Pricing an option with strike price 0

Consider a fictional world in which infinitely divisible containers of low-level nuclear waste are traded on a commodities market. On some days the oil price is high and one can turn a profit by purifying the waste and using it as fuel. On other days oil is so cheap that the container is just a liability. Suppose that at the moment the price of a container is 0. What is the correct price of an option with strike price 0 expiring 2t days from now if on each day the price of a container changes by either +1 (uptick) or -1 (downtick)?

The purpose of this post is to value this option by a dynamic hedging strategy. I know this is a standard result, and my solution is surely not the most efficient. But as usual, I’m very happy to spend a day messing around with binomial coefficients to save an hour reading the literature.

Preliminaries

From now on time means time-to-expiry. We take the point of view of the option writer and construct a hedge by a mixed portfolio of stock (i.e. a short or long position in a fraction of a waste container) and cash. Let S denote the stock price at time r and let (a_r(S), c_r(S)) be the hedge of stock and cash held at time r. If S \ge r then the option is certain to be exercised, so the hedging portfolio is (1,0). In this case, say that the option is deterministic. A small variation on the standard put-call-parity argument proves the following lemma, which allows us to reduce to the case when S \ge 0.

Lemma 1

c_r(-S) = c_r(S) for all r and S and a_r(-S) = 1-a_r(S) for all r and S \not= 0.

Proof. Consider a portfolio consisting of a short call option, a long put option (both with strike price 0) and one unit of stock. The pay offs are shown in the graph below.

By no-arbitrage the option prices are the same. A long put option behaves exactly the same as a long call option on an stock with price -S. Hence the portfolio

-(a_r(S), c_r(S)) + (a_r(-S), c_r(-S)) + (1,0)

is value 0. The lemma follows.\quad\Box

When S=0 we can hold any amount of the stock without cost. We shall therefore assume that a_r(0) = \frac{1}{2} for all r, so that the previous lemma also holds for a_r(0).

The diagram below shows some values of (a_r(S),c_r(S)), computed iteratively. For example (a_2(0),c_2(0)) = (1/2,1/2) since on an uptick the option is deterministic and we must top-up our stock holding by buying 1/2 stock at price 1.

One interesting feature is that when S=1 the cash holding equals the value of the option on a downtick. (Since on a downtick it is impossible to vary our cash holding, because the only non-cash asset in our model is then costless.) Also we see that, for fixed S, our cash holding increases and the stock holding decreases with the time to expiry.

Stock changes

Let v_r(S) be the value of the option at time r. Clearly v_r(S) = Sa_r(S) + c_r(S). Let \Delta v_\pm = v_{r-1}(S\pm 1) - v_r(S) and let \Delta a_\pm = a_{r-1}(S\pm 1) - a_r(S). The following lemma can be thought of as a differential equation for the value of the option.

Lemma 2

We have \Delta v_+ = -\Delta v_- = a_r(S).

Proof. Let (a,c) = (a_r(S),c_r(S)). On an uptick our cash holding is worth the same and the stock holding increases in value by a. By assumption (a,c) hedges the option, so a is also the change in value of the option. The proof for a downtick is similar.\quad\Box

The next corollary says that the stock holding after a tick is the average of the stock holdings we might have held before the tick.

Corollary 3

We have \Delta a_+ = -\Delta a_- and so

a_{r+1}(S) = \frac{a_r(S+1) + a_r(S-1)}{2}

for all r and S.

Proof. We hold a = a_r(S) asset at time r. Let a_\pm = a_{r-1}(S\pm 1). By Lemma 2, an uptick followed by a downtick changes the value of the option by a - a_+, whereas a downtick followed by an uptick changes the value of the option by -a+a_-. Hence a-a_+ = -a+a_-. Negating we get \Delta a_+ = -\Delta a_-, as claimed.\quad\Box

Cash changes

Let \Delta c_\pm = c_{r-1}(S\pm 1) - c_r(S).

Lemma 4

We have \Delta c_\pm = -(S\pm 1)\Delta a_\pm .

Proof. This follows from Corollary 3, since on an uptick we buy \Delta a_+ stock at price S+1, and on a downtick we sell -\Delta a_- stock at price S-1.

Stock and cash holdings in a hedging portfolio

It is now most convenient to change units. For 0 \le u \le r/2 let b_r(u) = 1-a_r(r-2u). Thus b_r(u) is the amount of stock we need to buy to increase our holding to 1 unit, if exactly u upticks are needed for the option to become deterministic. Let d_r(u) = c_r(r-2u) be the cash holding at this time. Some values of (b_r(u),d_r(u)) are shown below.

For example (b_6(2),d_6(2)) = (3/16,5/8). Note that an uptick decreases u by 1 and on a downtick u is constant. Thus the second part of Corollary 3 can be restated as

b_{r+1}(u) = \frac{b_r(u-1) + b_r(u)}{2}.

Proposition 5

We have

b_r(u) = \frac{1}{2^{r-1}} \sum_{k=1}^u \binom{r-1}{k-1}.

