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  1. 009A stock trades at 100 and in one period will be either 120 or 80. Interest rates are zero. Price a call option struck at 100 by building a portfolio of shares and borrowing that copies it, and explain why the real-world probability of the up move does not appear in the price.Pricing, options and index mathsCoreOptions market makingQuant trading

    Try it first

    If you believe the stock goes up with probability 90%, what is the call worth?

    Show the worked solution

    The call is worth 10. It pays 20 if the stock goes to 120 and 0 at 80. Half a share pays 60 or 40, so half a share with a loan of 40 pays 20 or 0, exactly the call. That portfolio costs 50 - 40 = 10 today. If the call traded at any other price, you could buy the cheap one and sell the dear one for a riskless profit, so no probability is needed.

    How do you build the copy?

    Match the swing first. The call's payoff moves by 20 between the two states while the stock moves by 40, so the copy needs 20/40 = 0.5 of a share: that ratio is the option's deltaHow much an option's value changes for a one-unit change in the underlying price; here, the number of shares that copies the option.. Half a share is worth 60 or 40 at the end, which is 40 more than the call in both states. Borrow 40 today, repay 40 at the end with zero interest, and the copy pays exactly 20 or 0.

    Copy the payoff with shares and borrowing, and price the copyStock 100Call ?Stock 120Call pays 20Stock 80Call pays 0updownDelta = (20 - 0) / (120 - 80) = 0.5 shareThe copy: 0.5 share, borrow 40TodayUp (120)Down (80)0.5 share506040Loan-40-40-40Total10200Matches the call in both statesCall = cost of the copy = 10No probability of up or down was used
    A call struck at 100 on a stock that moves to 120 or 80 is copied by half a share and a loan of 40, which pays 20 or 0 exactly as the call does and costs 10 today, so the call is worth 10 with no probability used.

    Why does the chance of the up move not matter?

    Think of a shop selling a bundle of two items that you can also buy separately. The bundle's price is pinned by the parts, whatever you think about how useful the items are. The call is a bundle of half a share and a loan; the share price already reflects everyone's views about the up move, so the option inherits them and adds none of its own. A 90% view is a reason to hold the stock itself, not a reason to pay more for the call than its parts cost.

    The relationship
    Δ=Cu−CdSu−Sd=20−0120−80=0.5C0=ΔS0−B=50−40=10=q Cu+(1−q) Cd,  q=S0−SdSu−Sd=0.5\Delta = \frac{C_u - C_d}{S_u - S_d} = \frac{20-0}{120-80} = 0.5 \qquad C_0 = \Delta S_0 - B = 50 - 40 = 10 = q\,C_u + (1-q)\,C_d,\; q = \frac{S_0 - S_d}{S_u - S_d} = 0.5
    \Deltashares held in the copy
    Bthe amount borrowed, 0.5 x 80 - 0 = 40
    qthe risk-neutral weight on the up state, fixed by the prices, not by beliefs
    What it says in wordsThe copy's cost gives the price, and the same price is an average of the payoffs using weights set by today's stock price.

    What would you do if the call traded at 12?

    Sell the dear thing and buy the cheap one. Sell the call for 12, buy half a share for 50 and borrow 40, a net cash inflow of 2 today; at the end the portfolio pays exactly what you owe on the call in either state. The 2 is kept whatever happens. The weight q = 0.5 that reproduces the price is called the risk-neutral probability, but it is a pricing weight backed out of the stock price, not a forecast. Say that distinction; interviewers listen for it.

    Where candidates lose it

    The trap is pricing the call as an expected payoff under your own view: 90% of 20 is 18. That price can be arbitraged against the stock, so nobody could trade it for long, and the interviewer wants to hear that the copying portfolio pins the price.

    The second loss is getting 10 by assuming a 50% chance. The number is right by coincidence of the symmetric tree; ask yourself what happens with an up move to 130, and the risk-neutral weight changes to 1/2.5 = 0.4.

    What the interviewer asks next

    • Price the put struck at 100 and check put-call parity.
    • What changes if interest rates are 5% for the period?
    • The stock can go to 130 or 80 instead. Price the call again.
  2. 054With interest rate r and volatility sigma, check which of these satisfy the Black-Scholes equation: V = S, V = K e^(-r(T-t)), and V = S squared. Explain what the ones that pass are as trades, and fix the one that fails.Pricing, options and index mathsHardQuant researchOptions market making

    Try it first

    Which candidates pass?

