VRM 14: Binomial Trees

Derivatives are priced by a no-arbitrage argument, which simply means the price is pinned down by the requirement that nobody can build a riskless profit out of it. The law of one price is the familiar version of that requirement: two portfolios delivering the same cash flows on the same future dates must cost the same today, because otherwise a trader shorts the expensive one, buys the cheap one, and collects the difference with no exposure left behind. A binomial tree turns that principle into arithmetic you can actually do. The technique was set out by Cox, Ross and Rubinstein in 1979, and it remains the workhorse for American-style options, because it shows the option at every date in its life rather than only at expiry.

Begin with the smallest tree there is, a single step. A stock trades at USD 50, and over the next three months it will return either +20% or minus 20%, so the price at the end of the period is either USD 60 or USD 40. No real stock behaves like that, but the mechanics carry over unchanged to realistic trees. Assume no dividends. Against it write a call option that expires in three months and is struck at USD 52. A finish at 60 gives a payoff of 8, and a finish at 40 leaves the option worthless. The USD label is dropped for the rest of this section.

Figure 1: A single binomial step and the call payoffs at its two ends

The strike price of 52 sits between the two terminal prices, so exactly one branch finishes in the money.

Constructing the riskless portfolio

Build a portfolio that holds a long position of delta shares against one call written short, where delta is the Greek capital letter used throughout for that share position. In three months the portfolio comes to 60 multiplied by delta, less 8, if the stock rises, because the written call forces a share worth 60 to be handed over for 52. If the stock falls the option lapses, leaving 40 multiplied by delta. Setting those two expressions equal makes the uncertainty disappear.

The share position that makes both outcomes identical. Either branch then leaves the portfolio worth 16.

Example 1 · Worked

Over three months the risk-free rate is 3%, continuously compounded.

1. What is the call option worth today?

Solution. With delta equal to 0.4 the portfolio is worth 16 in three months whichever way the stock moves, since 60 multiplied by 0.4 less 8 is 16 and 40 multiplied by 0.4 is also 16. A certain 16 in three months is worth 16 discounted at 3% over 0.25 years, or 15.880 today. Any other price invites arbitrage: below 15.880 a trader buys the portfolio and earns more than 3% risk free, and above 15.880 a trader shorts it and borrows at less than 3%. Now price the portfolio in terms of the unknown option value f. The 0.4 shares are worth 0.4 multiplied by 50, or 20, and the written call subtracts f, so the portfolio costs 20 less f. Setting that equal to 15.880 gives f = 4.120, the price of an option on one share.

The probabilities of the up and down moves were never used, and the only market input beyond the tree itself was the risk-free rate. Fractional share positions are awkward in practice, so scale everything up: sell 1,000 options, which is ten contracts, and buy 400 shares. The price of an option on a single share is unaffected, provided the stock and the option both trade in liquid markets.

Check yourself

Why does the option price not depend on how likely the up move is?

The option is valued relative to the stock, not in isolation. Whatever probability makes the stock attractive or unattractive is already sitting inside the price of 50 that the hedge is built from, and the two returns move together closely enough that neither one has to be estimated.

Replace the numbers with symbols and the same argument produces a formula. A non-dividend paying stock is priced at S. Over a time T it moves either up to Su, a return of u less 1, or down to Sd, a return of d less 1. Some derivative written on it, which need not be an option at all, pays f_u after the up move and f_d after the down move. Build the portfolio as before: one unit of the derivative sold short, a position of delta held in the stock.

Equation 14.1. The share position that makes the hedged portfolio worth the same on both branches.

With that choice, the portfolio ends at the same value whichever branch is taken, so it earns the risk-free rate r over the period. The portfolio costs S multiplied by delta, less f, where f is the derivative price today. Imposing no arbitrage on that relationship and substituting for delta gives a compact expression.

Equations 14.2 and 14.3. The quantity p is fixed by u, d, r and T, and never by anyone’s view of the stock.

The earlier example fits the general form directly. The stock moved from 50 to either 60 or 40, so u is 1.2 and d is 0.8, with T equal to 0.25 and r equal to 3%. Feeding those into Equation 14.3 gives p = 0.5188, and therefore 1 less p = 0.4812.

