Is anybody able to give the solution to the following problem?

Suppose we have two assets, each of which follows a GBM process, and where $dW_S$ and $dW_X$ are correlated $(dW_SdW_X=\rho)$.

$dS=\mu_s S \hspace{0.5mm}dt +\sigma_s S \hspace{0.5mm}dW_s $

$dX=\mu_X X \hspace{0.5mm}dt +\sigma_X X \hspace{0.5mm}dW_X $

Determine the price of an european put option $V$ with payoffs depending on the terminal value of both assets described by

$ V_T=X_T\max\{K-S_T,0\}$

  • $\begingroup$ You would probably get better or quicker help if you could also describe how you have tried solving this problem and exactly where you have failed. $\endgroup$
    – Alper
    Nov 24, 2021 at 20:13
  • $\begingroup$ Hint : 1. Write $X$ as a GBM with drift. 2. Write $W_X=\rho W_S+\sqrt{1-\rho^2}W_S^\bot\,.$ 3. Realise that $X$ is essentially a Radon-Nikodym derivative that gives you a new probability measure. 4. Calculate the expectation of the payoff under this new measure. $\endgroup$
    – Kurt G.
    Nov 25, 2021 at 2:10
  • 1
    $\begingroup$ Think of X as a annuity and S as the swap rate. You're valuing a swaption. Go to the annuity measure and price it. $\endgroup$
    – user121416
    Nov 25, 2021 at 8:22
  • $\begingroup$ @user121416 Not exactly, in the sense that for swaption pricing, $S \times X = P\left(0, T_0\right) - P\left(0, T_N\right)$, which is a traded asset (long-short of ZC bonds). Here, we have nothing to believe that $S \times X$ is a traded asset, so things are a bit more complicated. $\endgroup$
    – siou0107
    Nov 25, 2021 at 13:45
  • $\begingroup$ @siou0107 good point. As I understand, change of measure to annuities is just a mathematical convenience and it is still mathematically applicable, regardless of tradeability of $S*X$. $\endgroup$
    – user121416
    Nov 25, 2021 at 14:42

1 Answer 1


Your statement of the problem is not very detailed. Are $\mu_{S/X}$ constant ? What about interest rates? In the classic exchange option problem, where the payoff is $(X_T - S_T)^+$, they actually do not matter since all the risk is related to $S$ and $X$, no cash is involved (up to selling your replicating portfolio at maturity to pay immediately the requested amount, which does not involve interest rates); here, without specific assumption you do have interest rate risk, and your market is incomplete in general, which involves having to pick a risk-neutral measure based on some parameterisation. I will try to give you a general methodology for such pricing problems, explaining which assumption I make at each step. You should then be able to adapt the reasoning to your specific problem.

First thing to remark is that since $X$ and $S$ are assets, their discounted values must be martingales under any risk-neutral measure $\mathbb{Q}$ by the first Fundamental Theorem of Asset Pricing, if they do not pay dividends. If they do, you can just normalise by some dividend factor $D_t^{S/X}$ that represents the reinvestment of dividends, e.g. $\exp{-\int_0^t{q_s^{S/X} \mathrm{d}s}}$ if the dividend stream is continuous at rate $q^{S/X}$.

Let us assume for now that $D_T \equiv 1$, i.e. that $S$ and $X$ are non-dividend paying. We can thus write : \begin{align} & dX_t = r_t X_t dt + \sigma_X X_t dW_t^X \\ & dS_t = r_t S_t dt + \sigma_S S_t dW_t^S = r_t S_t dt + \sigma_S S_t \left(\rho \, dW_t^X + \sqrt{1 - \rho^2} dW_t^\perp\right) \end{align} The drift has to be $r_t$ by the FTAP I. Note that these dynamics induce that both $S$ and $X$ are positive at all times.

Picking a specific risk-neutral measure $\mathbb{Q}$, the corresponding price for your derivative is $$ V_0 = \mathbb{E}^\mathbb{Q} \left[e^{-\int_0^T{r_t\mathrm{d}t}} X_T \left(K - S_T\right)^+\right] $$ Now, observe the following : since $X e^{-\int_0^\cdot{r_t\mathrm{d}t}}$ is a martingale under $\mathbb{Q}$, $\frac{X_T}{X_0} e^{-\int_0^T{r_t\mathrm{d}t}}$ is a positive random variable with expected value 1. It can thus define a new probability $\mathbb{Q}^X$ equivalent to $\mathbb{Q}$ by $$ \frac{\mathrm{d}\mathbb{Q}^X}{\mathrm{d}\mathbb{Q}} := \frac{X_T}{X_0} e^{-\int_0^T{r_t\mathrm{d}t}} $$

