Options and Derivatives · Lesson 3
Pricing without a crystal ball: the binomial tree
Price an option by building something that pays exactly what it pays, whichever way the share goes. How likely "up" is never enters.
Know the basics? Read the lesson notesA share with two futures
A made-up share sells for €100 today. Suppose only two things can happen by next year: it rises to €119 or falls to €84. Every number in this lesson is made up.
A call is the right, but not the duty, to buy one share at a fixed price, the strike, on a set last day. Take a call with a strike of €105 and a last day a year from now. What it pays then is its payoff: the share's price minus €105, or 0 if the share is below €105, since nobody pays €105 for a share worth less.
If the share ends at €119, what does the call pay, in euros?
Hint: the call lets you buy at €105 a share that is worth €119.
One year, two outcomes (made-up prices)
Each box: the share's price, and what the call pays a year from now.
The question for the whole lesson: what is this call worth today, before anyone knows which way the share goes?
A first guess
Suppose you think the share has a 75% chance of going up (a made-up figure). Money can be borrowed or lent for a year at an interest rate of 5%: €1 today grows to €1.05 in a year, so €1.05 a year from now is worth €1 today. Dividing a later amount by 1.05 gives what it is worth today, its present value.
On average the call pays 0.75 × 14 + 0.25 × 0 = €10.50 in a year, and €10.50 ÷ 1.05 = €10 today. What would you say the call is worth today?
Copy the call's swing
Another way to value the call: build something you can price today that pays exactly what the call pays, in both cases. Start with shares. Between the two cases the call's payoff swings by 14 − 0 = €14; one share swings by 119 − 84 = €35. Drag the number of shares (fractions allowed) until your shares swing by €14 too.
What your shares and the call are worth in a year (made-up prices)
Each row runs from its value if the share falls (left dot) to its value if it rises (right dot).
0.4 of a share swings by 0.4 × 35 = €14, like the call: it is worth €33.60 if the share falls and €47.60 if it rises.
That number of shares, the call's swing divided by the share's swing, is the call's delta (written Δ): (14 − 0) ÷ (119 − 84) = 0.4.
Borrow the rest
The 0.4 shares will be worth €33.60 or €47.60; the call pays €0 or €14. In both cases the shares are worth €33.60 more than the call. So borrow money today and repay €33.60 in a year: then the shares minus what you owe pay €0 or €14, exactly what the call pays.
At 5% a year, how much can you borrow today, if you will repay €33.60 in a year?
Hint: what you borrow grows by 5% to €33.60, so divide €33.60 by 1.05.
The copy: 0.4 shares, minus €33.60 owed (made-up prices)
The bottom row is the copy. Its dots sit exactly on the call's.
The copy costs 0.4 × €100 = €40 for the shares, of which the loan pays €32: €40 − €32 = €8 of your own money today.
Holdings that pay what the call pays in every case are a replicating portfolio (a portfolio is a set of holdings). Two things that pay the same in every case must cost the same today, or someone collects money for nothing (a few steps on shows why). So the call is worth €8: the first guess's answer was C, less than €10.
The 75% never came in
The copy's cost used the share's price today, its two possible prices, the strike and the interest rate. The 75% chance of up appeared nowhere. Drag the chance of up and compare the two numbers.
The call's average payoff at today's value, against the copy's cost (made-up)
Across: the chance that the share goes up. Grey: the average payoff divided by 1.05. Orange: the copy's cost.
The average moves with the chance of up; the copy's cost stays at €8, because the copy pays what the call pays in both cases, whichever is more likely.
Beliefs still matter to the share: what people expect is part of why it sells for €100 today. Given that price, the call's price follows. The two lines cross at one chance of up, 60%. The next screen shows why there.
A shortcut: the risk-neutral chance
Call q the chance of up that would make the share grow on average exactly as fast as money at interest, from €100 to €105: q × 119 + (1 − q) × 84 = 105, so q = (105 − 84) ÷ (119 − 84).
Work out q, as a decimal.
Hint: (105 − 84) ÷ (119 − 84).
Average the payoff with q and take its present value: (0.6 × 14 + 0.4 × 0) ÷ 1.05 = 8.40 ÷ 1.05 = €8, the copy's cost again.
q is called the risk-neutral probability. It is not the real chance of up and not a forecast: it is a weight worked out from the share's prices and the interest rate, and would be the chance of up only in a world where nobody asked for extra return for taking risk, hence the name. Why it always matches: with q as the weight, the share's own average grows at the interest rate, and a loan grows at that rate too, so anything built from the share and a loan, the copy included, has a q-average a year from now equal to its cost today × 1.05; dividing by 1.05 gives back that cost. It says nothing about whether the share will rise. (The theory card further on writes q in general form.)