Example 2 · Worked

Keep the same one-step tree, but replace the call with a put option struck at 52.

1. Value the put, then confirm that put-call parity holds.

Solution. The put pays nothing if the stock rises to 60, so f_u = 0, and it pays 52 less 40, or 12, if the stock falls, so f_d = 12. Equation 14.2 gives 0.5188 multiplied by 0, plus 0.4812 multiplied by 12, discounted at 3% over 0.25 years, which comes to 5.731. Put-call parity for European options on a non-dividend paying stock with matching strike and maturity says the call price plus the present value of the strike price equals the put price plus the stock price. That present value is 52 discounted at 3% over 0.25 years, or 51.611. The left side is 4.120 plus 51.611 and the right side is 5.731 plus 50, both 55.731, so parity is satisfied.

Notice how little the formula asks for. Once u, d, r and T are known, the hedge ratio and the price both follow mechanically, and the same pair of equations prices a call, a put, a forward or any other payoff that can be written at the two terminal nodes.

Check yourself

A one-step tree has u = 1.2, d = 0.8 and a derivative paying 12 on the down branch and 0 on the up branch. Without recomputing p, what is delta, and what does its sign tell you?

Delta is 0 less 12, divided by 60 less 40, which is minus 0.6. The negative sign says the derivative gains when the stock falls, so hedging a short position in it requires a short stock position rather than a long one. That is the expected result for a put option.

Equations 14.2 and 14.3 hide the single most important principle in derivatives pricing. A risk-neutral world is defined as one in which investors demand no extra return for bearing risk, so every tradable asset is expected to earn the risk-free rate and an investor is indifferent between a safe asset and a volatile one. Probabilities computed on that assumption are risk-neutral probabilities. The risk-neutral valuation principle says that a derivative priced under this artificial assumption comes out at the price that is correct in the real world too, where investors very much do care about risk.

Two observations show why p earns the name. Read p as the probability of an upward movement, and the bracket in Equation 14.2 becomes the expected payoff of the derivative at time T, which the equation then discounts at the risk-free rate. Apply the same reading to the stock itself: its expected price at time T is Su multiplied by p plus Sd multiplied by 1 less p, and substituting the definition of p collapses that to S multiplied by e raised to rT. The stock grows at the risk-free rate under p, which is the defining property of a risk-neutral world.

Under p the stock is expected to appreciate at exactly r, which identifies p as the risk-neutral probability of an upward movement and 1 less p as the risk-neutral probability of a downward movement.

None of this claims the world is risk neutral. It is a device: the derivative fetches the same price in the real world as it would in the imaginary one, so the imaginary one is a legitimate place to work.

What real-world probabilities would give

Return to the tree in Figure 1 and suppose the true probability of the up move is 0.6. Over the three months the stock is then expected to return 0.6 multiplied by 20%, less 0.4 multiplied by 20%, which is 4%. The call option, valued earlier at 4.120, has an expected payoff of 0.6 multiplied by 8, giving an expected three-month return of 16.5%. The gap is large, and the explanation is leverage: the stock carries positive systematic risk, and the call amplifies it.

The put option is stranger. Its expected payoff is 0.4 multiplied by 12, and against a price of 5.731 that is an expected return of minus 16.2%. A negative expected return on a risky security looks wrong until the systematic risk is examined. The put moves opposite to the stock, so its systematic risk is negative, which drags its expected return below the risk-free rate, and the leverage in the option widens that shortfall.

Check yourself

The risk-neutral probability in the example is 0.5188 while the assumed real-world probability is 0.6. Does using 0.5188 understate the option value?

No. The two probabilities go with two different discount rates. Under 0.6 the correct discount rate for the call is its real-world expected return of 16.5%, and discounting the expected payoff at that rate returns 4.120 again. Risk-neutral valuation substitutes a lower probability and a lower discount rate together, and the two changes cancel exactly.

