I am considering two risky assets following the usual correlated GBM given by

$$\frac{\mathrm{d}S^{(i)}_t}{S^{(i)}_t}=\mu_i\mathrm{d}t+\sigma_i\mathrm{d}W^{(i)}_t,\quad i\in\{1,2\}$$



Here's my current understanding: I know that in general, in a complete and no-arbitrage market setting, if I use $S^{(1)}_t$ as my numéraire, I can develop a measure $\mathbb{Q}_1$ such that $\tilde{S}^{(2)}_t=\frac{S^{(2)}_t}{S^{(1)}_t}$ is a martingale. By Girsanov's theorem, defining $\tilde{\sigma}=\sqrt{\sigma_1+\sigma_2-2\rho\sigma_1\sigma_2}$ and $\tilde{W}_t^{\mathbb{Q}_1}=\frac1{\tilde{\sigma}}\left(\sigma_2W^{(2)}_t-\sigma_1W^{(1)}_t\right)$, I obtain the driftless equation


I realise Margrabe’s formula is the end goal, but some texts (one example p. 35) I have read include a risk-free asset or bank account. Notably, to obtain the above equations, they switch from the physical measure $\mathbb{P}$ to the risk-free measure $\mathbb{Q}$ first, then to the measure $\mathbb{Q}_1$. Some others claim that no bank account must be enforced. After reading these texts, I am now confused — if there is no such risk-free asset/account s.t. all the money must be invested in these risky assets, then:

  1. What do I count as my risk-free interest rate (if any, given that I am working with two pure risky assets) under this numéraire?

  2. Is it correct to claim that I cannot price all payoff types because the market is not complete due to lack of riskless asset, but I am I able to price, say, a spread option because it just so happens we can hedge it? If so, why are we even able to use this change of numéraire in the first place? How are we able to detect whether a particular type of payoff can be hedged or not? Is $|\rho|<1$ a sufficient condition for completeness of the market? What conditions are necessary and/or sufficient?

  3. Is it possible to construct a riskless process $\mathrm{d}B_t=r_tB_t\mathrm{d}t$ via a self-financing portfolio replicated by the two assets? I have a feeling it is impossible in incomplete markets.

  • $\begingroup$ I am only aware of the Wikipedia entry claiming that no risk-neutral measure is needed. Could you please post a credible source for this claim? $\endgroup$ Dec 28, 2020 at 11:39
  • 2
    $\begingroup$ @p.vitzliputzli I found this assertion with some explanation in chapter 7 of “A Course in Derivative Securities by Kerry Back”, as well as “Minimization of Conditional Value at Risk for Spread Options under Capital Constraints” by Chingis Maksimov, “Making No-Arbitrage Discounting Invariant” by Daniel Bálint, which mention we cannot assume a bank account exists. $\endgroup$
    – user107224
    Dec 28, 2020 at 12:37
  • 1
    $\begingroup$ Thanks! In Back's book, section 7.11 describes this issue. Based on your second question, I guess you have read that already. I would agree with you that in this case, it just happens that you can price the spread option, because you are able to replicate it. With regard to "change of numeraire", you have to be careful in the sense that you are not changing to an equivalent risk-neutral probability measure, but just to an equivalent probability measure. That is something you can always do. A risk-free process would be e.g. $\tilde{S}_t^{(1)}$, because its value is always 1. $\endgroup$ Dec 28, 2020 at 13:18
  • $\begingroup$ @p.vitzliputzli thanks for the reply; to be honest I don’t think I’ve fully grasped what’s in the book, could you explain it further? Also, more fundamentally what is meant by an equivalent measure? I thought EMMs referred to a risk neutral measure, but I guess my definition is incorrect. $\endgroup$
    – user107224
    Dec 29, 2020 at 4:03
  • $\begingroup$ @p.vitzliputzli also I have updated my third question, I was leaning more to a replicating portfolio for a risk-free asset $\endgroup$
    – user107224
    Dec 29, 2020 at 4:04

2 Answers 2


Under the risk-neutral measure both stocks follow the GBMs \begin{align} S^{(i)}_t=S^{(i)}_0\exp\left((r-q_i)t+\sigma_iW^{(i)}_t-\frac{\sigma_i^2t}{2}\right)\,,\quad i=1,2\,, \end{align} where the constant $r$ is the riskless interest rate and the constant $q_i$ is stock $S^{(i)}$'s dividend rate. The Margrabe formula says that the value of the option to exchange stock $S^{(2)}$ for stock $S^{(1)}$ at time $T$ is \begin{align}\tag{1} V&=e^{-q_1T}S^{(1)}_0N(d_1)-e^{-q_2T}S^{(2)}_0N(d_2)\,,\\ d_1&=\frac{\ln(S^{(1)}_0/S^{(2)}_0)+(q_1-q_2+\sigma^2/2)T}{\sigma\sqrt{T}}\,,\\ d_2&=d_1-\sigma\sqrt{T}\,,\\ \sigma&=\sqrt{\sigma_1^2+\sigma_2^2-2\sigma_1\sigma_2\rho}\,. \end{align} It is true (and was mentioned by William Margrabe in his original paper [1]) that in this formula the riskless rate $r$ does not occur. He wrote that "this may seem puzzling". In fact we rightfully expect the Black Scholes formula with strike $K$ to