A wrong price is free money
A dealer (a firm that quotes prices to buy and sell; made up here) offers to buy this call from you for €9, a euro above the copy's cost: its quote. An arbitrage is a set of trades that ends with money in every case, with none of your own money put in. Which trades lock one in?
The sure profit from a quote (made-up prices)
Across: the quoted price. The dashed line is the copy's €8.
Below €8 it runs the other way: buy the call and sell the copy, by short selling 0.4 shares (selling shares you have borrowed, to hand back later) and lending €32. Only at €8 is there no sure profit, which is why €8 is called the call's no-arbitrage price.
Two steps: work back from the end
Now give the call two years, the share moving the same way each year (× 1.19 or × 0.84) and interest 5% a year. After two years the share is at 100 × 1.19 × 1.19 = €141.61, 100 × 1.19 × 0.84 = €99.96 (up then down, or down then up), or 100 × 0.84 × 0.84 = €70.56. A drawing of every path like this is a binomial tree, binomial because each step has two branches. With the strike still €105, only the top end pays: 141.61 − 105 = €36.61.
Value the call after an up year. From there it is a one-step problem: a year later the call pays €36.61 or €0, so it is worth (q × 36.61 + (1 − q) × 0) ÷ 1.05, with q = 0.6 as before. How many euros is that?
Hint: 0.6 × 36.61, then divide by 1.05.
The two-year tree, worked back from the end (made-up prices)
Each point: the share's price, and the call's value there.
After a down year, both later prices are below €105, so the call is worth €0 there. One more step back: (0.6 × 20.92 + 0.4 × 0) ÷ 1.05 = 12.552 ÷ 1.05 = €11.95 today.
Stepping back from the end one step at a time like this is called backward induction. Each point has its own delta: (20.92 − 0) ÷ (119 − 84) ≈ 0.60 of a share today, (36.61 − 0) ÷ (141.61 − 99.96) ≈ 0.88 after an up year. So the copy must be adjusted at every step.
Many steps: the price settles
Back to the one-year call. Two outcomes for a whole year is crude, so cut the year into n shorter steps. Each step still goes up or down, but by less, because a shorter step has less time to move: the sizes are set so that the year's spread of prices stays about the same whatever n is (the exact rule is in the lesson's "For the precise" box). With one step it is the first tree, worth €8.
As n goes from 1 to 200, what happens to the call's price?
The one-year call's price by number of steps (made-up inputs)
Each dot: the price from a tree with that many steps, worked back from the end. Dashed: the limit.
It settles, so the answer is B. One step gives €8.00, 12 steps €7.06, 100 steps €6.95. On the way it zigzags between odd and even numbers of steps, by more than 10 cents in the first 20, and the swings shrink, unevenly: from 100 steps on, every price is within 2 cents of €6.94.
That limit has a formula of its own, the Black–Scholes formula, which the Black–Scholes lesson builds from this tree. €8 and €6.94 rest on the same spread of prices over the year; they differ because one step allows only two outcomes.
Pricing by copying: the one-step binomial model
If shares and borrowing pay exactly what a contract (an option such as the call, or any deal whose payoff depends on the share) pays in every case, the contract's price is their cost today; any other price is an arbitrage. This is no-arbitrage pricing by replication; with two outcomes per step it is the binomial model (Cox, Ross and Rubinstein, 1979). The risk-neutral probability q gives the same price: average the payoffs with q, then divide by 1 + r.
- S
- the share's price at the start of the step
- u, d
- the up and down factors: the share ends the step at S·u or S·d
- r
- the interest rate for one step, for borrowing and lending
- Cu, Cd
- the contract's payoff (in a longer tree, its value) at the end of the step if up / if down
- Δ
- delta: the shares in the copy
- B
- what the copy borrows at the start (negative: lends)
- C
- the contract's no-arbitrage price at the start of the step
- q
- the risk-neutral probability: a weight from u, d and r, not the real chance of up
S = 100, u = 1.19, d = 0.84, r = 0.05, strike 105: Cu = 14, Cd = 0.
Δ = 14 ÷ (119 − 84) = 0.4. B = (0.4 × 84 − 0) ÷ 1.05 = 32. C = 0.4 × 100 − 32 = 8.
q = (1.05 − 0.84) ÷ (1.19 − 0.84) = 0.6. C = (0.6 × 14 + 0.4 × 0) ÷ 1.05 = 8.
- over each step the share can end only at S·u or S·d;
- you can buy or sell any amount of the share, fractions and short sales included, with no costs;
- you can borrow and lend any amount at the one rate r;
- d < 1 + r < u (otherwise buying the share with borrowed money, or short selling it and lending, can't lose and may gain, and q is not strictly between 0 and 1);
- over several steps, the copy is adjusted at every step.
