# Changing numeraire in Margrabes formula

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?

• $S_t^1/S_t^2$ 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. Jul 13, 2020 at 11:37
• @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 Jul 13, 2020 at 12:10
• 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. Jul 13, 2020 at 12:26