Equivalent Martingale Measure(EMM) of Inverse of Stock Price

Question

I met this question says how to price a vanilla call option $C(St,t,T,K) = \frac{1}{S_T}$which pays the inverse of a stock $V_{t} = \frac{1}{S_{t}}$ at maturity if the stock price follows a geometric Brownian motion $dS_{t}=\mu S_{t}dt+\sigma S_{t}dB_{s}$? I tried to use the risk-neutral measure approach, however, I cannot prove that if the option is discounted by a risk-free bond it becomes a martingale i.e. $\frac{V_{t}}{B_{0}e^{rt}}$ does not have a drift term. Is this a correct change of numeraire?

Answer 1

Let $dB_t = rB_t dt$. Now

\begin{equation} d\Big(\frac{1}{B_t S_t}\Big) = -\frac{dS_t}{B_t S_t^2} -\frac{dB_t}{B_t^2S_t} +\frac{2}{2}\frac{(dS_t)^2}{B_t S_t^3} = (-\mu-r+\sigma^2)\frac{1}{B_tS_t}dt-\sigma\frac{1}{B_tS_t} dW_t \end{equation}

Using the EMM given by $dW_t = \frac{r-\mu}{\sigma}dt +dW_t^\mathbb{Q}$ we get the $\mathbb{Q}$-dynamics

\begin{equation} d\Big(\frac{1}{B_t S_t}\Big) = (\sigma^2-2r)\frac{1}{B_tS_t}dt-\sigma\frac{1}{B_tS_t} dW_t^\mathbb{Q} \end{equation}

This is only a martingale in special case when $2r = \sigma^2$, hence unless that holds $V_t = \frac{1}{S_t}$ cannot be the price of a traded asset. But the price of a contingent claim $V_T = \frac{1}{S_T}$ at some maturity date $T$ is still $e^{-r(T-t)}E^\mathbb{Q}\Big[\frac{1}{S_T}\Big|\mathcal{F_t}\Big]$ which is obviously a martingale.