This corresponds to changing numéraire from the riskless asset $e^{\int_0^\cdot{r_t \mathrm{d}t}}$ to $X$, hence our notation. We then have \begin{align} V_0 & = X_0 \mathbb{E}^\mathbb{Q} \left[e^{-\int_0^T{r_t\mathrm{d}t}} \frac{X_T}{X_0} \left(K - S_T\right)^+\right] \\ & = X_0 \mathbb{E}^{\mathbb{Q}^X} \left[\left(K - S_T\right)^+\right] \end{align} Now, all you have to do is determine the distribution of $S_T$ under $\mathbb{Q}^X$. To do so, you must compute the drift change induced by the change of probability. The Girsanov theorem allows you to do so. Using Itō's lemma (or, equivalently, solving the SDE for $X$ that I have given above), you have that $$ X_T = X_0 e^{\int_0^T{\left(r_t - \frac{\sigma_X^2}{2}\right)\mathrm{d}t} + \int_0^T{\sigma_X\mathrm{d}W_t^X}} $$ hence $$ e^{-\int_0^T{r_t\mathrm{d}t}} \frac{X_T}{X_0} = e^{- \frac{\sigma_X^2}{2}T + \sigma_X W_T^X} $$ By the Girsanov theorem, $$ \widehat{W}_t^X := W_t^X - \sigma_X t $$ is a Brownian motion under $\mathbb{Q}^X$. The dynamics under $\mathbb{Q}^X$ are thus \begin{align} dX_t & = \left(r_t + \sigma_X^2\right) X_t dt + \sigma_X X_t d\widehat{W}_t^X \\ dS_t & = r_t S_t dt + \sigma_S S_t \left(\rho dW_t^X + \sqrt{1 - \rho^2} d W_t^\perp\right) \\ & = \left(r_t + \rho \sigma_X \sigma_S\right) S_t dt + \sigma_S S_t \left(\rho d\widehat{W}_t^X + \sqrt{1 - \rho^2} d W_t^\perp\right) \\ & = \left(r_t + \rho \sigma_X \sigma_S\right) S_t dt + \sigma_S S_t d\widehat{W}_t^S \end{align} Because $W^\perp$ is independent from $W^X$, it is also a $\mathbb{Q}^X$-Brownian motion independent from $\widehat{W}^S$, hence $d \langle \widehat{W}^X, \widehat{W}^S\rangle_t = \rho \, dt$.

Assuming that rates are deterministic, you have two sources of randomness ($W^S$ and $W^X$) and two assets to hedge them: you market is complete, and the replication price is an expectation under the unique risk-neutral measure $\mathbb{Q}$, or under the equivalent martingale measure $\mathbb{Q}^X$. By observing that the dynamics we just derived induces a lognormal distribution for $S_T$ under $\mathbb{Q}^X$, you can use a generalized Black formula (i.e. $\mathbb{E} \left[\left(K - X\right)^+\right]$ with $X$ lognormally distributed).

NOW, if $D_T^{X/S} \not \equiv 1$, i.e. $\mu^{X/S} \neq r_t$, the problem can quickly become more complex ; notably, as you have randomness on the dividends, you have a dividend risk and your market is not complete. One assumption that allows to solve the problem quite simply is that the dividend flows are continuous at deterministic rates $q^{S/X}$. Then, you have necessarily $\mu_X = r_t - q_t^X$ and $\mu_S = r_t - q_t^S$ under $\mathbb{Q}$, and the extension is straightforward.

  • $\begingroup$ Thank you very much siou0107 for sharing your knowledge in such a detailed and clear way. Your answer has helped me a lot. Regards $\endgroup$
    – Eastwood94
    Nov 29, 2021 at 9:19
  • $\begingroup$ Of course. Just one last thing. In the event that $\mu^{X/S}\neq r_t$ but all r, $\mu^X$ and $\mu^S$ are constant, I arrived to the following solution by following your guideline: $X_0 e^{(\mu^X-r) t}E^{Q^X}[(K-S_T)^+]$ . Can you see any flaws? Thanks a lot. $\endgroup$
    – Eastwood94
    Nov 29, 2021 at 9:55
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    $\begingroup$ Nope, seems fair to me. That would give you a Black-like formula $X_0 e^{-q^X T} \left[K \Phi \left(-d_2\right) - S_0 e^{-q^S T} \left(-d_1\right)\right]$, with $d_1$ and $d_2$ found by standard ways. $\endgroup$
    – siou0107
    Nov 29, 2021 at 17:39
  • $\begingroup$ Thanks a lot again. You are awesome! $\endgroup$
    – Eastwood94
    Nov 29, 2021 at 18:51

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