Proof. We work by induction on r. For the inductive step, we have

\begin{aligned}  b_{r+1}(u) &=   \frac{b_r(u-1) + b_r(u)}{2} \\  &= \frac{1}{2^r} \Bigl(  \sum_{k=1}^{u-1} \binom{r-1}{k-1} + \sum_{k=1}^u \binom{r-1}{k-1} \Bigr) \\  &= \frac{1}{2^r} \Bigl( \sum_{k=1}^{u-1}\Bigl( \binom{r-1}{k-1} + \binom{r-1}{k} \Bigr) + 1 \Bigr) \\  &= \frac{1}{2^r} \Bigl( 1 + \sum_{k=1}^{u-1} \binom{r}{k} \Bigr) \\  &= \frac{1}{2^r} \sum_{k=1}^u \binom{r}{k-1}  \end{aligned}

as required.\quad\Box

In particular, we get

b_r(u) - b_r(u-1) =   \frac{1}{2^{r-1}} \binom{r-1}{u-1}

Using the relation a_r(S) = b_r((r-S)/2), we get

a_r(S) = \frac{1}{2^{r-1}} \sum_{k=1}^{(r-S)/2} \binom{r-1}{k-1}.

The next proposition gives the analogous results for the cash holding.

Proposition 6

We have

d_r(u) = \frac{1}{2^{r-1}} \sum_{k=1}^u \binom{r-1}{k-1}(r-2k+1).

Proof. Suppose at time r the stock price is r-2(u-1). We reach this state from time r+1 and price r-2u-1 by an uptick; if instead there is a downtick at time r+1 we arrive at time r with stock price r-2u. The change in our stock holding is the same in both cases, by Corollary 3, and it is \frac{1}{2^r} \binom{r-1}{u-1} by Proposition 5. Hence

\begin{aligned}  d_r(u)-d_r(u-1) &= \frac{1}{2^r} \binom{r-1}{u-1}\bigl( (r - 2(u-1)) + (r-2u) \bigr) \\  &= \frac{1}{2^{r-1}} \binom{r-1}{u-1} ( r - 2u + 1 )  \end{aligned}

for each u \le r/2. The proposition follows by summing this equation, using d_r(0) = 0 for all r.\quad\Box

Using the relation c_r(S) = d_r((r-S)/2), this becomes

c_r(S) = \frac{1}{2^{r-1}} \sum_{k=1}^{(r-S)/2} \binom{r-1}{k-1} (r-2k+1).

Endgame

Somewhat remarkably the sum simplifies. Let u \ge 1. Then

\begin{aligned}   \frac{1}{2^{r-1}} {}&{} \sum_{k=1}^u \binom{r-1}{k-1}(r-2k+1) \\  &= \frac{1}{2^{r-1}} \sum_{\ell=0}^{u-1} \binom{r-1}{\ell}  (r-2\ell-1) \\  &= \frac{1}{2^{r-1}} \sum_{\ell=0}^{u-1}  \Bigl( \binom{r-1}{\ell}(r-1-\ell) - \binom{r-1}{\ell} \ell  \Bigr)\\  &= \frac{1}{2^{r-1}}\Bigl( \sum_{\ell=0}^{u-1}  \binom{r-2}{\ell} (r-1) -   \sum_{\ell=1}^{r-1} \binom{r-2}{\ell-1} (r-1)  \Bigr) \\  &= \frac{r-1}{2^{r-1}}   \Bigl( \sum_{\ell=0}^{u-1} \binom{r-2}{\ell} -   \sum_{\ell=0}^{r-2} \binom{r-2}{\ell} \Bigr) \\  &= \frac{r-1}{2^{r-1}} \binom{r-2}{u-1} \\  &= \frac{u}{2^{r-1}} \binom{r-1}{u}   \end{aligned}

Here several uses have been made of the identities

\binom{n}{k} k = \binom{n-1}{k-1} n = \binom{n}{k-1} (n-k).

Hence

c_r(S) = \frac{u}{2^{r-1}}\binom{r-1}{u}

where u = (r-S)/2. In particular, taking S=0 and r=2t we get that the value of the option is

c_{2t}(0) = \frac{t}{2^{2t-1}}\binom{2t-1}{t} =  \frac{t}{2^{2t}} \binom{2t}{t}.

Final remarks

As I was reminded by the setter of this exercise (Dr George Wellen, an occasional commenter on this blog), the identities used to simplify the sum have nice bijective proofs using the `team-and-leader’ model. For example, \binom{n}{k}k counts teams of k people who elect their leader from their members, while \binom{n-1}{k-1}n counts teams of k-1 people with a leader imposed on them. (I do many similar proofs in my Combinatorics course, but often wonder if this is wise: clearly I find them much easier and more appealing than many of the students.)

Another expression for c_{2t}(0) is given by valuing the option probabilistically: the probability it ends in the money at price 2(t-k) is \frac{1}{2^{2t}}\binom{2t}{k} hence

c_{2t}(0) = \frac{1}{2^{2t}} \sum_{k=0}^{t-1} \binom{2t}{k} 2(t-k).

This sum is easily simplified to give the same answer as above, either by a similar argument, or by using the identity

\sum_{k=0}^m \binom{n}{k} (n/2 - k) = \frac{m+1}{2} \binom{n}{m+1},

which in turn has a short inductive proof.

Further exercises

This is an aide-memoire of two further interesting exercises I’ve been set, and might get around to one day.

  1. Explain why this value of the option agrees with the ‘risk neutral’ valuation in which we assume, quite possibly knowing that this is incorrect, that an uptick and downtick have equal probability 1/2.
  2. Derive the Black–Scholes formula by passing to a continuous limit.
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