Risk-Neutral Pricing Methods for Exchange and Spread Options

# Risk Neutral Evaluation - Exchange/Spread Options


# Risk Neutral Evaluation - Exchange/Spread Options












I have two assets, $S_1$ and $S_2$, which follow geometric Brownian motion processes. This implies that both $S_1$ and $S_2$ have a lognormal distribution.

I'm trying to get the exchange option price formula through the risk neutral valuation, or, in other words $$C(t,S_1,S_2)=e^{-r(T-t)}E(\max\{S_1-S_2,0\})$$.

How do I calculate the expected value $E(\max\{S_1-S_2,0\})$??

## Answer by SRKX (score 1)

https://quant.stackexchange.com/a/8884

Your formula, as it stands, is incorrect, at least is if $E$ means the "expected value under real-world probabilities".

I wrote a blog post explaining the basic rationale behind risk-neutral pricing where you will see that if the Fundamental Theorem of Asset Pricing theorem holds, you can write:

Let $X_t=S_{1,t}-S_{2,t}$

$$e^{-rt} X_t = \mathbb{E}_\mathbb{Q}[e^{-rT} \max(X_T,0) | \mathcal{F}_t]$$

where $\mathbb{E}_\mathbb{Q}$ stands for "expectation under the risk-neutral probability".

From there, what you have to do is to define the risk-neutral dynamics of $S_1$ and $S_2$. When you say that the assets are following a lognormal distribution, I'll assume that they follow are independent Geometric Brownian Motion. In this case, their risk neutral dynamics are well know (see here):

$$dS_{1,t}= r S_{1,t} dt + \sigma_1 S_{1,t} dW^{\mathbb{Q}}_{1,t} $$ $$dS_{2,t}= r S_{2,t} dt + \sigma_2 S_{2,t} dW^{\mathbb{Q}}_{2,t} $$

where is the risk-free rate, and $dW^{\mathbb{Q}}_{i,t}=dW_{i,t}+\frac{\mu_i-r}{\sigma_i}dt$.

So, you have:

$$dX_t=dS_{1,t}-dS_{2,t}=r (S_{1,t} -S_{2,t})dt + + \sigma_1 S_{1,t} dW^{\mathbb{Q}}_{1,t} - dS_{2,t}= r S_{2,t} dt - \sigma_2 S_{2,t} dW^{\mathbb{Q}}_{2,t}$$

and:

$$X_t = e^{-r(T-t)} \mathbb{E}_\mathbb{Q}[\max(X_T,0) | \mathcal{F}_t]= e^{-r(T-t)} \mathbb{E}_\mathbb{Q}[\mathbb{1}_{X_T>0}X_T | \mathcal{F}_t]$$

Then you carry on manipulation the equation, I would suggest changing the numeraire...

## Answer by Brian B (score 0)

https://quant.stackexchange.com/a/9095

Let's generalize to the slightly more common case of an option on a spread of correlated assets $S_{1,2}$ with correlation coefficient $\rho$. That is to say, write $s = S_1-S_2$, and you want

$$ e^{-rT}\mathbb{E}\left[ (s_T-K)^+ \right] $$

If $S_{1,2}$ follow geometric brownian motions (GBM) then the terminal distribution for $s_T$ is not in any simple form. Let me note, however, that it is common among professionals to simply assume $s$ follows an ordinary brownian motion with first two moments that match the corresponding moments of the "true" distribution, in which case the terminal distribution is a gaussian, and we can derive the option value

$$ V=e^{-rT}(S_2-S_1) N\left( \frac{e^{-rT}(S_2-S_1)-K e^{-rT}}{v} \right) + v \cdot n\left( \frac{e^{-rT}(S_2-S_1)-K e^{-rT}}{v} \right) $$

where

$$ v := e^{rT}\sqrt{ S_1^2(e^{\sigma_1^2 T}-1)-2S_1S_2(e^{\rho \sigma_1\sigma_2 T}-1) + S_2^2 (e^{\sigma_2^2 T}-1)} $$

To get an "exact" solution, you can instead observe that we write this as

$$ \frac{dS_1}{S_1} = r dt + \sigma_1 dW^{(1)} \\ \frac{dS_2}{S_2} = r dt + \sigma_2 \left( \rho dW^{(1)} + \sqrt{1-\rho^2} dW^{(2)}\right) $$

in which case the expectation integral becomes

$$ e^{-rT}\int_{z_1=-\infty}^{\infty} \int_{z_2=-\infty}^{\infty} (S_1(z_1)-S_2(z_1,z_2)-K)^+ n(z_1) n(z_2) dz_1 dz_2 $$

which we can change to

$$ e^{-rT}\int_{z_1=S_1^{-1}(K)}^{\infty} P( S_2^0, S_1(z_1)-K, r^\prime(z_1), \sigma^\prime(z_1) ) n(z_1) dz_1 $$

for adjusted drift $r^\prime(z_1)$ and adjusted vol $\sigma^\prime(z_1)$ in the Black-Scholes put option pricing formula $P(\cdot)$. Here for clarity $S_2^0$ indicates the initial value of $S_2$, while $S_{1,2}(\cdot)$ indicate the usual terminal value functions for GBMs using the customary volatilities and drifts.

This integral is one-dimensional and can be efficiently evaluated using numerical quadrature, such as a trapezoid rule scheme.

## Answer by wsw (score 0)

https://quant.stackexchange.com/a/9535

The solution can be found in, say, Bjork's book. Assume that the two Brownian motions are uncorrelated. I'm also swapping the symbols $S_{1,T}$ and $S_{2,T}$ in the payoff function.

We use the stock $S_1$ as numeraire. Note that $$ \frac{V_t}{S_{1,t}} = \mathbb{E}^{Q^1} \left[ \frac{S_{1,T} \max \left(S_{2,T}/S_{1,T} - 1 \right)}{S_{1,T}} \middle\vert \mathcal{F}_t \right] = \mathbb{E}^{Q^1} \left[ \max \left(\frac{S_{2,T}}{S_{1,T}} - 1 \right) \middle\vert \mathcal{F}_t \right] \; . $$ Let $Z_t \equiv S_{2,t}/S_{1,t} $. Since $Z_t$ is a martingale under the $Q^1$ measure by definition, $$ dZ_t = Z_t (\sigma_2 - \sigma_1) \; dW^{Q^1} \; . $$ We are essentially pricing a European call with strike $K=1$ for a stock $Z_t$ that has no drift but has volatility $\sigma = | \sigma_2 - \sigma_1 |$. The price is $$ V_t = S_{1,t} \left\{ Z_t N(d_1) - N(d_2) \right\} = S_{2,t} N(d_1) - S_{1,t} N(d_2) \; , $$ where $d_1 = \frac{\log Z_t + \frac{1}{2} \sigma^2 (T-t)} {\sigma \sqrt{T-t}}$ and $d_2 = d_1 - \sigma \sqrt{T-t}$.

See also Margrabe's formula [ http://en.wikipedia.org/wiki/Margrabe's_formula ].