New answers tagged martingale
0
$E_0[Y_{\lambda,t}] = 1\,\, \forall t$, hence $Y_t$ is a martingale.
Hint: Look at the arithmetic moments section of this wiki page on lognormal distribution
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We define the process $Y_t=Y(t,S_t)$ as follows:
$$Y_t=\left(\frac{S_t}{S_0}\right)^\lambda \exp\left\{-\left(r\lambda-\lambda(1-\lambda)\frac{\sigma^2}{2}\right)t\right\}$$
Let:
$$\alpha=\lambda\left(r-(1-\lambda)\frac{\sigma^2}{2}\right)$$
Then by Itô's Lemma:
$$\text{d}Y_t=-\alpha Y_t\text{d}t+\frac{\lambda}{S_t}Y_t\text{d}S_t+\frac{1}{2}\frac{\lambda(\...
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