3
$\begingroup$

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\}$

$\endgroup$
7
  • $\begingroup$ 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

6
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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