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?

  • $\begingroup$ Except for the first line I cannot seem to find any vanilla call options, would you mind clarifying? $\endgroup$
    – htd
    Commented Feb 5, 2018 at 13:11
  • $\begingroup$ I added the payoff... sorry I missed your comment. $\endgroup$
    – Owls
    Commented Feb 5, 2018 at 23:14

1 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.

  • $\begingroup$ Thanks for the great answer! I am a beginner on Q dynamics. I am confused about two points. One is why $V_{t}$ is a tradable asset only when $2r = \sigma^2$. The other is why the Q-measure pricing formula still hold? $dV_{t} = (\sigma^2 -r)Vdt - \sigma Vd^{Q}W_{s} $ is not a martingale right? $\endgroup$
    – Owls
    Commented Feb 5, 2018 at 17:44
  • $\begingroup$ Please understand that we're only computing the dynamics of $\frac{1}{S_t}$ in order to find the distribution of the payoff $\frac{1}{S_T}$, it doesn't need to be a martingale nor should you expected it to be one. The price is as always $e^{-r(T-t)} E^\mathbb{Q}\Big[\frac{1}{S_T}\Big|\mathcal{F_t}\Big] = \frac{1}{S_t}e^{(\sigma^2-2r)(T-t)}$. The factor $e^{(\sigma^2-2r)(T-t)}$ corrects the drift of $\frac{1}{S_t}$ such that the discounted price process is a $\mathbb{Q}$-martingale. $\endgroup$
    – Freelunch
    Commented Feb 6, 2018 at 8:51
  • $\begingroup$ If $\sigma^2=2r$ then obviously $\frac{1}{S_t}e^{(\sigma^2-2r)(T-t)} = \frac{1}{S_t}$ hence the price if simply $\frac{1}{S_t}$, but this is just a special situation. In general (with $\sigma^2\neq 2r$) the price can't be $\frac{1}{S_t}$, since otherwise we would be able to construct a risk-free portfolio with a drift different from $r$ and hence violate no-arbitrage (this is probably a useful exercise). $\endgroup$
    – Freelunch
    Commented Feb 6, 2018 at 9:18
  • $\begingroup$ What would the arbitrage portfolio be? Looks I'm confused by basic concepts of risk-neutral. Any suggestion about a good textbook? $\endgroup$
    – Owls
    Commented Feb 8, 2018 at 7:14

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