Unit 7, lesson 56

Multiasset derivatives and exchange options

How correlated diffusions give the effective volatility of a price ratio, how a change of numeraire prices the exchange option exactly, and why basket options rely on lognormal moment matching.

  • Video 56
  • Path-dependent and multiasset options

A fund holds the right to hand over one share of one company in a year and receive one share of another, with no cash changing hands. What is that right worth, and what does its price depend on beyond the two volatilities? This lesson builds the joint model of the two prices, finds the volatility of their ratio, prices the right exactly by measuring values in units of the delivered asset, and explains why a call on a basket of the same assets has only approximate formulas of this kind.

Key terms

Correlated diffusions

Definition Two positive asset prices follow correlated geometric Brownian motions under a pricing measure Q\mathbb{Q} with money-market account Bt=ertB_t = e^{rt} when

dSti=(r−qi)Sti dt+σiSti dWti,Q\begin{aligned} dS^i_t &= (r - q_i) S^i_t \, dt \\[0.4em] &\quad + \sigma_i S^i_t \, dW^{i,\mathbb{Q}}_t \end{aligned}

for i=1,2i = 1, 2, with a constant rate rr, constant continuous dividend yields qiq_i, constant volatilities σi≥0\sigma_i \ge 0, and Q\mathbb{Q}-Brownian motions with d[W1,Q,W2,Q]t=ρ dtd[W^{1,\mathbb{Q}}, W^{2,\mathbb{Q}}]_t = \rho \, dt for a constant −1≤ρ≤1-1 \le \rho \le 1. One construction takes W2,Q=ρW1,Q+1−ρ2 ZW^{2,\mathbb{Q}} = \rho W^{1,\mathbb{Q}} + \sqrt{1 - \rho^2}\, Z with ZZ a Q\mathbb{Q}-Brownian motion independent of W1,QW^{1,\mathbb{Q}}. The covariation of the prices is

d[S1,S2]t=ρσ1σ2St1St2 dt,\begin{aligned} &d[S^1, S^2]_t \\[0.4em] &\quad = \rho \sigma_1 \sigma_2 S^1_t S^2_t \, dt, \end{aligned}

and for σ1,σ2>0\sigma_1, \sigma_2 > 0 the pair (ln⁡ST1,ln⁡ST2)(\ln S^1_T, \ln S^2_T) is bivariate normal with correlation ρ\rho. With dd assets the matrix (ρjk)(\rho_{jk}) of driver correlations must be symmetric with unit diagonal and positive semidefinite, and pairwise values between −1-1 and 11 do not guarantee this. The law of each price on its own does not depend on ρ\rho, but the joint law does, and so does the price of a payoff such as (ST1−ST2)+(S^1_T - S^2_T)^{+}.

In plain words Each asset on its own follows the Black-Scholes model, and the new input is how their random shocks line up. Correlation one means the two drivers are the same Brownian motion, correlation zero means here that they are independent, and a set of correlations for several assets has to be consistent as a whole.

Example For three assets the correlations ρ12=ρ13=0.9\rho_{12} = \rho_{13} = 0.9 and ρ23=−0.9\rho_{23} = -0.9 each lie in [−1,1][-1, 1], yet the matrix has eigenvalues −0.8-0.8, 1.91.9 and 1.91.9. The combination W1−W2−W3W^1 - W^2 - W^3 would have variance (3−1.8−1.8−1.8)t=−2.4t(3 - 1.8 - 1.8 - 1.8)t = -2.4t, so no model has these correlations. For two constructed assets with S01=S02=100S^1_0 = S^2_0 = 100, zero dividends, σ1=0.2\sigma_1 = 0.2, σ2=0.3\sigma_2 = 0.3 and one year, the exchange option priced in a later card is worth about 19.741319.7413, 14.306514.3065, 10.524310.5243 and 3.98783.9878 for ρ=−1\rho = -1, 00, 0.50.5 and 11, while the law of each price is the same in every case.

