Derivatives Foundation puzzles, solved step by step

Puzzles: 100
Traced to a firm: 66
Topics: 12
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005A stock trades at 100. Each day for three days it moves up 10 or down 10, each with probability one half, and interest rates are zero. What is a call struck at 100 and expiring after the third day worth?Option pricing intuitionCore
Try it first
Before drawing the tree: how many distinct end prices are there, and how many equally likely paths?
Show the worked solution
7.50. Three moves of plus or minus 10 end at 130, 110, 90 or 70, reached by 1, 3, 3 and 1 of the eight equally likely paths. The 100 call pays 30 at 130, 10 at 110 and nothing below. Its value is the average payoff: (1 x 30 + 3 x 10) over 8, which is 60 over 8, or 7.50. With zero rates and symmetric moves, the real-world probabilities are already the pricing probabilities, so no discounting and no adjustment is needed.
Why is counting paths all the tree needs?
Think of three coin tosses where you get a sweet for each head. The chance of exactly two heads is not one in four; it is three in eight, because there are three orders in which two heads can arrive. The tree recombines, so the value of an end node is its payoff weighted by how many of the eight paths reach it, and the path counts are the binomial coefficients 1, 3, 3, 1. Nothing else in the problem carries information: the step size fixes the end prices and the counts fix the weights.
From 100, three moves of plus or minus 10 reach 130, 110, 90 or 70 by 1, 3, 3 and 1 of the eight equally likely paths, the call pays 30 and 10 at the two upper nodes and nothing below, and the average payoff (1 x 30 + 3 x 10) over 8 gives a price of 7.50.
The relationship
$C = \frac{1}{8}\sum_{k=0}^{3}\binom{3}{k}\,\max(100 + 10(2k-3) - 100,\,0) = \frac{1 \times 30 + 3 \times 10 + 3 \times 0 + 1 \times 0}{8} = 7.50$
k the number of up days out of three
C(3, k) the number of paths with k up days: 1, 3, 3, 1
100 + 10(2k - 3) the end price after k ups and 3 minus k downs
What it says in wordsThe call is the payoff at each end price, weighted by the share of paths that reach it.
Where does the risk-neutral machinery go?
In a general tree you would replace the real probabilities with the risk-neutral ones, chosen so that the stock's expected growth equals the interest rate. Here rates are zero and the moves are symmetric, so the stock already has zero expected drift and the risk-neutral probability is the same one half you were given. Say that out loud: it shows you know the shortcut is a coincidence of the setup, not a rule. If the up move were 10 and the down move 5, one half would no longer price the stock and you would have to solve for the probability that does.
What sanity checks do you say before the number?
Two quick ones. The call cannot be worth more than the expected value of the stock above the strike ignoring the max, which is zero here, so the call is worth exactly the expected positive part, and that is what 7.50 is. And put-call parity with zero rates says the 100 put must also be 7.50, which you can confirm from the lower nodes: (3 x 10 + 1 x 30) over 8. Giving the put price unprompted, and showing it matches, is the cheapest way to prove the tree was right.
Where candidates lose it
The common error is to treat the four end prices as equally likely, which gives (30 + 10) over 4 = 10. The outer nodes are reached by one path each and the inner ones by three; the weights are 1, 3, 3, 1, not 1, 1, 1, 1.
The second loss is reaching for a risk-neutral formula and getting lost in it. With zero rates and symmetric moves, the given probabilities already price the stock. Say why, then count.
What the interviewer asks next
Now the up move is 10 and the down move 5. What probability prices the stock, and what is the call worth?
Price the 110 call and the 90 put on the same tree.
Four days instead of three: what is the 100 call worth, and why does it rise?
068Your book is delta neutral with gamma of 2,000 shares per rupee on a stock trading at Rs 500. The stock jumps Rs 10. Roughly what is your P&L before you rehedge, and how many shares do you now need to trade?Option pricing intuitionCoreEquity derivativesVolatility trading
Try it first
Delta zero, gamma 2,000 shares per rupee, a Rs 10 jump. P&L?
Show the worked solution
