Proving a process is martingale under the Risk Neutral Measure

Question

Show that for any $\lambda \in \Re$, the process $Y_{\lambda,t}$ defined as:

$$Y_{\lambda,t} = (S_t/S_0)^\lambda e^{-(r\lambda-\lambda(1-\lambda)\sigma^2/2)t}$$

is a martingale under the risk neutral measure $Q$.

I was thinking that I could apply Ito's Lemma, in which to show that the $\text{d}t$ term will be zero. However, after doing the partial derivatives, the terms do not cancel each other out.

Would really appreciate all the help I can get!

Answer 1

We define the process $Y_t=Y(t,S_t)$ as follows: $$Y_t=\left(\frac{S_t}{S_0}\right)^\lambda \exp\left\{-\left(r\lambda-\lambda(1-\lambda)\frac{\sigma^2}{2}\right)t\right\}$$

Method 1

Let: $$\alpha=\lambda\left(r-(1-\lambda)\frac{\sigma^2}{2}\right)$$ Then by Itô's Lemma: $$\text{d}Y_t=-\alpha Y_t\text{d}t+\frac{\lambda}{S_t}Y_t\text{d}S_t+\frac{1}{2}\frac{\lambda(\lambda-1)}{S_t^2}Y_t\text{d}\langle S_t,S_t\rangle$$ Assuming $S_t$ follows a Geometric Brownian Motion with drift $\mu$ and diffusion $\beta$: $$\text{d}S_t=\mu S_t\text{d}t+\beta S_t\text{d}W_t$$ Then: $$\text{d}Y_t=\left(\lambda\mu+\lambda(\lambda-1)\frac{\beta^2}{2}-\alpha\right)Y_t\text{d}t+\lambda\beta Y_t\text{d}W_t$$ Hence for $Y_t$ to be a (local) martingale we need: $$\lambda\mu+\lambda(\lambda-1)\frac{\beta^2}{2}=r\lambda+\lambda(\lambda-1)\frac{\sigma^2}{2}$$ This is true if: $$\begin{align} \text{C.1}\quad\mu&=r\\ \text{C.2}\quad\beta&=\sigma \end{align}$$

Method 2

Note also that, if the above conditions $\text{C.1}$ and $\text{C.2}$ hold: $$\begin{align} Y_t&=\left(\frac{S_t}{S_0}\right)^\lambda \exp\left\{-\left(r\lambda-\lambda(1-\lambda)\frac{\sigma^2}{2}\right)t\right\} \\ &=\exp\left\{\left(r\lambda-\lambda\frac{\sigma^2}{2}\right)t+\lambda\sigma W_t\right\}\exp\left\{-\left(r\lambda-\lambda(1-\lambda)\frac{\sigma^2}{2}\right)t\right\} \\ &=\exp\left\{-\lambda^2\frac{\sigma^2}{2}t+\lambda\sigma W_t\right\} \end{align}$$ Let $0<s<t$. Then: $$\begin{align} \mathbb E^Q\left[Y(t,S_t)|\mathcal{F}_s\right]&= \exp\left\{-\lambda^2\frac{\sigma^2}{2}s+\lambda\sigma W_s\right\}\mathbb E^Q\left[\exp\left\{-\lambda^2\frac{\sigma^2}{2}(t-s)+\lambda\sigma (W_t-W_s)\right\}|\mathcal F_s\right] \\ &= \exp\left\{-\lambda^2\frac{\sigma^2}{2}s+\lambda\sigma W_s\right\}\mathbb E^Q\left[\exp\left\{-\lambda^2\frac{\sigma^2}{2}(t-s)+\lambda\sigma (W_t-W_s)\right\}\right] \\ &= \exp\left\{-\lambda^2\frac{\sigma^2}{2}s+\lambda\sigma W_s\right\} \\[7pt] &= Y_s \end{align}$$

Answer 2

$E_0[Y_{\lambda,t}] = 1\,\, \forall t$, hence $Y_t$ is a martingale.

Hint: Look at the arithmetic moments section of this wiki page on lognormal distribution