Let us define $$X_t:=e^{-t\lambda}B_{\frac{e^{2\lambda t}-1}{2\lambda}},$$
where $B$ is a Brownian motion and $\lambda>0$
How can I calculate quadratic variation $<X>_t$?
The first thing that comes to mind is to use Ito's formula and then calculate the quad. var. of the stochastic integral, as other parts are cont. differentiable and their quad. var is $0$. But the problem is, that I do not know how to apply it to this situation, as I know it only for functions with non-time changed BM's.
Is there a theorem that applies to this situation, should I consider replacing the changed time with a new variable?
Best Answer
Since $B_t$ is a standard Brownian motion, we always have $$ \left<B\right>_t=t $$ for all forms of $t$, even if $t$ is written as a function of other variables. More precisely, if $$ Y_t=B_{g(t)} $$ for some strictly monotone function $g$ with $g(0)=0$, we have $$ \left<Y\right>_t=g(t). $$ Now, note that this $Y$ is a martingale, i.e., $$ \mathbb{E}\left(Y_t|Y_s\right)=\mathbb{E}\left(B_{g(t)}|B_{g(s)}\right)=B_{g(s)}=Y_s $$ for all $t\ge s\ge 0$. Hence by martingale representation theorem, there must exist some predictable process $\sigma_t$, such that $$ {\rm d}Y_t=\sigma_t\,{\rm d}W_t, $$ where, in case of any ambiguity, $W_t$ here denotes another standard Brownian motion. Note that by this formula, it is straightforward that $$ {\rm d}\left<Y\right>_t=\sigma_t^2\,{\rm d}t. $$ Hence, combined with the above $\left<Y\right>_t=g(t)$, it is obvious that $$ \sigma_t=\sqrt{g'(t)}, $$ for which $$ {\rm d}Y_t=\sqrt{g'(t)}\,{\rm d}W_t. $$
Thanks to this result, we may find Ito's formula for $$ X_t=f(t)\,B_{g(t)}=f(t)\,Y_t $$ with a general function $f$. We have $$ {\rm d}X_t=f'(t)\,Y_t\,{\rm d}t+f(t)\,{\rm d}Y_t=f'(t)\,Y_t\,{\rm d}t+f(t)\sqrt{g'(t)}\,{\rm d}W_t, $$ which indicates that $$ {\rm d}\left<X\right>_t=f^2(t)\,g'(t). $$ Therefore, $$ \left<X\right>_t=\int_0^tf^2(s)\,g'(s)\,{\rm d}s. $$
Hope this explanation could be somewhat helpful for you.