Suppose a Gaussian process $\{B_t\}$, apparently $2^{B_t}$ is not a martingale. Can someone teach me how to change the measure so that $2^{B_t}$ is a martingale?
Thanks.
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Is $(B_t)_t$ supposed to be a Brownian motion or indeed an arbritrary Gaussian process? – saz Aug 18 '13 at 10:42
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Sorry. Should be the former, i.e $N(0,\sigma^2)$. – user90846 Aug 18 '13 at 11:01
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If $B=(B_t)_{t\geqslant0}$ is a standard Brownian motion, then each $M^{(a)}_t=\mathrm e^{aB_t-a^2t/2}$ defines a martingale. For every $c$, $\mathrm e^{cB_t}\cdot M^{(a)}_t=\mathrm e^{(c+a)B_t-a^2t/2}$ also defines a martingale if and only if $a^2=(c+a)^2$, that is, $a=-c/2$. Thus, $$ \left.\frac{\mathrm dQ}{\mathrm dP}\right|_{\mathcal F_t}=M^{(-c/2)}_t=\mathrm e^{-cB_t/2-c^2t/8} $$ defines a probability measure $Q$ such that $X_t=\mathrm e^{cB_t}$ defines a $Q$-martingale $X$. Use this for $c=\ln2$.
Just (a special case of) Girsanov theorem.
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