Gerry

I’ve not been able to contact Ian about getting out a bit of code for this sequential option you spoke of so I’ve knocked up the following. Here are the assumptions I’ve made.

1 Pricing the option

The time interval is [0, T] and there are intermediate times, 0 < T_{1} < T_{2} < T_{3} < T, at each of the times, 0, T_{1}, T_{2}, T_{3}, one has to make a payment, K_{0}, K_{1}, K_{2}, K_{3}, (respectively) in order to stay in the game for the ensuing time period. One has the option to bale out at times T_{1}, T_{2}, T_{3}. I’ve not considered the case of a decision at time zero, but one could. Finally I’ve assumed that one gets to purchase the underlying for an amount K_{4} at time T. At times T_{1} to T_{3} one can decide to carry on or not. This decision must be made upon some basis. I decided that the rational way to do this was to assume that the decision would be made upon the current value of the underlying; if it exceeds a certain level one goes on, and if not one bales out. This stock level can be arrived at in zillions of different ways. For example, it might be set by the requirement that, the probability that the final value of the underlying exceeds the costs incurred in purchasing it, is always greater than one half. Accordingly I’ve called these levels L_{1}, L_{2}, L_{3}.

If you think about how the space splits up into various sets depending on whether the option is continued or abandoned then one can write down the payoff in a straightforward manner. Some notation follows. Let r denote the risk-free rate of interest. Let E_{i} = {S_{T_{i}} ≤ L_{i}}, 1 ≤ i ≤ 3, and for any set E, let E′ denote the complement of E. The payoff of the option can be written down directly

(S_{T} − K_{4})I_{E′_{1}}I_{E′_{2}}I_{E′_{3}}I_{{S_{T} >K_{4}}} −
K_{0}e^{rT} I_{E_{1}} −
(K_{0}e^{rT} + K_{1}e^{r(T −T_{1})})I_{E′_{1}}I_{E_{2}} −
(K_{0}e^{rT} + K_{1}e^{r(T −T_{1})} + e^{r(T −T_{2})}K_{2})I_{E′_{1}}I_{E′_{2}}I_{E_{3}} −

(K_{0}e^{rT} + K_{1}e^{r(T −T_{1})} + e^{r(T −T_{2})}K_{2} + K_{3}e^{r(T −T_{3})})I_{E′_{1}}I_{E′_{2}}I_{E′_{3}}I_{{S_{T} ≤K_{4}}}.

The price of the option is given by “risk-neutral” pricing. Let Q be the riskneutral measure. The difficulty is, of course, evaluating the expectations of the terms above once there are the indicators of more than one set involved. However, we can write down many of the expectations in terms of joint densities.

1.

E^{Q}(I_{E_{1}}) = Q({S_{T_{1}} ≤ L_{1}})
= Q(\{S_{0}e^{σW_{T_{1}}+(r− \frac{σ^{2}}{2} )T_{1}} ≤ L_{1}\})
= Q(\{W_{T_{1}} ≤ \frac{log(\frac{L_{1}}{S_{0}} ) − (r − \frac{σ^{2}}{2} )T_{1}}{σ} \})
= N(\frac{log(\frac{L_{1}}{S_{0}} ) − (r − \frac{σ^{2}}{2} )T_{1}}{σ\sqrt{T_{1}}} )

2.

E^{Q}(I_{E′_{1}}I_{E_{2}}) = Q({S_{T_{1}} > L_{1}, S_{T_{2}} ≤ L_{2}})
= Q(\{W_{T_{1}} > \frac{log(\frac{L_{1}}{S_{0}} ) − (r − \frac{σ^{2}}{2} )T_{1}}{σ} , W_{T_{2}} ≤ \frac{log(\frac{L_{2}}{S_{0}} ) − (r − \frac{σ^{2}}{2} )T_{2}}{σ} \})

We need to contract the notation: set

c_{i} = \frac{log(\frac{L_{i}}{S_{0}} ) − (r − \frac{σ^{2}}{2} )T_{i}}{σ}

then we can continue as follows

E^{Q}(I_{E′_{1}}I_{E_{2}}) = Q({W_{T_{1}} > c_{1}, W_{T_{2}} ≤ c_{2}})
= Q({W_{T_{2}} ≤ c_{2}}) − Q({W_{T_{1}} ≤ c_{1}, W_{T_{2}} ≤ c_{2}})
= N(\frac{c_{1}}{\sqrt{T_{1}}} ) − ∫_{−∞}^{c_{1}} ∫_{−∞}^{c_{2}} \frac{1}{\sqrt{4πT_{1}(T_{2} − T_{1})}}exp^{− \frac{x^{2}_{1}}{2T_{1}}} exp^{− \frac{(x_{2}−x_{1})^{2}}{2(T_{2}−T_{1})}} dx_{2}dx_{1}.

