Upper Bounds for Swing Options

So far, we looked at an upper bound for an option with a single exercise. In order to under-stand the extension for multiple exercises we should think of the value contribution of every extra exercise right to be similar to an additional American option value. Intuitively, we could then apply Rogers’ duality on the marginal value of the n-th exercise right of a swing option

∆C_t^∗(n) := C_t^∗(X_t, n) − C_t^∗(X_t, n − 1). (2.26) Meinshausen and Hambly [50] prove that this intuition is correct

∆C₀^∗(n) = inf

π0(n−1) inf

{M_t}_t∈M0

(

M0+ E0

"

max

t∈T\{τn−1,...,τ1}(Z_t− M_t(n))

#)

, (2.27) with T = {0, ..., T } and τk := τ_k(0, n − 1), k = n-1,...,1. Then {M_t(n)}_t is the marginal martingale process for the n-th exercise right. Note the definition of the domain space for the maximization that only considers not yet exercised stages up to n-1. Accordingly, the additionally introduced infimum is defined over all policies with n-1 exercise rights π₀(n − 1) (see also equation 2.1). Equation 2.27 does not only require the knowledge of the stopping times τ_k(0, n − 1), k = 1,...,n-1, but also of all marginal martingales in t = 1,...T and n = 1,..., N. Meinshausen and Hambly derive that analogously to the American option it is again the martingale component M_t^∗(n) of the optimal (marginal) value function ∆C_t^∗(n) that attains the infimum. Furthermore the authors show that this martingale can be derived recursively similar to the case of the American option. They present an extension of the Andersen and Broadie iterative method in order to use again the lower bound approximation M_t(n) and

∆C_t(n) as our best guess for M_t^∗ and ∆C_t^∗(n). This leads to the duality gap ∆D₀(n)

∆C₀(n) := ∆C₀(n) + E0[ max

t∈T\{τn−1,...,τ1}(Z_t− M_t(n))] = ∆C₀(n) + ∆D₀(n), (2.28) and thus to the upper bound value of the total swing option

C0(x₀, N ) =

N

X

n=1

∆C₀(n). (2.29)

The decisive extension of the iteration in equation 2.22 is the introduction of the index k to find the martingales M_t(n). Meinshausen and Hambly define k as the largest natural number with t ≤ τ_k and 0 ≤ k < n. Recall that τ_k(0, n − 1) is the optimal stopping time of the k-th

exercise right for a swing option with n − 1 (!) exercise rights available at t = 0⁴ The definition of the iteration for computing the martingale process in conjunction with the option exercise process l_t is analogous to equation 2.22. It is even identical to equation 2.22 in case of n=1. Now it only applies for the marginal instead of the total figures. As the authors do not elaborate the formulas and algorithm in detail, especially their sub indices are not thoroughly applied and therefore cause confusion, we give a detailed description of the method in the next paragraphs. In context of the current stage t, τ_k(0, n − 1) is the next available stopping time from t onwards in the sequence of optimal stopping times according to π₀(n − 1) (see also equation 2.1). So, we do not look at the local policy π_t(n − 1), but always at the global one π₀(n − 1). Also, the local maximum number of exercise rights n during the marginal martingale calculation should not be confused with the global number of exercise rights N of the entire swing option (see also discussion around Figure 2.4). These martingales will then be used to calculate the expectation in equation 2.27 and henceforth C₀(x₀, N ). Note that M_t(n) is a random variable. Thus M_t(n) varies with every realized path. We therefore rewrite the previous two equations in terms of realizations of the random price variable and the approximation of the lower bound values with the help of our estimates for the continuation value. For l_t−1(X_t−1, k + 1) = 1, this is

where ˜x^j_t are simulated prices with start value xⁱ_t−1 according to equation 2.7. And for l_t−1(X_t−1, k + 1) = 0 this is

4Again we skip the underline for M_t for better readability.

hence z_t−1ⁱ = 0. The two previous equations apply for t = 2, ..., T . For t = 0 and t = 1 we have mⁱ₀(n) = 0 and mⁱ₁(n) = ∆cⁱ₁(n). For t = T we set Y_{T +1} := 0 and consequently we get Y_T(k) = Z_T for k>1. Then, for k >1 almost all terms cancel out and we receive mⁱ_T(n) = mⁱ_{T −1}(n). For k = 0 we have Y_t(0) = 0 for t = 1, ..., T and we receive the same result as for the American option

mⁱ_T(n) :=

(mⁱ_{T −1}(n) + z_Tⁱ − y_Tⁱ(1) l_{T −1}ⁱ = 0

mⁱ_{T −1}(n) + z_Tⁱ −_J^{1 PJ}_j=1Z_T(x^j_T) otherwise. (2.33) Once we have computed the marginal martingales for the n-th exercise right along all hours t, we compute the individual upper bound ∆cⁱ₀(n) and approximate the expectation in equation 2.27 via averaging the individual upper bounds

∆cⁱ₀(n) = max

t∈T\{τ_n−1ⁱ ,...,τ₁ⁱ}

(z_tⁱ− mⁱ_t(n))

∆C₀(n) ≈ 1 I

I

X

i=1

∆cⁱ₀(n).

