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  1. 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?