A single step is far too crude for a stock price over a year. The fix is to chop the life of the option into many short steps, each handled exactly like the one-step tree already built. Write the length of one step as delta t, measured in years, and every formula so far survives with T replaced by delta t.

Equations 14.4 and 14.5, the one-step formulas rewritten for a single step of a multi-step tree.

That leaves the question of where u and d come from. They are not free choices: their job is to make the tree reproduce the volatility of the stock, and Cox, Ross and Rubinstein showed that the following pair does it.

Equation 14.6. Volatility per year is sigma, and delta t is the step length in years. Note that d is the reciprocal of u.

Volatility therefore enters in one place only, through the width of the branches. A stock with a higher sigma gets a larger u and a smaller d, so its terminal prices fan out further and more of them land deep in the money or deep out of the money. Shortening the step shrinks each branch, because the square root of delta t shrinks, and that is what holds the annualised volatility constant as steps are added.

Why these particular formulas match the volatility

Over a short interval delta t the variance of the stock return is sigma squared multiplied by delta t. On a binomial step the return is u less 1 with probability p and d less 1 with probability 1 less p, so its variance follows from the statistical definition, the expectation of the squared return less the square of the expected return.

The matching condition that u and d must satisfy. Substituting the risk-neutral probability p and simplifying reduces it to a condition on u and d alone.

Expanding the exponentials and discarding terms of order higher than delta t, so that e raised to r delta t becomes 1 plus r delta t and e raised to 2r delta t becomes 1 plus 2r delta t, leaves a condition the Cox, Ross and Rubinstein choices satisfy. One further result makes the construction usable. Girsanov’s theorem, published in 1960, establishes that volatility is identical in the risk-neutral world and the real world, so sigma can be estimated from historical price data and dropped straight into a risk-neutral calculation.

Check yourself

A tree is built with sigma of 25% and a step of half a year. If the step is cut to a quarter of a year with sigma unchanged, does u rise or fall, and why does that not change the volatility the tree represents?

u falls, from e raised to 0.25 times the square root of 0.5 down to e raised to 0.25 times the square root of 0.25. Each move is smaller, but there are twice as many of them, and because the exponent scales with the square root of the step length the variance accumulated over a year is unchanged.

A two-step tree is enough to show the whole procedure. A non-dividend paying stock trades at 29 with a volatility of 25%. Value a one-year European call option struck at 30 when the risk-free rate is 3% with continuous compounding. Splitting the year in half gives delta t = 0.5, and Equation 14.6 gives u = 1.1934 and d = 0.8380. So an up move returns 19.34% and a down move returns minus 16.20%. Equation 14.5 then gives p = 0.4984, leaving 0.5016 for a down move in the risk-neutral world.

Label the starting node A. After one step the stock sits at either 34.608, which is 29 multiplied by u, or 24.301, which is 29 multiplied by d. Call those nodes B and C. After two steps there are three prices rather than four, because the tree recombines: an up move then a down move lands where a down move then an up move lands. The terminal prices are 41.299, 29.000 and 20.363.

Figure 2: A recombining two-step tree and the direction of the calculation

Because an up move followed by a down move gives the same price as the reverse order, two steps produce three terminal prices, not four. Payoffs are known only at the right edge, so the calculation runs leftward.

Payoffs are known only at the end of the tree, so the calculation runs from right to left. That procedure is called backward induction, or rolling back through the tree: at each earlier node, apply Equation 14.4 to the two values one step ahead.

Example 3 · Worked

Roll the European call back through the two-step tree just described.

1. What is the option worth today?

Solution. Start at the terminal nodes. The highest price, 41.299, gives a payoff of 11.299 against the strike of 30, while 29.000 and 20.363 are both below 30 and pay nothing. Move back one step. At node B the values ahead are 11.299 and 0, so the option is worth 0.4984 multiplied by 11.299, plus 0.5016 multiplied by 0, discounted at 3% over half a year, or 5.548. Both values ahead of node C are zero, so node C is worth zero. At node A the values ahead are 5.548 and 0, giving 0.4984 multiplied by 5.548, discounted over half a year, or 2.724. The two-step tree values the call at 2.724.