(2.34)

The domain space for the maximization T shrinks with increasing exercise rights n as all previous stopping times will be extracted first. Recall that the sequence of stopping points⁵ τ₁ⁱ(0, n − 1), ..., τ_n−1ⁱ (0, n − 1) changes with every price scenario i. Let us illustrate the entire calculation with an example. We assume that our swing option has a time horizon of 12

Figure 2.1.: Illustration for the computation of the marginal martingales

stages and a max number of N = 10 exercise rights. We further assume that we already ran a backward iteration and therefore can compute an approximation for all continuation values Yt(n) for all t = 0, .., , 12 and n = 1, .., 10. Now, we want to compute the upper bound for the marginal option value of the 6-th exercise right for the first price trajectory ∆c¹₀(6). Hence, we use our optimal policy of the lower bound calculation to find the first five(!) stopping points for our price scenario 1. This is π₀¹(5) which is presented in Figure 2.1. From now on we skip the index 1 indicating realizations of price path 1 (x := x¹, z := z¹, ...) for readability. Note that the stopping points are always in strict descending order with respect to the remaining exercise rights. Now, we want to present the calculation of the 6th martingale m_t(6) for three different stages t = 3, 10 and 11. At (t = 3) the largest k is still the max number of 5 exercise rights like for any hour t ≤ 3. As l₂(6) = 0 we compute

m₃(6) = m₂(6) + ∆c₃(6) − ∆y₃(6)

= m₂(6) + max [z₃+ y₄(5), y₄(6)] − max [z₃+ y₄(4), y₄(5)] − y₃(6) + y₃(5). (2.35)

5We declare a stopping point τ_kⁱ(t, n) as the realization of a stopping time τk(t, n) for a specific price path i.

At (t = 10) the largest available number of exercises is 1 with τ₁= 10. This time the exercise

where ˜x^j₁₀ are simulated prices with start value xⁱ₉ according to equation 2.7. At (t = 11) all options are already exercised (k=0) and therefore the martingale value of the first (k+1=1) exercise right will be computed. We have to take into account that the first exercise right was executed at the previous hour l₁₀(1) = 1

m₁₁(6) = m₁₀(6) + ∆c₁₁(1) − E [∆C11(X₁₁, 1)|x₁₀]

where ˜x^j₁₁ are simulated prices with start value xⁱ₁₀ according to equation 2.7. Likewise we compute m_t(6) for all other t. Next, we calculate the upper marginal value function ∆c0(6).

For n = 6 we need to ignore the stopping points τ₁ to τ₅ and can only look for the maximum at the states 0,1, 2, 4, 6, 7, 11 and 12

∆c₀(6) = max

t∈{0,1,2,4,6,7,11,12}

(z_t− m_t(6)). (2.38)

The same way we proceed with all other scenarios 2,...,I. Then we can appxorimate the marginal upper option value according to equation 2.34. Likewise, we proceed with all other exercise rights n = 1,...,12. Again, recall that for every price trajectory and intermediate number of exercise rights n there will be a separate sequence of stopping points and hence a new Figure 2.1. Let us summarize the algorithm:

1. Run a backward iteration to determine the coefficients for all approximated continuation values α_r,n,t according to equation 2.12.

2. Generate I new price paths.

3. For each exercise right n = 1,..,N:

a) For each price path xⁱ_t, i = 1, ..., I:

i. Find the stopping points π^∗,i₀ (n − 1) = {τ_n−1ⁱ (0, n − 1), ..., τ₁ⁱ(0, n − 1)} for all (n-1) exercise rights in a forward iteration (set π^∗,i₀ (0) = {}).

ii. Set mⁱ₁(n) = ∆cⁱ₁(n) and mⁱ₀(n) = 0.

iii. For each stage t = 2, ..., T :

A. Find in π₀^∗,i(n − 1) the highest exercise right k whose stopping point τ_kⁱ(0, n − 1) lies within the remaining delivery period from t to T (set k = 1 for π₀^∗,i(0)).

B. Check whether the (k+1)-th option was exercised at the previous stage (lⁱ_t−1(k + 1) = 1).

C. if it was, then generate another set of J prices ˜x^j_t starting from xⁱ_t−1 and calculate mⁱ(n) according to equation 2.31

D. if it was not (lⁱ_t−1(k + 1) = 0), then calculate mⁱ(n) according to equation 2.32 without any extra loop

iv. Compute the upper value functions ∆cⁱ₀(n) according to equation 2.34, do not forget to skip all stopping points from π₀^∗,i(n − 1).

b) Average all ∆cⁱ₀(n) to retrieve an estimate for the marginal upper option values

∆C₀(n).

4. Sum up all marginal option values for the total upper option value C₀(x₀, N ).