Effective volatility

Definition In the correlated model the ratio Rt=St1/St2R_t = S^1_t/S^2_t satisfies

dRtRt=(q2−q1+σ22−ρσ1σ2) dt+σ1 dWt1,Q−σ2 dWt2,Q.\begin{aligned} \frac{dR_t}{R_t} &= \bigl(q_2 - q_1 + \sigma_2^2 \\[0.4em] &\qquad - \rho \sigma_1 \sigma_2\bigr)\, dt \\[0.4em] &\quad + \sigma_1 \, dW^{1,\mathbb{Q}}_t \\[0.4em] &\quad - \sigma_2 \, dW^{2,\mathbb{Q}}_t . \end{aligned}

To derive it, Itô’s lemma gives d(1/St2)=(1/St2)[(q2−r+σ22) dt−σ2 dWt2,Q]d(1/S^2_t) = (1/S^2_t)\bigl[(q_2 - r + \sigma_2^2)\, dt - \sigma_2 \, dW^{2,\mathbb{Q}}_t\bigr], where the second-order term contributes σ22\sigma_2^2, and the product rule for S1⋅(1/S2)S^1 \cdot (1/S^2) adds the covariation d[S1,1/S2]t=−ρσ1σ2Rt dtd[S^1, 1/S^2]_t = -\rho \sigma_1 \sigma_2 R_t \, dt. The Brownian part has quadratic variation σeff2 dt\sigma_{\text{eff}}^2 \, dt, where the effective volatility σeff≥0\sigma_{\text{eff}} \ge 0 is given by

σeff2=σ12+σ22−2ρσ1σ2=(σ1−σ2)2+2(1−ρ)σ1σ2,\begin{aligned} \sigma_{\text{eff}}^2 &= \sigma_1^2 + \sigma_2^2 \\[0.4em] &\quad - 2\rho \sigma_1 \sigma_2 \\[0.4em] &= (\sigma_1 - \sigma_2)^2 \\[0.4em] &\quad + 2(1 - \rho)\sigma_1 \sigma_2 , \end{aligned}

so d[ln⁡R]t=σeff2 dtd[\ln R]_t = \sigma_{\text{eff}}^2 \, dt. With σ1,σ2>0\sigma_1, \sigma_2 > 0, it is zero exactly when ρ=1\rho = 1 and σ1=σ2\sigma_1 = \sigma_2, and it equals σ12+σ22\sqrt{\sigma_1^2 + \sigma_2^2} only when ρ=0\rho = 0. The effective volatility is the same under every equivalent measure, since a change of measure alters drifts and not quadratic variation.

In plain words The ratio of two prices moves with the difference of their shocks, and the variance of a difference is the sum of the variances minus twice the covariance. Positive correlation lowers the risk of the ratio, but perfectly correlated drivers with different volatilities still leave a ratio that moves.

Example With the constructed values σ1=0.2\sigma_1 = 0.2, σ2=0.3\sigma_2 = 0.3 and ρ=0.5\rho = 0.5, the effective variance is 0.04+0.09−0.06=0.070.04 + 0.09 - 0.06 = 0.07, so σeff=0.07≈0.2646\sigma_{\text{eff}} = \sqrt{0.07} \approx 0.2646. At ρ=1\rho = 1 it is (0.3−0.2)2=0.01(0.3 - 0.2)^2 = 0.01 and σeff=0.1\sigma_{\text{eff}} = 0.1, not zero. At ρ=0\rho = 0 it is about 0.36060.3606, and at ρ=−1\rho = -1 it is 0.50.5.

Change of numeraire

Definition Let Nt=eq2tSt2N_t = e^{q_2 t} S^2_t be the value of one unit of asset 2 bought at time 00 with its dividends reinvested in the asset, a strictly positive traded price. In the correlated model, e−rtNt=S02exp⁡(σ2Wt2,Q−12σ22t)e^{-rt} N_t = S^2_0 \exp\bigl(\sigma_2 W^{2,\mathbb{Q}}_t - \tfrac12 \sigma_2^2 t\bigr) is a positive Q\mathbb{Q}-martingale, and the measure QN\mathbb{Q}^{N} on FT\mathcal{F}_T is defined by

dQNdQ=e−rTNTN0=exp⁡(σ2WT2,Q−12σ22T).\begin{aligned} \frac{d\mathbb{Q}^{N}}{d\mathbb{Q}} &= \frac{e^{-rT} N_T}{N_0} \\[0.4em] &= \exp\bigl(\sigma_2 W^{2,\mathbb{Q}}_T - \tfrac12 \sigma_2^2 T\bigr). \end{aligned}

