Quant puzzles, solved step by step

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.

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.

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.

Both are worth Rs 1. The call pays 0, 0, 0, 1, 2 and 3 on faces 1 to 6, which sum to 6, so its average payoff is 6/6 = Rs 1. The put pays 3, 2, 1, 0, 0 and 0, also summing to 6, so it is worth Rs 1 too. They match because the die is symmetric: face x and face 7 - x are equally likely, which turns one payoff into the other.

Why is the price just an average of payoffs?

If a friend offers you a game where you win the number of rupees shown on a die above 3, playing a hundred times earns about Rs 100: some rolls pay nothing, some pay 1, 2 or 3. With no interest and no risk premium in a dice game, the fair value of any payoff is its average over the equally likely outcomes. That is exactly how an option is priced in the simplest world: list the states, write the payoff in each, weight by the probabilities and add.

Why is 0.50 the tempting wrong answer?

Because it plugs the average face, 3.5, into the payoff: 3.5 - 3 = 0.5. An option's value is the average of the payoff, not the payoff at the average, and because the payoff is floored at zero, the average payoff is always at least the payoff at the average. The floor throws away the losing faces, which is the whole point of owning an option. That gap, here Rs 0.50, is what traders pay for, and it grows with how spread out the outcome is: this is convexityA payoff that curves upward, so averaging over uncertain outcomes gives more than the payoff at the average outcome. in a single roll.

Where is put-call parity in this?

Take a call and a put with the same strike, 3. Owning the call and selling the put pays max(X - 3, 0) - max(3 - X, 0) = X - 3 on every face, so its value must be the average of X - 3, which is 0.5. The put struck at 3 pays 2, 1, 0, 0, 0, 0, worth 1/2, and 1 - 0.5 = 0.5 as parity requires. Saying this unprompted shows the interviewer you see a dice game as a model of a real options book, which is why the question is asked.

Duration is 1/(r - g) = 25 years, so a one-point rise cuts the value by about a fifth. The Gordon price is D1/(r - g), and its percentage sensitivity to r is 1/(r - g) = 1/0.04 = 25. With a Rs 4 dividend the price moves from Rs 100 at 10% to Rs 80 at 11%, a 20% fall. The 25% duration estimate overshoots because the price curve is convex.

Why does a stock have a duration at all?

A promise of money in one year hardly changes in value when rates move; a promise of money in twenty five years changes a lot, because the rate is compounded over every one of those years. A stock is a stream of dividends stretching forever, and when the dividends grow, most of its value sits in cash flows far in the future, so it behaves like a very long bond. Duration measures exactly that: the percentage price change for a change in the discount rate.

Why does the estimate say 25% when the price falls 20%?

Duration is the slope at one point, and the price curve bends. For a one-point rise the tangent predicts a 25% fall, but the exact move from Rs 100 to Rs 80 is 20%, because the curve is convexCurving upward, so it always sits above any of its tangent lines. and flattens as r rises. The same bend makes a one-point fall worth more than 25%: at 9% the price is Rs 133.3, up 33.3%. For a big rate move, reprice exactly instead of trusting the slope. Duration here also equals price over dividend, 100/4, which is a quick way to say it: one over the dividend yield.

For precision, the Macaulay durationThe present-value weighted average time at which cash flows arrive. is (1 + r)/(r - g) = 27.5 years, and dividing by 1 + r gives the modified duration of 25. A 10-year bond paying 10% at a 10% yield has a modified duration of about 6.1. The gap r - g is what matters, which is why high-growth stocks carry the most duration: the same stock with 2% growth would have a duration of 12.5 years. The limitation is that the Gordon model holds growth fixed while rates move; in practice both shift together.

Yes: sell the 95 call at 9.80 and buy the 100 call at 4.50, collecting 5.30 for a position that can never cost more than 5. A 95/100 call spread pays between 0 and the strike gap of 5 at expiry, so its price must sit between 0 and 5. The market lets you sell it for 5.30, which locks in at least 0.30 per spread, more if the stock ends below 100.

What is a 95/100 call spread worth at most?

Think of two coupons for the same shirt: one lets you buy it for Rs 950, the other for Rs 1,000. The first is worth more, but never by more than Rs 50, because the most it can save you over the second is the Rs 50 difference in price. Long the 95 call and short the 100 call pays the stock's rise above 95, capped once it reaches 100, so at expiry it is worth between 0 and the strike gap of 5. Anything that is certain to pay no more than 5 cannot be worth more than 5 today; with interest it is worth at most 5 discounted, slightly less.

Which side of each quote do you trade at?

