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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!

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    $\begingroup$ What dynamics does $S_t$ follow? What is $x$ and $\lambda$? Your exponential is unclear too. $\endgroup$ – Daneel Olivaw Nov 8 '19 at 12:25
  • $\begingroup$ I have amended your equation to what I believe is the proper expression, and provided an answer based on that. Can you confirm that my correction is correct? $\endgroup$ – Daneel Olivaw Nov 8 '19 at 13:30
  • $\begingroup$ @DaneelOlivaw Sorry, I made a mistake! It's supposed to be 2 and not $x$! $S_t$ follows the dynamics of the Black & Scholes model! $\endgroup$ – Math user Nov 8 '19 at 14:01
  • $\begingroup$ Ok, that is what I was wondering. You can check my answer below, I assume $S_t$ follows lognormal dynamics (i.e. as in the Black-Scholes model) with drift $r$ and diffusion/volatility $\sigma$ then? $\endgroup$ – Daneel Olivaw Nov 8 '19 at 14:20
  • $\begingroup$ Find the partial derivative with respect to t & S and plug it into Ito's lemma. Then use the fact that $S_t$ follows the black scholes model $dS_t=S_t(rdt+\sigma dW_t)$ and plug it back into $dS_t and (dS_t)^2$ should help in cancelling out the dt terms and you'll be left with only the $dW_t$ terms. $\endgroup$ – Anon Nov 9 '19 at 0:19
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$E_0[Y_{\lambda,t}] = 1\,\, \forall t$, 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: $$Y_t=\left(\frac{S_t}{S_0}\right)^\lambda \exp\left\{-\left(r\lambda-\lambda(1-\lambda)\frac{\sigma^2}{2}\right)t\right\}$$

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: $$\begin{align} \mathbb E^Q\left[Y(t,S_t)|\mathcal{F}_s\right]&= \exp\left\{-\lambda^2\frac{\sigma^2}{2}s+\lambda\sigma W_s\right\}\mathbb E^Q\left[\exp\left\{-\lambda^2\frac{\sigma^2}{2}(t-s)+\lambda\sigma (W_t-W_s)\right\}|\mathcal F_s\right] \\ &= \exp\left\{-\lambda^2\frac{\sigma^2}{2}s+\lambda\sigma W_s\right\}\mathbb E^Q\left[\exp\left\{-\lambda^2\frac{\sigma^2}{2}(t-s)+\lambda\sigma (W_t-W_s)\right\}\right] \\ &= \exp\left\{-\lambda^2\frac{\sigma^2}{2}s+\lambda\sigma W_s\right\} \\[7pt] &= Y_s \end{align}$$

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  • $\begingroup$ Do we assume the expected value under Q of the line just before the last one is 0 because expected value of $W_t - W_s$ is 0 ? If it the case, how does it cancel the all exponential ? $\endgroup$ – TmSmth Nov 9 '19 at 11:22
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    $\begingroup$ @TmSmth $W_t-W_s$ is a normally-distributed random variable with $0$ expectation and variance $\lambda^2\sigma^2(t-s)$ thus its Laplace transform is $\exp\{(\lambda^2\sigma^2(t-s))/2\}$, hence terms cancel and we get $e^0=1$. $\endgroup$ – Daneel Olivaw Nov 19 '19 at 18:28

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