The exponent is a constant times a Brownian motion, so Novikov’s condition holds, and Girsanov’s theorem with this positive sign gives the QN\mathbb{Q}^{N}-Brownian motions

Wt2,N=Wt2,Q−σ2t,Wt1,N=Wt1,Q−ρσ2t,\begin{gathered} W^{2,N}_t = W^{2,\mathbb{Q}}_t - \sigma_2 t, \\[0.4em] W^{1,N}_t = W^{1,\mathbb{Q}}_t - \rho \sigma_2 t, \end{gathered}

for 0≤t≤T0 \le t \le T, still with correlation ρ\rho. For a claim XX paid at TT with EQ[∣X∣]<∞\mathbb{E}^{\mathbb{Q}}[\lvert X \rvert] < \infty, Bayes’ formula with the density process Zt=e−rtNt/N0Z_t = e^{-rt} N_t/N_0 gives

Vt=e−r(T−t) EQ[X∣Ft]=Nt EQN[X/NT∣Ft].\begin{aligned} V_t &= e^{-r(T - t)}\, \mathbb{E}^{\mathbb{Q}}[X \mid \mathcal{F}_t] \\[0.4em] &= N_t \, \mathbb{E}^{\mathbb{Q}^{N}}\bigl[X/N_T \mid \mathcal{F}_t\bigr]. \end{aligned}

Substituting dWt1,Q=dWt1,N+ρσ2 dtdW^{1,\mathbb{Q}}_t = dW^{1,N}_t + \rho \sigma_2 \, dt and dWt2,Q=dWt2,N+σ2 dtdW^{2,\mathbb{Q}}_t = dW^{2,N}_t + \sigma_2 \, dt into the ratio dynamics adds ρσ1σ2−σ22\rho \sigma_1 \sigma_2 - \sigma_2^2 to the drift, so

dRtRt=(q2−q1) dt+σ1 dWt1,N−σ2 dWt2,N\begin{aligned} \frac{dR_t}{R_t} &= (q_2 - q_1)\, dt \\[0.4em] &\quad + \sigma_1 \, dW^{1,N}_t - \sigma_2 \, dW^{2,N}_t \end{aligned}

under QN\mathbb{Q}^{N}, with the same effective volatility. The general change-of-numeraire theorem is developed later, in Numeraires and forward measures, and this card uses only the explicit Gaussian density above.

In plain words Measuring every value in units of the second asset, with its dividends reinvested, reduces the exchange problem to a problem about the ratio alone. The new measure gives more weight to paths on which asset 2 does well, exactly enough that traded values expressed in these units become martingales. This shifts the drifts of the Brownian motions and leaves volatilities and correlation unchanged.

Example With the constructed values σ1=0.2\sigma_1 = 0.2, σ2=0.3\sigma_2 = 0.3, ρ=0.5\rho = 0.5 and zero dividends, the shifts are Wt2,N=Wt2,Q−0.3tW^{2,N}_t = W^{2,\mathbb{Q}}_t - 0.3t and Wt1,N=Wt1,Q−0.15tW^{1,N}_t = W^{1,\mathbb{Q}}_t - 0.15t. The ratio drift is 0.09−0.03=0.060.09 - 0.03 = 0.06 under Q\mathbb{Q} and 00 under QN\mathbb{Q}^{N}. With S01=S02=100S^1_0 = S^2_0 = 100 and T=1T = 1, the probability that ST1>ST2S^1_T > S^2_T is about 0.44740.4474 under QN\mathbb{Q}^{N} and about 0.53760.5376 under Q\mathbb{Q}. Both are pricing weights for different numeraires, and neither is a forecast under P\mathbb{P}.