3. It is becoming clear that because of the particular form of the joint density function for (W_{t_{1}}, W_{t_{2}}, …, W_{t_{n}}) when (W_{t}) is a Brownian motion that we will be served well by specifying that T_{0} = 0, T_{4} = T and L_{4} = K_{4} so that E_{4} = {S_{T} ≤ K_{4}} and c_{4} means the obvious thing. So, letting juxtapositioning represent intersection of events, and writing x_{0} = 0 and t_{0} = 0,

Q(E′_{1}E′_{2}E_{3}) = ∫_{c_{1}}^{∞} ∫_{c_{2}}^{∞} ∫_{−∞}^{c_{3}} ∏_{i=1}^{3} \frac{1}{\sqrt{2π(T_{i} − T_{i−1})}}exp\{−\frac{(x_{i} − x_{i−1})^{2}}{2(T_{i} − T_{i−1})}\}dx_{3}..dx_{1}

and we can write

4.

Q(E′_{1}E′_{2}E′_{3}E_{4}) = ∫_{c_{1}}^{∞} ∫_{c_{2}}^{∞} ∫_{c_{3}}^{∞} ∫_{−∞}^{c_{4}} ∏_{i=1}^{4} \frac{1}{\sqrt{2π(T_{i} − T_{i−1})}}exp\{−\frac{(x_{i} − x_{i−1})^{2}}{2(T_{i} − T_{i−1})}\}dx_{4}..dx_{1}

This takes care of all the terms which arise from abandonment of the option before time T or from one declining to exercise the option at time T. The remaining term we must evaluate is

E^{Q}((S_{T} − K_{4})I_{E′_{1}}I_{E′_{2}}I_{E′_{3}}I_{E′_{4}}).

Naturally we split this into the part containing S_{T} and the part containing K. The term

Q(E′_{1}...E′_{4}) = E^{Q}(I_{E′_{1}}I_{E′_{2}}I_{E′_{3}}I_{E′_{4}})

is taken care of by the appropriate modification to the last multiple integral above. The difficult term is the one involving S_{T}. Now, multiplying S_{T} is the discount factor, e^{−rT}. While S_{T} looks like a log-normal with mean rate of return equal to r and volatility σ, the discounted stock looks like a lognormal with mean rate of return equal to zero and volatility σ. Based on this, here is a suggestion; Let R be the probability measure given by

E ⟶ ∫_{E} e^{σW_{T} − \frac{1}{2} σ^{2}T} dQ.

Under this measure (W_{t} − σt) is a standard R-Brownian Motion. Now, for example,

E′_{i} = {S_{T_{i}} > L_{i}}
= \{W_{T_{i}} > \frac{log(\frac{L_{i}}{S_{0}} ) − (r − \frac{σ^{2}}{2} )T_{i}}{σ} \}
= \{W_{T_{i}} − σT_{i} > \frac{log(\frac{L_{i}}{S_{0}} ) − (r − \frac{σ^{2}}{2} )T_{i}}{σ} − σT_{i}\}
= \{W_{T_{i}} − σT_{i} > \frac{log(\frac{L_{i}}{S_{0}} ) − (r + \frac{σ^{2}}{2} )T_{i}}{σ} \}.

So E′_{i} is simply the event that the R-Brownian motion W^{R} = (W_{t} − σt) is

greater than the number d_{i} = ^{\frac{log( \frac{L_{i}}{S_{0}} )−(r+ \frac{σ^{2}}{2} )T_{i}}{σ}} at time T_{i}. And the term

E^{Q}(S_{T} I_{E′_{1}}I_{E′_{2}}I_{E′_{3}}I_{E′_{4}}) = S_{0}R({W_{T_{i}}^{R} > d_{1}}..{W_{T_{4}}^{R} > d_{4}}},

which is given by the appropriate modification to the last multiple integral above.