    Show the worked solution

    V = S and V = K e^(-r(T-t)) satisfy it; V = S squared does not. The first is the stock itself and the second is a zero-coupon bond paying K at T, both traded assets that must earn r. S squared has gamma 2 and leaves (r + sigma squared) S squared unbalanced. Multiplying by e^((r + sigma squared)(T-t)) fixes it, which is the price of a claim paying S squared at expiry.

    What is the equation actually saying?

    Think of a household budget rule that any fair arrangement must obey: over one day, what you hold must earn the same as the same money in a savings account, once the risk has been hedged away. The Black-Scholes equation says that for a delta-hedged position, time decay plus the gamma term plus the financing of the hedge equals r times the value. Written in {term('greeks', 'Theta is the change in value with time, delta with the stock price, and gamma is the change in delta with the stock price.')}, it is theta + half sigma squared S squared gamma + r S delta = r V. A candidate price passes only if its greeks balance that line.

    The relationship
    ∂V∂t+12σ2S2∂2V∂S2+rS∂V∂S−rV=0\frac{\partial V}{\partial t} + \tfrac12\sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + rS\frac{\partial V}{\partial S} - rV = 0
    dV/dttheta, the change in value as time passes
    d2V/dS2gamma, how fast delta changes
    dV/dSdelta, the hedge ratio
    rthe interest rate
    What it says in wordsA hedged position's decay, convexity and financing must add up to exactly the interest the money would earn.
    Substitute each candidate into theta + half sigma^2 S^2 gamma + r S delta - r VCandidate VThetaDeltaGammaLeft side of equationPasses?Sthe stock010rS - rS = 0yesK e^(-r tau)a zero-coupon bondrV00rV - rV = 0yesS^2not a price02S2(r + sigma^2) S^2noS^2 ff = e^((r + sigma^2) tau)-(r + sigma^2) V2S f2 f0yestau = T - t. The failing S^2 leaves a surplus; the time factor f is exactly what cancels it, pricing a claim paying S_T^2.At S = 100, r = 5%, sigma = 20%, one year: the S^2 claim is worth 10,942, not 10,000.
    Substituting each candidate's theta, delta and gamma, the stock and the zero-coupon bond balance the equation exactly, S squared leaves a surplus of (r + sigma squared) S squared, and S squared times e^((r + sigma squared)(T - t)) balances it again.

    Why do the two that pass make sense as trades?

    Anything that is itself a traded, self-financing asset must satisfy the equation, because the equation is only the statement that no hedged position earns more than r. V = S is just holding the stock: delta 1, no gamma, no decay, and the financing term rS matches rV. V = K e^(-r(T-t)) is a zero-coupon bond: it does not depend on S at all, and its value grows at exactly r as it approaches T. The stock and the bond are also the two pieces of the call price formula, which is why the check is worth a minute.

    Why does S squared fail, and how do you repair it?

    S squared has gamma 2, so the half sigma squared S squared gamma term adds sigma squared S squared, the delta term adds 2rS squared, and subtracting rV leaves (r + sigma squared) S squared with nothing to cancel it. A convex payoff gains from every move, so a fair price for it has to decay over time to pay for that gain, and S squared on its own has no decay. Try V = S squared times f(t): the equation forces f' = -(r + sigma squared) f, so the price of a claim paying S squared at T is S squared e^((r + sigma squared)(T - t)). At S = 100, r = 5%, sigma = 20% and one year, that is about 10,942, not 10,000, and the extra is the value of volatility.

    Where candidates lose it

    Candidates often say every function of S and t is a solution, or differentiate correctly and then fail to say what the passing solutions are. The question asks for the trades: the stock and a bond. Naming them turns a calculus check into finance.

    The second trap is the sign of theta for the bond. Its value rises as t approaches T, so theta is +rV; getting that sign wrong makes the bond appear to fail.

    What the interviewer asks next

    • Which power of S, S to the a, satisfies the equation with no time factor?
    • What is the price today of a claim paying log S at expiry?
    • Why does the drift of the stock not appear anywhere in the equation?