Check yourself

Why does the middle terminal price come out at exactly 29.000, the starting price?

Because d is the reciprocal of u under Equation 14.6. Multiplying 29 by u and then by d returns 29 exactly, which is what makes the tree recombine and keeps the number of terminal nodes down to one more than the number of steps.

Keep the stock, the strike price of 30, the volatility and the rate as before, and switch to a put option. The stock prices on the tree are unchanged, and so are u, d and p. Only the payoffs differ. The top terminal node at 41.299 is above the strike, so it pays nothing. The middle node at 29.000 pays 1, and the bottom node at 20.363 pays 9.637.

Rolling back works the same way. Node B looks ahead to 0 and 1, which discounts to 0.494. Node C looks ahead to 1.000 and 9.637, giving 0.4984 multiplied by 1.000 plus 0.5016 multiplied by 9.637, discounted over half a year, or 5.252. Node A then combines 0.494 and 5.252 to give 2.838, the European put value from this tree.

Testing for early exercise at every node

An American option can be exercised at any time up to maturity, so a second calculation is needed wherever the tree offers the choice. Every node carries two figures, the value on immediate exercise and the value on holding, and takes whichever is larger. The exercise value is the intrinsic value, which for a put is the greater of the strike price less the stock price and zero.

The American test applied at every node of the tree, with K the strike price and S the stock price at that node.

Example 4 · Worked

Make the put option American, leaving everything else unchanged.

1. What is the option worth today, and what is the American feature worth?

Solution. The terminal values are unchanged, since exercise at maturity is exercise at maturity. At node B the stock is at 34.608, so exercising would give 30 less 34.608, a loss of 4.608, and the option is held. Node B stays at 0.494. Node C is the interesting one: holding is worth 5.252, while exercising is worth 30 less 24.301, or 5.699. Since 5.699 beats 5.252, early exercise is optimal there and node C takes the value 5.699. Roll back to node A with that revised figure: 0.4984 multiplied by 0.494 plus 0.5016 multiplied by 5.699, discounted over half a year, gives 3.058. Exercising at node A would yield only 1.000, so the option is held. The American put is worth 3.058 against 2.838 for the European put, so the early exercise right is worth 0.22.

One subtlety about the roll back: the value at any node already accounts for exercise opportunities further down the tree as well as the immediate one, because those later decisions were settled before the calculation reached that node.

Check yourself

Node C had a holding value of 5.252 and an exercise value of 5.699. What would happen to the option value at node A if the early exercise test were skipped?

Node A would use 5.252 rather than 5.699 and return 2.838, the European value. Skipping the test does not merely lose accuracy at node C, it propagates a low value backwards and understates the option at every earlier node.

Equation 14.1 has been in the background since the first section and deserves a name of its own. Delta measures how strongly a derivative price responds to the price of its underlying asset, and it doubles as the stock position that hedges one unit of the derivative sold short, which is why it is the first of the Greek letters a trading desk watches. On a tree it is read straight off two adjacent nodes: the change in the option price divided by the change in the stock price between them.

Delta from any two nodes reachable in one step. Strictly this is the sensitivity at time delta t rather than at time zero, but on a tree with small steps the difference is negligible.

Figure 4: Delta as the slope of the chord between two adjacent nodes

The slope of the chord is delta. A steeper chord means the option responds more strongly to the stock.

The four-step American put tree shows how delta is computed and how quickly it moves. At the two nodes ending the first step, option prices of 1.336 and 4.896 sit against stock prices of 32.861 and 25.592, so delta is minus 0.490. The sign is negative because a put gains as the stock falls, and the magnitude says a small move in the stock changes the put price by about 49% of that amount.

Delta does not stay put. Two steps into the same tree, one pair of nodes has option prices of 0.247 and 2.439 against stock prices of 37.237 and 29.000, giving a delta of minus 0.266. A lower pair, with option prices of 2.439 and 7.415 against stock prices of 29.000 and 22.585, gives minus 0.776. The put is far more sensitive after the stock has fallen than after it has risen, which is why a delta hedge has to be rebalanced rather than set once.

