Derivatives Foundation puzzles, solved step by step

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  1. 042An option lets you decide in six months whether it becomes a call or a put, both struck at 100 and expiring in one year. The stock is at 100, pays no dividend, and rates are zero. Express it as a portfolio of plain options.Option pricing intuitionHardExotics tradingStructured products

    Try it first

    Which portfolio of plain options replicates the chooser?

    Show the worked solution

    A one-year call struck at 100 plus a six-month put struck at 100. At the decision date you take the larger of the call and the put. Write that as the call plus max(P - C, 0). Put-call parity with zero rates and no dividend says P - C = 100 - S at that date, so the extra piece is max(100 - S, 0) paid at six months: a six-month put. At 20% volatility that is 7.97 + 5.64 = 13.60, against 15.93 for a one-year straddle.

    Why is a chooser worth less than a straddle but more than a single option?

    Booking a restaurant table for a date six weeks away while keeping the right, at four weeks, to switch it to a different restaurant is worth more than a fixed booking but less than holding two bookings to the end. The chooser is the same: it beats any one plain option because you choose with six months of news in hand, but it is cheaper than a straddle because you must give up one leg at six months. A straddle keeps both the call and the put for the full year; the chooser keeps only whichever looks better at the half-way point, and then lives or dies with it. At 20% volatility the one-year straddle is 15.93, a one-year call alone is 7.97, and the chooser sits in between at 13.60. The question is whether you can find that middle number without a model of the choice itself.

    How does put-call parity turn the choice into plain options?

    At six months you hold max(C, P), where C and P are the values then of the call and the put with six months left. Split it: max(C, P) = C + max(P - C, 0). Now use parity. With no dividend and zero rates, a put minus a call with the same strike and expiry is worth the strike minus the stock, so P - C = 100 - S at the decision date. The awkward piece max(P - C, 0) becomes max(100 - S, 0), which is exactly the payoff of a put struck at 100 that expires at six months, so the chooser is a one-year call plus a six-month put. The call is always in the portfolio because you either keep it or, by parity, the put you switch into is that call plus a short forward, and the short forward's value is what the six-month put pays you when you switch.

    The relationship
    Vt=max⁡(Ct,Pt)=Ct+max⁡(Pt−Ct,0)=Ct+max⁡(K−St,0)⇒Chooser0=C(K,T)+P(K,t)V_{t} = \max(C_t, P_t) = C_t + \max(P_t - C_t, 0) = C_t + \max(K - S_t, 0) \quad \Rightarrow \quad \text{Chooser}_0 = C(K, T) + P(K, t)
    tthe decision date, six months
    Tthe final expiry, one year
    C t, P tthe values at the decision date of the call and the put struck at K
    K - S tput minus call by put-call parity, with zero rates and no dividend
    What it says in wordsThe choice is a call you always own plus a put on the decision date that pays the switch.
    A chooser is a long-dated call plus a put that expires on the decision dateChooserat 6 months pick call orput: K 100, expiry 1 year1-year call, K 100always held: if you pickthe call, you keep it6-month put, K 100pays 100 - S at 6 monthsif the put is the better pick=+At 6 months, r = 0: max(C, P) = C + max(P - C, 0), and parity gives P - C = 100 - SValues at 20% volatility, stock 100, rates zero1-year call7.976-month put5.64chooser = sum13.601-year straddle15.93The straddle costs 2.33 more than the chooser: it keeps the put alive six months longer
    The chooser equals a one-year call plus a six-month put, because at the decision date the larger of call and put is the call plus max(100 - S, 0), and at 20% volatility the two pieces add to 13.60, below the 15.93 one-year straddle.

    How do you check the answer, and what changes with rates and dividends?

    Test the two ends of the decision date. If the choice had to be made today, the chooser is just whichever is dearer now, and at the money with zero rates the call and put cost the same, 7.97; the formula gives a call plus a put expiring today at the money, which is worth zero, so it agrees. If the choice is made at one year, you simply take whichever pays, which is a straddle, and the formula gives a call plus a one-year put, which agrees too. A simulation of 40,000 paths to six months, taking the better of call and put on each, gives 13.61 against 13.60. With rates or dividends the same split works, but parity puts a discounted strike on the put: it is struck at K times the discount factor from six months to a year, adjusted for the dividends paid in that window. The limitation to say: the decomposition relies on European options and on parity holding exactly; an American chooser, or one on a stock with a borrow cost, needs a model.

    Where candidates lose it

    The common loss is answering a straddle. A straddle keeps both legs for a year; the chooser forces you to give one up at six months, so it must cost less, and the difference is the put's last six months of life.

    The second is getting the put's expiry wrong: writing a one-year put alongside the call, or a six-month call alongside the put. The switch happens at six months, so the piece that pays for it expires at six months. Write max(C, P) = C + max(P - C, 0) and parity does the rest.

    What the interviewer asks next

    • What is the chooser worth if the decision date moves to nine months?
    • Rates are 6% instead of zero. What strike does the six-month put carry?
    • The stock pays a dividend of 3 at month nine. How does the decomposition change?
    • The call and the put have different strikes, 95 and 105. Is there still a closed form?