The price can land anywhere else in a step (three or more outcomes, or a jump): then no mix of shares and a loan copies the call in every case. Costs, limits on borrowing or short selling, and moves of a different size from the tree's leave a gap, as the hedging simulation on the next screen shows.
What is the call's no-arbitrage price today, in euros (to the cent)?
Hint: u = 60 ÷ 50 = 1.2 and d = 45 ÷ 50 = 0.9; the call pays 60 − 52 = 8 if up and 0 if down. Find q, then the price.
Hedging with the tree, in two worlds
A seller of the call can cancel its risk by holding the copy, adjusted at every step to the tree's delta. Holding something that cancels a risk you carry is hedging. In the tree, the hedged seller should end with no surplus and no shortfall.
A Monte Carlo simulation (many random paths, then counts) tests this on 10,000 made-up paths of 12 months. The seller sells the one-year call for the 12-step tree's price, €7.06, and adjusts monthly. Each month is up with a made-up chance of 55%, not the tree's q. In the tree's own world a month is up 5.15% or down 4.90%, as the tree assumes. In the "bigger moves" world the same ups and downs are about half as big again, up 7.83% or down 7.27%, but the seller still prices and hedges with the tree. Look at both.
The hedged seller's result after a year, 10,000 simulated paths
What is left after paying the call's payoff, before costs. Each bar counts paths; 0 means the copy paid exactly what the call owed.
In this simulation, in the tree's own world the hedged seller ended within a cent of 0 on all 10,000 paths, whichever way each went. With bigger moves, the same hedge lost money on all 10,000 paths, €3.64 on average, before costs.
The seller charged for moves of the tree's size, and bigger moves cost more to hedge, because the shares you hold are set for the tree's size of move, not the real one, so each month's adjustment leaves a small loss, whichever way the move goes. That is why every path lost. These counts belong to two made-up worlds with two outcomes a month; they can't show how often real prices move more than assumed, or what costs or jumps would do.
What the tree assumes
The tree's price is exact, given its assumptions. Turn them over one at a time.
- Two outcomes per step.
A real price can end anywhere. Shorter steps bring the tree closer, but each step still has only two outcomes, and a sudden jump fits none of them.
- No costs, one interest rate.
Buying, selling and short selling are free, in any amount, and borrowing and lending happen at the same rate. Real trading has fees, a gap between buying and selling prices, and limits on borrowing and short selling.
- Adjusting at every step.
As the steps shrink, the copy must be adjusted more and more often, in the limit without pause. Real hedgers adjust at intervals, and pay costs each time.
- The size of the moves is an assumption.
u and d come from an assumed spread of prices, its volatility (how widely returns typically spread: roughly, the typical size of a move). Nobody knows next year's. In the simulation, moves bigger than assumed made the hedge lose on every path. The tree is a model: its price follows exactly from its assumptions, and no further.
A dealer's quote
What sure profit, in euros today, can you lock in per call (to the cent)?
Hint: price the call first. u = 100 ÷ 80 = 1.25, d = 68 ÷ 80 = 0.85, payoffs 20 and 0. Then compare with €10.
A confident friend
Which is right?
Lesson done
When a portfolio of shares and a loan pays exactly what an option pays in every case, as it does in a binomial tree, the option's price is that portfolio's cost today; any other price lets someone make a sure profit. The real chance of up never enters: the risk-neutral probability q is a pricing weight, not a forecast. With many short steps, moves shrinking as the steps do, the tree's price settles toward the Black–Scholes value, and it is only as good as the tree's assumptions about the moves.
Key ideas
- Binomial tree
- a model of a price that, at each step, goes up by a factor u or down by a factor d, and nothing else
- Replicating portfolio
- shares and borrowing (or lending) that pay exactly what an option pays in every outcome of the model
- Delta (Δ)
- the shares in the replicating portfolio: the option's swing between up and down divided by the share's
- Arbitrage
- trades that end with money in every case, with none of your own money put in
- No-arbitrage price
- the only price that allows no arbitrage: in a binomial tree, the replicating portfolio's cost
- Short selling
- selling shares you have borrowed, to hand back later
- Risk-neutral probability (q)
- in a binomial tree, the weight on "up" that makes the share grow on average at the interest rate; used for pricing, it is not the real chance of up
- Backward induction
- valuing a tree from its last step back to today, one one-step problem at a time
Δ = (Cu − Cd) ÷ (S·u − S·d) · B = (Δ·S·d − Cd) ÷ (1 + r) · C = Δ·S − B
q = ((1 + r) − d) ÷ (u − d) · C = [q·Cu + (1 − q)·Cd] ÷ (1 + r)