8
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ @Gordon we are curious ;) $\endgroup$ – Ric Oct 22 '15 at 12:01
5
$\begingroup$

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 $t$.

$\endgroup$
  • $\begingroup$ Of course, it's a mistake in the question, let me adjust. $\endgroup$ – SRKX Oct 23 '15 at 2:19
  • $\begingroup$ Ok, so my option will depend on $\mu_{A,t}$ this time, right? $\endgroup$ – SRKX Oct 23 '15 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 '15 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.