)?

Answer 1

Yes, your approach is okay. Not sure why you need (1)-(3) because $\langle X,X \rangle_t = \langle X \rangle_t$ is really just the definition.

How to get from (6) to (7)? If you expand the squre in (6), then you get 3 terms; one of them is the term which shows up in (7) and we need to discuss away the other two terms.

Term 1:$$\mu^2 \sum (t_{i+1} \wedge t - t_i \wedge t)^2 $$

Clearly, $$\mu^2 \sum (t_{i+1} \wedge t - t_i \wedge t)^2 \leq \mu^2 |t_{i+1}-t_i| \underbrace{\sum (t_{i +1}-t_i)}_{t}.$$

If the mesh size $|\Pi^n|$ of your partition $\Pi^n$ tends to zero, then the right-hand side converges to zero.

Term 2: $$\mu \sigma \sum (t_{i+1} \wedge t-t_i \wedge t) (B_{t_{i+1} \wedge t}-B_{t_i \wedge t})$$

We have $$\sum (t_{i+1} \wedge t-t_i \wedge t) (B_{t_{i+1} \wedge t}-B_{t_i \wedge t}) \leq \sup_{|u-v| \leq |\Pi^n|, u,v \in [0,t]} |B_u-B_v| \underbrace{\sum (t_{i+1}-t_i)}_{=t}.$$ Since the sample paths of Brownian motion are continuous (with probability 1), they are uniformly continuous on compact intervals, and so the right-hand side converges to $0$ almost surely as $|\Pi^n| \to 0$.

In summary, this gives

$$\mathbb{Q} \left( \lim_{n \to \infty} \sum_{t_i \in \Pi^n} (X_{t_{i+1} \wedge t}-X_{t_i \wedge t})^2 = \sigma^2 t\right)=1,$$

i.e. the event

$$N=\left\{ \omega \in \Omega; \lim_{n \to \infty} \sum_{t_i \in \Pi^n} (X_{t_{i+1} \wedge t}-X_{t_i \wedge t})^2 \neq \sigma^2 t\right\}$$

has measure zero with respect to $\mathbb{Q}$. If $\mathbb{P}$ is another probability measure and $\mathbb{P}$ is absolutley continuous w.rt.t $\mathbb{Q}$, then $\mathbb{Q}(N)=0$ implies $\mathbb{P}(N)=0$, and so

$$\mathbb{P} \left( \lim_{n \to \infty} \sum_{t_i \in \Pi^n} (X_{t_{i+1} \wedge t}-X_{t_i \wedge t})^2 = \sigma^2 t\right)=1,$$

i.e. $\langle X \rangle_t = \sigma^2 t$ for any such measure.