Consider a Black Scholes market with constant coefficients, a bond and two risky assets: $$dB_{t}=r B_{t}dt \\ dS_{t}^{i}=S_{t}^{i}(b_{i}dt+\sigma_{i,1}dW_{t}^{1}+\sigma_{i,2}dW_{t}^{2})$$ where $i=1,2$ and $W_{t}^{1},W_{t}^{2}$ are two independent Brownian Motions.

I want to determine the fair price of the option with payoff at maturity $T$: $$X(T)=(S_{T}^{1}-S_{T}^{2})^{+}$$

I have applied Girsanov´s Theorem twice, first to find a risk-neutral measure $Q$, then to find another measure $Q'$ such that the quotient process $S^{1}/S^{2}$ is a martingale under $Q'$.

Using risk-neutral pricing, I know that the fair price must be: $$E_{Q}[X(T)]e^{-rT}$$ I now want to change numeraire, to arrive at: $$E_{Q}[X(T)]e^{-rT}=E_{Q''}[(\frac{S^{1}_{T}}{S_{T}^{2}}-1)^{+}]S_{0}^{2}$$ which is the Black-Scholes price of a call on $S^{1}/S^{2}$ with strike price $K=1$ and maturity $T$ (and for which I have the classical formula). But this of course only holds, if $Q''$ is a risk-neutral measure for $S^{1}/S^{2}$.

And unfortunately $Q' \neq Q''$ ... how can I conclude?

  • $\begingroup$ $S_t^1/S_t^2$ should emerge as a $Q''$-martingale by definition, i.e. it has zero drift (if no divs and no repo are assumed). Also, since it's the ratio of 2 lognormals, it remains a lognormal, with volatility given by $\sigma^2 = (\sigma_1^2 + \sigma_2^2 - 2 \rho \sigma_1 \sigma_2)$, since only the drift is altered per a change of measure, not volatility. $\endgroup$
    – Quantuple
    Jul 13, 2020 at 11:37
  • $\begingroup$ @Quantuple Change of numeraire means $\frac{dQ''}{dQ} = e^{rT} \frac{S^{2}_{0}}{S_{T}^{2}}$ - how does this imply that $S_{t}^{1}/S_{t}^{2}$ is indeed a $Q''$-martingale? I have a measure $Q'$ under which it is, but I got that from Girsanov and it is not $Q''$ as defined above $\endgroup$ Jul 13, 2020 at 12:10
  • $\begingroup$ Under the measure $\Bbb{N}$ associated to the numéraire $N_t$, for any self-financing strategy $V_t$, $V_t/N_t$ is a $\Bbb{N}$-martingale by definition, i.e. $$ V_0 = \Bbb{E}^\Bbb{N} \left[ N_0/N_T V_T \right] $$ So, if you use the numéraire $S_t^2$ $$ V_0 = S_0^2 \Bbb{E}^{\Bbb{S}^2} \left[ V_T^+/S_T^2 \right] $$ Noting that trading $V_t$ (spread option) and $S_t^1$ are both self-financing strats then $\Bbb{S}^2$ is such that $S_t^1/S_t^2$ is a martingale and same for $V_t/S_t^2$ which gives $$V_t = S_0^2 \Bbb{E}^{\Bbb{S}^2}\left[ (S_T^1/S_T^2 - 1)^+ \right] $$ so Q' = Q'' for you. $\endgroup$
    – Quantuple
    Jul 13, 2020 at 12:26


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