This is where the question is really won or lost. You sell at the bid and buy at the offer, so the spread you can sell is worth 9.80 - 4.50 = 5.30 to you, not the mid of 5.60. 5.30 is still above 5, so the bound is broken at prices you can actually deal at. Buying the spread would cost 10.20 - 4.30 = 5.90 for something worth at most 5, a certain loss, so only one direction works. Check the stock price cases: below 95 both calls expire worthless and you keep 5.30; at 97 you owe 2 on the short call and keep 3.30; at 100 or above you owe exactly 5 net and keep 0.30.

What could stop the arbitrage from paying?

If the calls are American and the short 95 is exercised early, exercise the 100 call too: you pay the stock price minus 95 and receive the stock price minus 100, a net 5, and you already hold 5.30. The real frictions are fees, the margin the short call ties up, and the risk that the quote vanishes after you trade one leg. Trade both legs together as a spread order. On a real screen a 0.30 bound violation lasts seconds, which is why the interviewer is testing whether you can see it fast and name the side, not whether such quotes are common.

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.

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.

Macaulay duration is 1.943 years and modified duration is 1.833. At par the yield equals the coupon, 6%. The year 1 coupon is worth 6 / 1.06 = 5.660 and the final payment 106 / 1.06^2 = 94.340, adding to 100. Weight each time by its share: 0.0566 x 1 + 0.9434 x 2 = 1.943. Divide by 1.06 for modified duration: a 1 point rise in yield cuts the price by about 1.83%.

What is Macaulay duration actually measuring?

Think of a seesaw with weights placed along it at the dates money arrives, each weight equal to what that payment is worth today. Macaulay duration is the point where that seesaw balances: the average time to your money, weighted by present value. A zero-coupon bond has all its weight at maturity, so its duration equals its maturity. Any coupon puts a little weight earlier and pulls the balance point forward. Here the coupon is small and the bond is short, so the pull is small, and that is the whole shape of the answer before any arithmetic.

How do you get the numbers fast in your head?

At par, the yield is the coupon, so discount at 6%. The coupon's present value, 6 / 1.06, is about 5.66, and the rest of the price, 94.34, sits at year 2, so duration is 2 minus the coupon's weight: 2 - 0.0566 = 1.943. That shortcut works for any two-period bond: start at the maturity and subtract the early weight times the time it saves. For a par bond there is also a closed form, (1 + y)/y times (1 - 1/(1 + y)^n), which gives the same 1.9434.

Why are there two durations, and how good is the estimate?

Macaulay duration is a time; modified duration is a slope. Modified duration is the percentage price change for a one point change in yield, so this bond loses about 1.83% if yields rise from 6% to 7%, about Rs 0.018 per Rs 100 for each basis point. Repricing exactly gives 98.192 at 7%, a fall of 1.808%, and 101.859 at 5%, a rise of 1.859%. The rise is bigger than the fall, which is convexityThe curvature of the price against yield: prices rise more when yields fall than they drop when yields rise by the same amount.; for a two-year bond it is a small correction, for a thirty-year bond it is not.

The new divisor is 2. Before the split the prices sum to 300, and 300 / 3 is 100. After a 3 for 1 split the 150 stock trades at 50, the sum is 200, and only a divisor of 2 keeps the index at 100. A cap-weighted index needs no adjustment at all, because a split triples the share count as it cuts the price to a third, leaving market value unchanged.

Why must the divisor change when nothing about the company changed?

Cut a pizza into twelve slices instead of four and you have not made more pizza. A stock split does the same to a company: three times the shares, each worth a third. A price-weighted index adds up share prices, so a split drops the sum even though no value was lost, and the divisor must be cut to stop a fake fall in the index. Solve for it by keeping the index level fixed: 200 divided by the new divisor must equal 100, so the divisor is 2.

What else changes in a price-weighted index after the split?

The weights. In a price-weighted index a stock's weight is its price over the sum of prices, so the expensive stock dominates whatever the size of the company. Before the split C carried 50% of the index; after it, C carries only 25% and B, untouched, jumps to 50%. A 10% rise in C used to add 5 index points; now it adds 2.5. Nothing about C's business changed; the index simply started caring less about it, which is the main criticism of price weighting.

How do the other common methods treat the split?

A market-cap-weighted index sums price times shares, and a 3 for 1 split multiplies shares by 3 while dividing price by 3, so the stock's market value, its weight and the index are all unchanged; no divisor adjustment is needed for a split. Cap-weighted divisors still change for events that alter total market value without a price move, such as share issuance, buybacks or a constituent being replaced. An equal-weighted index is also untouched by a split, since weights are reset to equal at each rebalance, but it has to trade at every rebalance to get back to equal, which costs money.