Exchange option

Definition An exchange option gives its holder the right at TT to receive one unit of asset 1 by delivering one unit of asset 2, so it pays (ST1−ST2)+(S^1_T - S^2_T)^{+}, with no cash strike. In the correlated model with τ=T−t>0\tau = T - t > 0 and σeff>0\sigma_{\text{eff}} > 0, its value at tt is

Vt=St1e−q1τ Φ(d1)−St2e−q2τ Φ(d2),\begin{aligned} V_t &= S^1_t e^{-q_1 \tau}\, \Phi(d_1) \\[0.4em] &\quad - S^2_t e^{-q_2 \tau}\, \Phi(d_2), \end{aligned}d1=ln⁡(St1/St2)+b τσeffτ,d2=d1−σeffτ,\begin{gathered} d_1 = \frac{\ln(S^1_t/S^2_t) + b\,\tau}{\sigma_{\text{eff}}\sqrt{\tau}}, \\[0.4em] d_2 = d_1 - \sigma_{\text{eff}}\sqrt{\tau}, \end{gathered}

with the carry term b=q2−q1+σeff2/2b = q_2 - q_1 + \sigma_{\text{eff}}^2/2. This is Margrabe’s formula. To derive it, note that NT=eq2TST2N_T = e^{q_2 T} S^2_T, so the payoff in units of the numeraire is (ST1−ST2)+/NT=e−q2T(RT−1)+(S^1_T - S^2_T)^{+}/N_T = e^{-q_2 T}(R_T - 1)^{+}. Writing EtN\mathbb{E}^{N}_t for the conditional expectation under QN\mathbb{Q}^{N} given Ft\mathcal{F}_t, the pricing identity gives

Vt=Nte−q2T EtN[(RT−1)+]=St2e−q2τ EtN[(RT−1)+].\begin{aligned} V_t &= N_t e^{-q_2 T}\, \mathbb{E}^{N}_t\bigl[(R_T - 1)^{+}\bigr] \\[0.4em] &= S^2_t e^{-q_2 \tau}\, \mathbb{E}^{N}_t\bigl[(R_T - 1)^{+}\bigr]. \end{aligned}

Given Ft\mathcal{F}_t, the variable ln⁡RT\ln R_T is normal under QN\mathbb{Q}^{N} with mean ln⁡Rt+(q2−q1−σeff2/2)τ\ln R_t + (q_2 - q_1 - \sigma_{\text{eff}}^2/2)\tau and variance σeff2τ\sigma_{\text{eff}}^2 \tau, and the lognormal call computation with unit strike gives the formula. The rate rr does not appear, because both legs are traded assets and no cash is paid. Here Φ(d2)=QN(ST1>ST2∣Ft)\Phi(d_2) = \mathbb{Q}^{N}(S^1_T > S^2_T \mid \mathcal{F}_t). If σeff=0\sigma_{\text{eff}} = 0, the ratio is deterministic and Vt=(St1e−q1τ−St2e−q2τ)+V_t = \bigl(S^1_t e^{-q_1 \tau} - S^2_t e^{-q_2 \tau}\bigr)^{+}, and at τ=0\tau = 0 the value is the payoff. Since (x)+−(−x)+=x(x)^{+} - (-x)^{+} = x, the option to receive asset 1 for asset 2 and the option to receive asset 2 for asset 1 differ in value by St1e−q1τ−St2e−q2τS^1_t e^{-q_1 \tau} - S^2_t e^{-q_2 \tau}.

In plain words The holder gets asset 1 and gives up asset 2, so the formula looks like Black-Scholes with the second asset in the place of the discounted strike and σeff\sigma_{\text{eff}} in the place of σ\sigma. The interest rate drops out because nothing is paid in cash, while the dividend yields stay, because the holder of the option receives neither asset’s dividends before TT.