We can begin to collect the various parts of our pricing formula together. We are simply using risk-neutral valuation; the price is the discounted riskneutral expectation of the payoff. We will need to introduce some notation for the terms which have appeared above: Write

N^{1}(t, x) = Q{W_{t} ≤ x}.

For n ≥ 1, a vector of time points, t = (t_{1}, …, t_{n}) and a vector of real values, x = (x_{1}, …, x_{n}), write

N^{n}(t, x) = Q{W_{t_{1}} ≤ x_{1}, …, W_{t_{n}} ≤ x_{n}}.

Now, if any of the numbers, x_{1}, …, x_{n} is written with a prime, for example, x′_{n}, so that x = (x_{1}, …, x′_{n}) then

N^{n}(t, x) = Q{W_{t_{1}} ≤ x_{1}, …, W_{t_{n}} > x_{n}}.

Our time ‘vectors’ will always be one of T^{1} = (T_{1}) ≡ T_{1}, T^{2} = (T_{1}, T_{2}), T^{3} = (T_{1}, T_{2}, T_{3}) and finally T^{4} = (T_{1}, T_{2}, T_{3}, T_{4}). Refering back to the expression for the payoff at the start of this note and multiplying by the discount factor e^{−rT} we get

S_{0}N^{4}(T^{4}, (d′_{1}, d′_{2}, d′_{3}, d′_{4})) −
K_{0}N^{1}(T^{1}, c_{1}) −
(K_{0} + K_{1}e^{−rT_{1}})N^{2}(T^{2}, (c′_{1}, c_{2})) −
(K_{0} + K_{1}e^{−rT_{1}} + e^{−rT_{2}}K_{2})N^{3}(T^{3}, (c′_{1}, c′_{2}, c_{3})) −
(K_{0} + K_{1}e^{−rT_{1}} + e^{−rT_{2}}K_{2} + K_{3}e^{−rT_{3}})N^{4}(T^{4}, (c′_{1}, c′_{2}, c′_{3}, c_{4})) −
e^{−rT} K_{4}N^{4}(T^{4}, (c′_{1}, c′_{2}, c′_{3}, c′_{4})).

So this is the price of the option (so long as I’ve not made any mistakes). To compute this number for the various possible values of the parameters involved is a job for Mathematica or MatLab. I could write a program to do it but it would be a (very long) waste of my time. Someone, somewhere, must have coded those multivariate joint distribution functions. I bet Nicos knows who and where.

2 Pricing a basket option

As a preliminary let us consider the case of two underlyings driven by independent Brownian Motions. For simplicity assume that at times T_{1}, … T_{3} we can elect to abandon the entire basket or continue. Continuing comes at the (single) cost of K_{0}, .., K_{3} and is decided on the basis of S^{i} exceeding level L_{j}^{i} at time T_{j}. One might call this an all or nothing basket option. Another possibility is to allow assets to fall out of the basket at times T_{1}, etc. This is very complicated, even for just two assets (but I think I can price it!) so I’ll continue the discussion for an “all or nothing” basket. To give an idea of the kinds of terms that will emerge consider abandonment at time T_{1}. Continuation occurs only if S_{T_{1}}^{1} > L_{1}^{1} and S_{T_{1}}^{2} > L_{1}^{2}, so abandonment occurs upon the complement of the intersection of these sets;

Ω \ {S_{T_{1}}^{1} > L_{1}^{1}} ∪ Ω \ {S_{T_{1}}^{2} > L_{1}^{2}}

For the case of independent driving noise for the assets in the basket we can evaluate this probability (it’s just one minus the probability of the original intersection). But if we allow even a simple time independent correlation of the driving noises we are back to multivariate distribution functions. As one looks at the possibilities of abandonment and continuation at later time stages these kinds of problems multiply. On a more general note: The usual way to deal with a basket option is to use an approximation based upon replacing the arithmetic average with the geometric average. It might be possible to carry this technique over to the case of the sequential options, I simply don’t know if it works. But I would caution you against assuming that there is an easy outcome because it would involve taking products of terms like the payoff to the single asset sequential option on page 1. The number of terms then generated is not the problem. We would have to have at our disposal the required multivariate distribution functions (coded, because I think a nice closed form might not emerge). This might involve some bespoke numerical programming, unless I take a much closer look I cant say. That would take a lot more than a week!

Chris