  1. contain the riskless rate,

  2. be a special case of the Margrabe formula (1) when $\sigma_2=0$ and $q_2=0\,.$

In fact this is the case: When $\sigma_2=0$ and $q_2=0$ then $$ S^{(2)}_T=S^{(2)}_0e^{rT}\,. $$ When $S^{(2)}_T$ equals the Black-Scholes strike $K$ then

\begin{align} S_0^{(2)}=e^{-rT}K\,,\quad\quad d_1=\frac{\ln(S^{(1)}_0/K)+(q_1+r+\sigma^2/2)T}{\sigma\sqrt{T}}\,,\quad\quad\sigma=\sigma_1 \end{align} and (1) becomes $$ V=e^{-q_1T}S^{(1)}_0N(d_1)-e^{-rT}KN(d_2) $$ as expected.

Further remarks:

  1. The model with two stocks, each following a GBM is complete. This means that every payoff that depends on the two stocks at maturity $T$ can be replicated with a self financing strategy in the two stocks. See [1].

  2. The fact that we don't need the riskless asset $B_t=e^{rt}$ in the replication of the payoff $(S^{(1)}_T-S^{(2)}_T)^+$ does not mean that we are 'lacking' it. It is not needed. That's all.

  3. An example of an incomplete model would be one where the stocks follow more complicated processes such as having stochastic volatility, or jumps with stochastic jump sizes.

[1] W. Margrabe, The value of an option to exchange one asset for another. Journal of Finance, Vol. 33, No. 1 (March 1978), 177-186.


Since you have asked in your title specifically about the case with no bank account, I would like to provide a slightly different answer than Kurt G. You have correctly inferred that then the model is incomplete. The $T$-claim $\mathcal{X} = 1$, i.e. a zero-coupon bond, cannot be replicated. Alternatively, note that there is no unique martingale measure. You can move the drift term from the normalized asset $\tilde{S}^{(2)}$ into either $W^{(1)}$ or $W^{(2)}$ to make $\tilde{S}^{(2)}$ a martingale under the corresponding changed measure $\mathbb{Q}$, in fact you have whole line $\{(\varphi_1, \varphi_2)\} \subset \mathbb{R}^2$ of possible such Girsanov kernels.

However, there is still a family of claims that can be priced (e.g. linear combinations of $S^{(1)}_T, S^{(2)}_T$). The spread option turns out to be one such claim. Namely, with $\mathcal{X} = \left(S^{(1)}_T - S^{(2)}_T\right)^+$, we have the discounted claim $$ \tilde{\mathcal{X}} := \frac{\mathcal{X}}{S^{(1)}_T} = \frac{\left(S^{(1)}_T - S^{(2)}_T\right)^+}{S^{(1)}_T} = \left(1 - \tilde{S}^{(2)}_T\right)^+. $$

Replicating $\mathcal{X}$ is equivalent to replicating $\tilde{\mathcal{X}}$ in the discounted economy $(\tilde{S}^{(1)}, \tilde{S}^{(2)}) = (1, \tilde{S}^{(2)})$. There, $\tilde{\mathcal{X}}$ is just a put option with strike $1$, so it can be replicated by the results for the Black-Scholes model. Its price is of course $$ \Pi_0\left(\tilde{\mathcal{X}}\right) = \mathbb{E}^{\mathbb{Q}_1}\left[\left(1 - \tilde{S}^{(2)}_T\right)^+\right], $$ and the price of $\mathcal{X}$ is obtained by multiplying by $S^{(1)}_0$.

The answer to question 2 is now straightforward. Any claim such that the discounted claim can be hedged in the discounted economy is replicable. Specifically, if the claim is simple and a function of $\tilde{S}^2_T$ and such that the corresponding Black-Scholes PDE with this function as the boundary condition emits a $C^{1, 2}$-solution, the hedging portfolio can be constructed as in the Black-Scholes framework.

  • $\begingroup$ This answer is not correct. First of all the zero-coupon bond does not exist in this economy because there is no bank account, therefore it cannot be used as an argument regarding completeness. Secondly letting $W:=(W^{(1)},W^{(2)})$ it is clear there is a unique change of measure that allows to shift from the $(1)$-measure to the $(2)$-measure. $\endgroup$ Oct 23, 2023 at 14:02
  • $\begingroup$ It is an example of something that does not exist, precisely like you are saying. Indeed that can be used as an argument, as I was arguing for incompleteness. The change is not unique. Consider changes of measure from $\mathbb{Q}$. Any Girsanov kernel of the form $(\varphi, \frac{\sigma_1}{\sigma_2}\varphi)$ will correspond to another measure $\mathbb{Q}_\varphi$ that is also a martingale measure, since the change will cancel out in $\tilde{W}_t^{\mathbb{Q}_1}=\frac1{\tilde{\sigma}}\left(\sigma_2W^{(2)}_t-\sigma_1W^{(1)}_t\right)$. Thus there are infinitely many martingale measures. $\endgroup$ Oct 24, 2023 at 10:17

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