Example For the constructed data St1=St2=100S^1_t = S^2_t = 100, zero dividends, σ1=0.2\sigma_1 = 0.2, σ2=0.3\sigma_2 = 0.3, ρ=0.5\rho = 0.5 and τ=1\tau = 1, the effective volatility is 0.07≈0.264575\sqrt{0.07} \approx 0.264575. Equal prices and yields give d1=σeff/2≈0.132288d_1 = \sigma_{\text{eff}}/2 \approx 0.132288 and d2=−d1d_2 = -d_1, so

Vt=100(Φ(0.132288)−Φ(−0.132288))≈10.5243.\begin{aligned} V_t &= 100\bigl(\Phi(0.132288) \\[0.4em] &\qquad - \Phi(-0.132288)\bigr) \\[0.4em] &\approx 10.5243 . \end{aligned}

Taking asset 2 to be a zero-coupon bond that pays KK at TT, with St2=Ke−rτS^2_t = K e^{-r\tau}, σ2=0\sigma_2 = 0 and q2=0q_2 = 0, gives σeff=σ1\sigma_{\text{eff}} = \sigma_1 and returns the Black-Scholes call formula, with rr entering through the bond price. For instance St1=100S^1_t = 100, K=95K = 95, r=0.03r = 0.03, q1=0.01q_1 = 0.01, σ1=0.25\sigma_1 = 0.25 and τ=0.75\tau = 0.75 give about 11.848511.8485 in both formulas.

Basket option

Definition For assets S1,…,SdS^1, \dots, S^d in the correlated model with correlation matrix (ρjk)(\rho_{jk}) and fixed quantities wjw_j, the basket is Bt=∑jwjStjB_t = \sum_j w_j S^j_t, and a basket call with strike K>0K > 0 pays (BT−K)+(B_T - K)^{+} at TT. When Bt≠0B_t \ne 0, the value shares are aj(t)=wjStj/Bta_j(t) = w_j S^j_t / B_t and the instantaneous proportional variance of the basket is

σB2(t)=∑j,kajakρjkσjσk,\sigma_B^2(t) = \sum_{j,k} a_j a_k \rho_{jk} \sigma_j \sigma_k ,

so that d[B]t=σB2(t)Bt2 dtd[B]_t = \sigma_B^2(t) B_t^2 \, dt, with every covariance term included. This number is exact at the current composition, but the shares aj(t)a_j(t) change as prices move, so σB2(t)\sigma_B^2(t) is random and BB is in general not a geometric Brownian motion. A sum of lognormal prices is in general not lognormal, so BTB_T has no lognormal law to integrate against. Dividing by one asset does not help either, since for two assets (w1ST1+w2ST2−K)+/ST2=(w1RT+w2−K/ST2)+(w_1 S^1_T + w_2 S^2_T - K)^{+}/S^2_T = (w_1 R_T + w_2 - K/S^2_T)^{+} still depends on ST2S^2_T.

In plain words A basket adds prices instead of dividing them, so there is no single ratio whose law settles the price. Its current volatility combines all variances and covariances with today’s value shares, and that number drifts as the mix of the basket changes.

Example Take two constructed assets at 100100 with w1=w2=0.5w_1 = w_2 = 0.5, σ1=σ2=0.2\sigma_1 = \sigma_2 = 0.2, ρ=0.5\rho = 0.5, zero dividends and r=0r = 0. Both shares are 0.50.5, so

σB2=0.25⋅0.04+0.25⋅0.04+2⋅0.25⋅0.5⋅0.04=0.03,\begin{aligned} \sigma_B^2 &= 0.25 \cdot 0.04 \\[0.4em] &\quad + 0.25 \cdot 0.04 \\[0.4em] &\quad + 2 \cdot 0.25 \cdot 0.5 \cdot 0.04 \\[0.4em] &= 0.03, \end{aligned}

and σB≈0.1732\sigma_B \approx 0.1732, below 0.20.2 because the assets are not perfectly correlated. Conditioning on the driver of the first asset and integrating numerically gives the one-year call with strike 100100 as about 6.90696.9069. The Black-Scholes formula with σ=0.1732\sigma = 0.1732 gives about 6.90136.9013, close but not equal.

