 Solution

The deterministic clock is $a(t)=e^{2\lambda t}$, so $a(0)=1$. For $s\le t$, the martingale property of Brownian motion givesIntegrability follows from the Gaussian second moment. Thus $X$ is a martingale in the time-changed filtration.

Its increments have variance $a(t)-a(s)$ and are independent of ${\mathcal{G}}_s$. Equivalently, $X_t^2-a(t)$ is a martingale. Since a quadratic variation starts at zero, its compensator must be normalized at $t=0$:The initial random variable $X_0=B_1$ does not contribute to quadratic variation. The expression $e^{2\lambda t}$ alone would violate $[X{]}_0=0$.