I recently asked this question regarding how to price an option with payoff:

$$\text{Payoff}_T = (A_TR_T - A_T \lambda)^+ $$

Let's assume for generality that $A_t$ and $R_t$ are GMB's:

$$dA_t = \mu_{A,t} A_t dt + \sigma_{A,t} A_t dW_{A,t}$$ $$dR_t = \mu_{R,t} R_t dt + \sigma_{R,t} R_t dW_{R,t}$$

This comment mentioned that because the payoff can be seen as

$$\text{Payoff}_T = A_T(R_T - \lambda)^+ $$

we could use a change of numeraire approach to price it with Black's formula.

I'm not really familiar with this method, could somebody show me how it would work?

  • 4
    $\begingroup$ @Gordon we are curious ;) $\endgroup$
    – Richi Wa
    Oct 22, 2015 at 12:01

1 Answer 1


Note: your previous question assumed log-normality instead of normality.

By Cholesky decomposition, we assume that, under measure $P$, \begin{align*} \frac{dR_t}{R_t} &= \mu_{R,t} dt + \sigma_{R,t}\, dW^1_t\\ \frac{dA_t}{A_t} &= \mu_{A,t} dt + \sigma_{A,t}\, d\left(\rho W^1_t + \sqrt{1-\rho^2} W^2_t\right), \end{align*} where $W^1$ and $W^2$ are two independent standard Brownian motions. Here, we assume that $\mu_{R,t}$, $\mu_{A,t}$, $\sigma_{R,t}$, and $\sigma_{A,t}$ are deterministic or constants. Define the probability measure $\widetilde{P}$ such that we have the Radon Nykodym derivative \begin{align*} \frac{d\widetilde{P}}{dP}\big|_t &= \frac{A_t}{A_0}\frac{1}{e^{\int_0^t\mu_ {A,s}ds}}\\ &= \exp\left(-\frac{1}{2}\int_0^t\sigma_{A,s}^2 ds + \int_0^t\sigma_{A,s}\,d\left(\rho W^1_s + \sqrt{1-\rho^2} W^2_s\right)\right). \end{align*} By Girsanov theorem, \begin{align*} \widetilde{W}^1_t &= W_t^1 - \rho \int_0^t \sigma_{A,s} ds \,\, \mbox{ and}\\ \widetilde{W}^2_t &= W_t^2 - \sqrt{1-\rho^2} \int_0^t \sigma_{A,s} ds \end{align*} are two independent standard Brownian motions under $\widetilde{P}$. Moreover, under $\widetilde{P}$, \begin{align*} \frac{dR_t}{R_t} &= \left(\mu_{R,t}+\rho \sigma_{A,t} \sigma_{R,t}\right) dt + \sigma_{R,t}\, d\widetilde{W}^1_t. \end{align*} Note also that \begin{align*} \frac{dP}{d\widetilde{P}}\big|_t &= \frac{A_0}{A_t}e^{\int_0^t\mu_ {A,s}ds}. \end{align*} Then, \begin{align*} E_P(A_T(R_T - \lambda)^+) &= E_{\widetilde{P}}\left(\frac{dP}{d\widetilde{P}}\big|_T A_T (R_T - \lambda)^+ \right)\\ &= A_0\, e^{\int_0^T\mu_{A, s} ds }\,E_{\widetilde{P}}\left( (R_T - \lambda)^+ \right), \end{align*} which reduces to Black-Scholes' formula. Some details are omitted here.

If $\mu_{A,t}$ is the short interest rate at time $t$, then $e^{\int_0^t\mu_ {A,s}ds}$ is the money-market account value at time $t$.

  • $\begingroup$ Of course, it's a mistake in the question, let me adjust. $\endgroup$
    – SRKX
    Oct 23, 2015 at 2:19
  • $\begingroup$ Ok, so my option will depend on $\mu_{A,t}$ this time, right? $\endgroup$
    – SRKX
    Oct 23, 2015 at 4:09
  • $\begingroup$ @SRKX: Yes, your option will depend on $\mu_{A, t}$. You may also check whether you can reach the same formula as before for the case where $\mu_{A, t} = 0$. $\endgroup$
    – Gordon
    Oct 23, 2015 at 12:35

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