Lognormal moment matching

Definition Lognormal moment matching approximates BTB_T by a lognormal variable L=exp⁡(μL+sLZ)L = \exp(\mu_L + s_L Z) with Z∼N(0,1)Z \sim N(0, 1) that has the same first two Q\mathbb{Q}-moments. With forward prices Fj=S0je(r−qj)TF^j = S^j_0 e^{(r - q_j)T} these moments are exact,

M1=EQ[BT]=∑jwjFj,M2=EQ[BT2]=∑j,kwjwkFjFkeρjkσjσkT.\begin{aligned} M_1 &= \mathbb{E}^{\mathbb{Q}}[B_T] = \sum_j w_j F^j, \\[0.4em] M_2 &= \mathbb{E}^{\mathbb{Q}}[B_T^2] \\[0.4em] &= \sum_{j,k} w_j w_k F^j F^k e^{\rho_{jk} \sigma_j \sigma_k T}. \end{aligned}

The matching needs M1>0M_1 > 0 and M2>M12M_2 > M_1^2, and then sL2=ln⁡(M2/M12)s_L^2 = \ln(M_2/M_1^2) and μL=ln⁡M1−sL2/2\mu_L = \ln M_1 - s_L^2/2. The approximate call price is

CB≈e−rT[M1Φ(d1)−KΦ(d2)],\begin{aligned} C_B &\approx e^{-rT}\bigl[M_1 \Phi(d_1) \\[0.4em] &\qquad - K \Phi(d_2)\bigr], \end{aligned}d1=ln⁡(M1/K)+sL2/2sL,d2=d1−sL.\begin{gathered} d_1 = \frac{\ln(M_1/K) + s_L^2/2}{s_L}, \\[0.4em] d_2 = d_1 - s_L . \end{gathered}

This is an approximation, not a statement about the law of BTB_T. Two moments do not fix the third moment or the tails, the method carries no general error bound, and its accuracy has to be checked numerically, for example against Monte Carlo with its standard error. With negative weights, as in a spread, the basket can be zero or negative with positive probability, which no lognormal variable can match, and the mean M1M_1 can be zero or negative.

In plain words Replace the awkward sum by the lognormal variable with the same mean and variance, and price that variable with a Black-Scholes-type formula. The two agree on the first two moments and in general on nothing more, so the method can be very accurate when the basket is close to lognormal and poor when its components behave very differently.

Example For the basket of the previous card, M1=100M_1 = 100 and M2=104(0.5e0.04+0.5e0.02)≈10305.06M_2 = 10^4(0.5 e^{0.04} + 0.5 e^{0.02}) \approx 10305.06, so sL2≈0.030050s_L^2 \approx 0.030050 and sL≈0.17335s_L \approx 0.17335. The approximate call is about 6.90706.9070, against 6.90696.9069 from numerical integration. The third moments already differ, about 1094339.51094339.5 for BTB_T against 1094338.41094338.4 for the matched lognormal. For a constructed basket with σ1=0.1\sigma_1 = 0.1, σ2=0.6\sigma_2 = 0.6 and ρ=−0.3\rho = -0.3 and the same other data, sL≈0.3115s_L \approx 0.3115, and the approximation gives about 12.378712.3787 at strike 100100 and 1.71731.7173 at strike 150150, while numerical integration gives 11.373911.3739 and 2.35602.3560. The approximation overprices the first call by about 9%9\% and underprices the second by about 27%27\%.

Check yourself

  1. Two assets have volatilities 0.250.25 and 0.150.15. Find the effective volatility of their ratio for the correlations 0.60.6, 11 and −1-1. A colleague says that at correlation 11 the ratio carries no risk. Is that right?
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The effective variance is

σeff2=0.0625+0.0225−0.075ρ=0.085−0.075ρ.\begin{aligned} \sigma_{\text{eff}}^2 &= 0.0625 + 0.0225 \\[0.4em] &\quad - 0.075\rho \\[0.4em] &= 0.085 - 0.075\rho . \end{aligned}

At ρ=0.6\rho = 0.6 it is 0.040.04, so σeff=0.2\sigma_{\text{eff}} = 0.2. At ρ=1\rho = 1 it is 0.010.01, so σeff=0.1\sigma_{\text{eff}} = 0.1, and at ρ=−1\rho = -1 it is 0.160.16, so σeff=0.4\sigma_{\text{eff}} = 0.4. The colleague is wrong. At correlation 11 both prices respond to the same shock, but with different sizes, so the ratio still moves with volatility ∣σ1−σ2∣=0.1\lvert \sigma_1 - \sigma_2 \rvert = 0.1. The ratio is riskless only when the correlation is 11 and the volatilities are also equal.