About Rs 1,00,000 profit, and you need to sell about 20,000 shares. P&L from gamma is one half of gamma times the move squared: 0.5 x 2,000 x 10 squared = Rs 1,00,000. The delta picked up during the move is gamma times the move, 2,000 x 10 = 20,000 shares long, which you sell to get back to neutral. A Rs 10 fall would earn the same amount and leave you 20,000 shares short to buy back.
Why does a delta-neutral book make money on a move?
A cyclist at the bottom of a valley is on flat ground, but every metre up either slope gets steeper. Delta neutral means the P&L is flat at the current price only; gamma is how fast the slope changes, so as the stock moves the book acquires delta in the direction of the move and earns on it the whole way. With gamma of 2,000 shares per rupee, after the first rupee you are 2,000 shares long, after the fifth 10,000, after the tenth 20,000. The P&L is the area under that rising delta, a triangle with base 10 and height 20,000, which is 1,00,000.
A delta-neutral book with gamma of 2,000 shares per rupee earns one half of gamma times the move squared, Rs 1,00,000 on a Rs 10 move in either direction, and at the new price its slope is gamma times the move, 20,000 shares long, which is what must be sold to be flat again.
What is the arithmetic, and where does the one half come from?
Expand the book's value as a Taylor series in the stock price. The first-order term is delta times the move, zero here; the second-order term is one half of gamma times the move squared, 0.5 x 2,000 x 100 = Rs 1,00,000; and the new delta is the derivative of that, gamma times the move, 20,000 shares. The one half is the same one half as in the area of a triangle: delta started at zero and finished at 20,000, so on average it was 10,000 shares over the Rs 10 move. The limitation is that a jump also changes implied volatility and burns a day of theta, both ignored here.
The relationship
$\Delta P\&L \approx \Delta\cdot\delta S + \tfrac{1}{2}\Gamma\,(\delta S)^2 = 0 + \tfrac{1}{2}\times 2000\times 10^2 = 1{,}00{,}000, \qquad \Delta_{\text{new}} = \Gamma\,\delta S = 20{,}000$
delta the book's share-equivalent exposure, zero before the move
Gamma the change in delta per rupee of stock move, 2,000 shares
delta S the stock move, Rs 10
What it says in wordsThe profit is half of gamma times the move squared, and the delta to be hedged afterwards is gamma times the move.
What happens if you rehedge and the stock comes back?
You sell 20,000 shares at Rs 510. If the stock then falls back to Rs 500, the options give back their Rs 1,00,000 but the short stock earns 20,000 x Rs 10 = Rs 2,00,000, so you net Rs 1,00,000 from the round trip. That is what long gamma means in practice: each rehedge locks in half of gamma times the move squared, and a stock that moves a lot and comes back pays you twice. The cost is theta, the daily decay you pay for holding the options, and the trade only works if realised movement is larger than the implied volatility you paid for.
Where candidates lose it
Candidates say zero because the book is delta neutral, or they give gamma times the move squared without the one half and double the answer. Say the triangle: delta climbs from zero to 20,000, average 10,000, times Rs 10.
The second loss is confusing the two numbers. The P&L is in rupees and uses the move squared; the delta to trade is in shares and uses the move once.
What the interviewer asks next
The stock falls Rs 10 instead. What is the P&L and what do you trade?
You rehedge at Rs 510 and the stock returns to Rs 500. What have you made on the round trip?
What daily theta would make this book break even on a Rs 10 move per day?
The book is short gamma instead. Describe the same Rs 10 move.
093A stock at 100 will be 120 with probability 70% or 90 with probability 30% in one period, and interest rates are zero. Price a call struck at 100. Why does the 70% not appear in your answer?Option pricing intuitionCoreQuant trading
Try it first
What is the call worth?
Show the worked solution
The call is worth 20/3, about 6.67, and the 70% does not matter because the call can be copied with stock and cash. Hold 2/3 of a share and borrow 60: at 120 the copy is worth 80 - 60 = 20, at 90 it is worth 60 - 60 = 0, matching the call in both states. The copy costs 66.67 - 60 = 6.67, so the call must too. The real-world odds are already in the stock price, which the copy uses.
How do you copy the call?
