Let $B_t$ be the Brownian Motion.
Find the quadratic variation of a martingale $ M_t = B_t^2-t$.
My solution:
By Ito's formula for $f(t, x) = x^2-t$, we know $$d(B_t^2-t) = 2B_t dB_t$$ thus $\langle M \rangle_t = 4 \int_0^t B_s^2 ds$.
My question is:
- Is this correct?
- Can I simplify the last integration?
- Any other solutions? (For instance, can I calculate this directly by $$\sum_{i=1}^n (M_{t_i} – M_{t_{i-1}})^2$$ using the fact that this sum converges in probability to the quadratic variation?)
Best Answer
• Yes, it's correct.
• No.
• The quadratic variation of $M$ will be the same as that of $B^2$, as these two processes differ by $t$. Write $$ (B_{t_{i}}^2 - B_{t_{i-1}}^2)^2 = (B_{t_{i}} + B_{t_{i-1}})^2(B_{t_{i}} - B_{t_{i-1}})^2. $$ Now sum over $i$ and let $n\to\infty$. By continuity of the Brownian path, the factor $(B_{t_{i}} + B_{t_{i-1}})^2$ will be (uniformly in time) close to $(2B_{t_{i-1}})^2=4B_{t_{i-1}}^2$.