  1. With τ=0.5\tau = 0.5, St1=105S^1_t = 105, St2=100S^2_t = 100, q1=0.02q_1 = 0.02, q2=0q_2 = 0 and σeff=0.2\sigma_{\text{eff}} = 0.2, price the option to receive asset 1 for asset 2. Then find the price of the option to receive asset 2 for asset 1 without a second use of the formula. Does either answer depend on rr?
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The carry term gives bτ=(q2−q1+σeff2/2)τ=(−0.02+0.02)⋅0.5=0b\tau = (q_2 - q_1 + \sigma_{\text{eff}}^2/2)\tau = (-0.02 + 0.02) \cdot 0.5 = 0, so

d1=ln⁡1.050.20.5≈0.3450,d2≈0.3450−0.1414=0.2036.\begin{aligned} d_1 &= \frac{\ln 1.05}{0.2\sqrt{0.5}} \approx 0.3450, \\[0.4em] d_2 &\approx 0.3450 - 0.1414 \\[0.4em] &= 0.2036 . \end{aligned}

With Φ(d1)≈0.6350\Phi(d_1) \approx 0.6350 and Φ(d2)≈0.5807\Phi(d_2) \approx 0.5807, the value is

V=105e−0.01Φ(d1)−100 Φ(d2)≈7.9408.\begin{aligned} V &= 105 e^{-0.01} \Phi(d_1) \\[0.4em] &\quad - 100\, \Phi(d_2) \\[0.4em] &\approx 7.9408 . \end{aligned}

The two payoffs differ by ST1−ST2S^1_T - S^2_T, so the reverse option is worth 7.9408−(105e−0.01−100)≈7.9408−3.9552=3.98567.9408 - (105 e^{-0.01} - 100) \approx 7.9408 - 3.9552 = 3.9856. The formula with the roles of the assets swapped gives the same number. Neither price depends on rr, because both legs are traded assets and no cash changes hands.

  1. A desk wants a call on the spread ST1−ST2S^1_T - S^2_T with strike 55, using the data of the constructed exchange example, and plans to approximate the spread by a lognormal variable with matched moments. Why does this plan fail, and which spread contract does have an exact price in this model?
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The spread is a basket with weights 11 and −1-1. With zero rates and dividends its mean is M1=100−100=0M_1 = 100 - 100 = 0, so ln⁡(M2/M12)\ln(M_2/M_1^2) is undefined. More basically, the spread is negative with positive probability. Under Q\mathbb{Q}, with R0=1R_0 = 1 and T=1T = 1, the variable ln⁡RT\ln R_T is normal with mean q2−q1+(σ22−σ12)/2=0.025q_2 - q_1 + (\sigma_2^2 - \sigma_1^2)/2 = 0.025 and variance 0.070.07. Since 0.025/0.07≈0.09450.025/\sqrt{0.07} \approx 0.0945,

Q(ST1<ST2)=Φ(−0.0945)≈0.4624,\begin{aligned} &\mathbb{Q}(S^1_T < S^2_T) \\[0.4em] &\quad = \Phi(-0.0945) \approx 0.4624, \end{aligned}

while a lognormal variable is always positive. The contract with strike 00 is the exchange option, which the ratio and the asset-2 numeraire price exactly at about 10.524310.5243. With strike 55 the payoff divided by ST2S^2_T is (RT−1−5/ST2)+(R_T - 1 - 5/S^2_T)^{+}, which still depends on ST2S^2_T, so the law of the ratio alone does not price it, and it needs a numerical method or an approximation whose error is checked.

How this lesson connects