Find the value of the real $\alpha$ for which exp($2B(t) - \alpha t$) is a martingale.

Question

I tried to answer using the three conditions to be a martingale (measurability, integrability, and martingality), validating the integrability condition, which is $$ E |e^{2B(t) - \alpha t} | < \infty $$ I found that $$ \alpha = 2 $$ This result is validated verifying the martingality condition.

Anyway, are there other ways to find the value of $\alpha$?

Answer 1

$$ Y_t = \mathrm{e}^{2B_t-\alpha t} $$ using Ito $$ dY_t = -\alpha Y_t dt + 2Y_tdB_t + \frac{4}{2}Y_t dB_t^2 = -\alpha Y_t dt + 2Y_tdB_t + 2Y_t dt = (2-\alpha)Y_tdt + 2Y_tdB_t $$ thus martingale requires a driftless SDE thus $\alpha = 2$. Was this the way you performed it?

Answer 2

Yeah, thanks. It was quite simple. My solution was to verify the martingality of y(t) following the three conditions:
1 – Measurability Condition. y(t) $\in$ Ft natural filtration since y(t) is a transformation of Bt which is a Ft-martingale for the probabilistic properties of Brownian Motion.

2 – Integrability Condition, $$ E |e^{2B_t - \alpha t}| < \infty \\ e^{-\alpha t} E | e^{2 B_t} | = e^{-\alpha t} E | e^{N(0,4t)} | = e^{-\alpha t} e^{0 + \frac{1}{2} 4t} = e^{-\alpha t + 2t} < \infty $$

which is a.s. finite if and only if $\alpha = 2$.

3 – Martingality Condition using the found $\alpha = 2$, with s ≤ t $$ E[e^{2B_t - 2t}| F_s] = e^{-2t}E[e^{2B_t}|F_s] = e^{-2t}E[e^{2(B_t - B_s + B_s}|F_s] = e^{-2t}E[e^{2(B_t - B_s)2B_s} | F_s] = e^{2B_s -2t}E[e^{2(B_t - B_s)} | F_s] = e^{2B_s -2t}E[e^{2(B_t - B_s)}] = e^{2B_s -2t}E[e^{N(0,4(t-s)}] = e^{2B_s -2t}e^{2(t - s)} = e^{2B_s - 2s} $$

Then also the measurability condition is proved, so y(t) is a martingale for $\alpha = 2$. Maybe that's quite wrong, but with the same result.