If a shop sells a gift box of two items for more than the items cost separately, you buy the items and skip the box; the box's price is pinned by what goes in it. Options work the same way. Find a mix of stock and cash that pays exactly what the call pays in every state, and the call must cost what the mix costs, whatever anyone believes about the odds. The call pays 20 or 0, a swing of 20, while the stock swings from 120 to 90, a swing of 30. So hold 20/30 = 2/3 of a share. At 120 that is 80, which is 60 too much, and at 90 it is 60, also 60 too much: borrow 60 today and repay it in either state.
Two thirds of a share financed with 60 of borrowing pays 20 when the stock goes to 120 and 0 when it goes to 90, exactly like the call, so the call costs what that portfolio costs, 2/3 x 100 - 60 = 6.67, and the real-world 70% chance of an up move, which would give an average payoff of 14, never enters.
Where did the 70% go?
It is in the stock price. A stock that goes up 70% of the time to 120 and is still priced at 100 today is one the market demands a return on, because it is risky. The copy buys the stock at that price, so it inherits whatever the market thinks of the odds and the risk. The option is priced relative to the stock, not relative to anyone's forecast, so the real-world probability cancels out of the answer. What does appear is a different probability, q, the one that makes the stock earn the risk-free rate: 100 = q x 120 + (1 - q) x 90, so q = 1/3. Discounting the call's payoff at q gives 1/3 x 20 = 6.67, the same answer by another route.
The relationship
$\Delta = \frac{20 - 0}{120 - 90} = \frac{2}{3}, \qquad C = \Delta S - B = \tfrac{2}{3}(100) - 60 = 6.67, \qquad q = \frac{100 - 90}{120 - 90} = \tfrac{1}{3}$
Delta shares held in the copy, the call's swing over the stock's swing
B cash borrowed, 60, so the copy pays nothing in the down state
q the risk-neutral probability of the up move, not a forecast
What it says in wordsHold two thirds of a share, borrow sixty, and you have built the call for 6.67; the risk-neutral probability of a third gives the same number.
Then show the arbitrage, since that is what makes the answer binding. If the call traded at 8, sell it and buy the copy for 6.67: you pocket 1.33 today and the two positions cancel in both states. If it traded at 5, do the reverse. The limitation is the one-step world: real prices take many values, so the copy has to be rebalanced as the stock moves, which is where the Black-Scholes model comes from, and where trading costs and jumps make the copy imperfect.
Where candidates lose it
The common loss is answering 14, the call's expected payoff under the stated odds. It is the natural first move and the one the question is built to catch. Expected payoff under real-world odds is not a price unless everyone is indifferent to risk.
The second loss is getting q = 1/3 and then calling it the true chance of an up move. It is not a forecast. Say what it is: the probability that makes the stock's expected return equal the risk-free rate.
What the interviewer asks next
Price the put struck at 100 in the same tree, and check put-call parity.
If interest rates were 5% for the period, what is q and what is the call worth?
The stock's up probability rises to 90% but its price stays at 100. What happens to the call price, and why?

k	the number of up days out of three
C(3, k)	the number of paths with k up days: 1, 3, 3, 1
100 + 10(2k - 3)	the end price after k ups and 3 minus k downs

delta	the book's share-equivalent exposure, zero before the move
Gamma	the change in delta per rupee of stock move, 2,000 shares
delta S	the stock move, Rs 10

Delta	shares held in the copy, the call's swing over the stock's swing
B	cash borrowed, 60, so the copy pays nothing in the down state
q	the risk-neutral probability of the up move, not a forecast

Try it first

Why is counting paths all the tree needs?

Where does the risk-neutral machinery go?

What sanity checks do you say before the number?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a delta-neutral book make money on a move?

What is the arithmetic, and where does the one half come from?

What happens if you rehedge and the stock comes back?

Where candidates lose it

What the interviewer asks next

Try it first

How do you copy the call?

Where did the 70% go?

Where candidates lose it

What the interviewer asks next