Example 5 · Worked

A stock trades at 40 and will be worth either 42 or 38 in one month. Take an annual risk-free rate of 4% under continuous compounding, and a call struck at 39 that expires in one month.

1. Value the option and state the hedge for a short position in it.

Solution. Here u = 42 divided by 40, or 1.05, and d = 38 divided by 40, or 0.95. Equation 14.5 with delta t = 0.0833 gives p = 0.5334. The payoffs are 3 and 0, so the option is worth 0.5334 multiplied by 3, discounted at 4% over one month, which is 1.595. Delta is 3 less 0, divided by 42 less 38, or 0.75. So the hedge is long 0.75 of a share against each short option, and that portfolio is worth 28.5 on both branches.

Check yourself

Why is the delta of a put option always negative, and what does a delta close to zero indicate?

A put pays more when the stock is lower, so the numerator of Equation 14.1 has the opposite sign to the denominator. A delta close to zero means the put is deep out of the money: both nodes ahead give a payoff of zero or nearly zero, so a small stock move changes almost nothing.

Everything so far assumed a non-dividend paying stock. The bridge to other underlying assets runs through a stock paying a continuous dividend yield at rate q, so that the dividend over a short interval delta t equals the stock price multiplied by q multiplied by delta t. Real stocks pay discrete dividends, but several assets that underlie derivatives behave just like this idealised one.

Only one thing changes. In a risk-neutral world the total return must be r, and the dividend supplies q of it, so the price itself can only grow at r less q. Replacing the growth condition rewrites the risk-neutral probability.

Equation 14.7. The calculation of u and d and the whole roll back procedure are untouched.

Stock indices

An option on a stock index is handled by setting q to the average dividend yield expected on the index over the life of the option. Take an index at 2,500 with a dividend yield of 2%, a risk-free rate of 3% and a volatility of 15% per annum. A three-step tree over six months gives delta t = 0.1667, so u = 1.0632 and d = 0.9406, and Equation 14.7 gives p = 0.4983. A European call struck at 2,500 rolls back to 119.579.

Currencies

A foreign currency earns interest at the foreign risk-free rate, so it is an asset providing a yield, and q is set equal to that foreign rate. Take a one-year American call giving its holder the right to buy one unit of a foreign currency at 0.8000. The exchange rate today is 0.7800 and its volatility is 12%, while the domestic risk-free rate is 2% and the foreign one is 6%. A four-step tree has delta t = 0.25, giving u = 1.0618 and d = 0.9418. The foreign rate exceeds the domestic rate, so the drift is negative and p works out at 0.4021. The tree values the option at 0.0188, with early exercise optimal at two nodes.

Futures

Entering a futures contract costs nothing, so in a risk-neutral world the futures price must have zero expected growth. That is the same condition as a stock whose dividend yield equals the risk-free rate, so set q equal to r and the exponential in Equation 14.7 collapses to 1.

The risk-neutral probability for a tree built on a futures price.

For a nine-month American put on a futures contract with a current futures price of 38 and a strike price of 40, a volatility of 20% and a risk-free rate of 4%, a three-step tree gives delta t = 0.25 and u = 1.1052. Then d = 0.9048, so p = 0.4750 and the option is worth 3.828.

Check yourself

For an index, a currency and a futures contract, what plays the role of the dividend yield q?

A stock index uses the dividend yield on the index itself, a currency uses the foreign risk-free rate, and a futures price uses the domestic risk-free rate, which is what reduces the probability to 1 less d over u less d.

How a no-arbitrage argument prices an option on a one-step binomial tree

Constructing the riskless portfolio

The general one-step formula for delta and the value of a call or put option

What risk-neutral valuation means and why it gives the correct option price

What real-world probabilities would give

How the binomial model captures volatility through the formulas for u and d

Why these particular formulas match the volatility

How a two-step binomial tree values a European call option

Valuing a European put option and an American put option on the same tree

Testing for early exercise at every node

How the binomial value converges as the number of time steps is increased

What delta is and how to calculate the delta of a stock option from a tree

Altering the binomial model for dividends, stock indices, currencies and futures

Stock indices

Currencies

Futures