Binomial Trees

Staff who price options use a binomial model tree to value an American option, which can be exercised before expiry. Traders use the delta that comes out of the tree (how much the option price moves when the underlying price moves) as a guide to the size of their hedge.

Replicating a call in a one-period binomial model

The binomial model assumes that after one period, the price of the underlying asset will be one of two fixed values. It then builds the two possible payoffs of an option from a combination of the underlying and borrowing or lending money. The cost of the combination that produces the same payoffs is the price of the option¹. The idea that a price is set by a combination producing the same payoffs (replication) is covered on the page No-Arbitrage and Replication.

Consider a hypothetical example in which the underlying is the shares of Company B, which pays no dividends. The underlying price today is 100, and after one period it will be either 110 or 90 (the price units are omitted). A call option (call) to buy these shares at a strike price of 100 pays 10 at the end of the period if the underlying is 110, and 0 if it is 90. The one-period risk-free rate, a rate with no credit risk or similar risks, is 2%. An amount one period from now is converted to today's value (discounted) by dividing it by 1.02.

You can build the payoff of this call by buying 0.5 units of the underlying and borrowing 44.12. The 0.5 units come from dividing the difference between the two payoffs (10 − 0 = 10) by the difference between the two underlying prices (110 − 90 = 20). If the underlying falls to 90, the 0.5 units are worth 45, so a loan that repays 45 at the end of the period leaves you with 0, the same as the call. The amount you borrow today is 45 ÷ 1.02 = 44.12.

The cost of building the combination today is the 50 paid for the underlying (0.5 × 100) minus the 44.12 borrowed, which is 5.88. The combination pays the same as the call in both cases at the end of the period, so the price of the call is also 5.88.

If the price of the call (the premium) were 6.00, you could sell the call and build the combination for 5.88, keeping 0.12 today. At the end of the period, the payoff of the combination exactly covers what you owe on the call, so this 0.12 is a profit with no chance of a loss. If instead the price were 5.50, you could buy the call, short sell 0.5 units of the underlying, and lend 44.12, keeping 0.38 today. The price that removes such arbitrage opportunities is 5.88². Transaction costs and restrictions on short selling are ignored here.

The 0.5 units of the underlying used for replication are also the delta, which shows how much the call price moves when the underlying price moves. A seller of the call can offset the gains and losses from moves in the underlying by holding 0.5 units of it³.

The price is set without probabilities

The calculation of 5.88 in the previous section uses neither the probability that the underlying rises to 110 nor the probability that it falls to 90. The cost of replication depends only on the two prices at the end of the period, the strike price, and the interest rate⁴. Whether you expect a rise to be likely or unlikely, building the same payoffs costs 5.88.

Even so, you can rewrite the price of 5.88 as the payoffs weighted by the probability $p$ below, divided by 1.02. With the up factor $u$ (1.1 in this example), the down factor $d$ (0.9), and the one-period interest rate $r$ (0.02), $p$ is given by the following formula⁵.

$$p = \frac{(1 + r) - d}{u - d}$$

In this example, $p$ = (1.02 − 0.9) ÷ (1.1 − 0.9) = 0.6. The price of the call is (0.6 × 10 + 0.4 × 0) ÷ 1.02 = 5.88, which matches the cost of replication. $p$ depends only on the two underlying prices and the interest rate, not on the strike price or the payoff of the option⁴. With the same $p$, you can price other options on the same underlying by the same steps.

$p$ is called the risk-neutral probability, but it is not the probability that the underlying actually rises. Why the price can be written as an expected value, and in what kind of world $p$ is a probability, are covered on the page Risk-Neutral Valuation.

$p$ falls between 0 and 1 when the down factor is below 1.02 and the up factor is above 1.02⁶. For example, if the down factor were 1.03, borrowing money to buy the underlying would leave 103 − 102 = 1 even if the price went down, a profit with no chance of a loss. The binomial model assumes two factors that leave no such arbitrage opportunity.

Deep diveGeneral formulas for one-period replication

Let the underlying price today be $S$, the price at the end of the period be $uS$ or $dS$, and the payoffs of the option in each case be $C_u$ and $C_d$. The number of units of the underlying used for replication, $\Delta$, is as follows⁵.

$$\Delta = \frac{C_u - C_d}{uS - dS}$$

The amount lent or borrowed, $B$ (negative for borrowing), follows from the condition that the combination matches the payoff when the price goes down.

$$B = \frac{u C_d - d C_u}{(1 + r)(u - d)}$$

Substituting the two formulas into the option price $C = \Delta S + B$ and rearranging gives the following form in terms of $p$.

$$C = \frac{p C_u + (1 - p) C_d}{1 + r}$$

With the values from the example in the previous section ($S = 100$, $u = 1.1$, $d = 0.9$, $C_u = 10$, $C_d = 0$, $r = 0.02$), $\Delta = 0.5$, $B = -44.12$, and $C = 5.88$.

Working back through a multi-period tree from expiry

A binomial tree is a diagram of the branching underlying prices, drawn as nodes and branches at each step. In a tree that divides the time to expiry into many steps, you start from the payoffs at expiry and repeat the one-period calculation one step at a time until you reach today's value⁷. The value at each node is found by weighting the values at the two nodes it branches into at the next step by $p$ and $1 - p$, adding them, and dividing by the interest factor for one step (1.02 in this example).

As in the previous section, take a two-step tree in which the underlying price rises by a factor of 1.1 or falls by a factor of 0.9 at each step, and the interest rate for one step is 2%. Use it to value a put option (put) with a strike price of 100. Here the put is a European option, which can be exercised only at expiry. Whether the price goes up and then down, or down and then up, the underlying price ends at 99, so the branches merge and the second step has three nodes. $p$ is 0.6, as in the previous section.

Figure 1 draws this calculation as a two-step tree. The upper number at each node is the underlying price and the lower number is the put value.

Figure description

The tree starts from an underlying price of 100, and at each step the price is multiplied by 1.1 or 0.9. Step 1 has 110 and 90, and step 2 has 121, 99 and 81. An up move followed by a down move gives the same 99 as a down move followed by an up move, so there are three nodes. The put with a strike of 100 is worth 0, 1 and 19 at expiry, step 2. At step 1 it is worth 0.39 at the 110 node and 8.04 at the 90 node, and at time 0 it is worth 3.38. The interest rate per step is 2% (a discount factor of 1.02), and the up probability is 0.6. The figure shows that discounting back one step at a time from the expiry values gives the value at time 0.

The price of the put at time 0 is 3.38. This calculation is equivalent to building a one-period replication at each node. The number of units of the underlying used for replication, however, differs from node to node. At time 0, it is a short position of 0.3824 units. As in the one-period case, it is the difference between the values at the two nodes of the next step divided by the difference between the underlying prices ((0.3922 − 8.0392) ÷ (110 − 90)). If the price moves to the 110 node at step 1, you rebalance to a short position of 0.0455 units, and if it moves to the 90 node, to a short position of 1 unit. Replicating an option in a multi-period tree requires you to keep rebalancing the combination as the underlying price moves⁸.

Deciding on early exercise of an American put

When you value an American option (one that can also be exercised before expiry) with a tree, you compare the value of exercising now with the value of holding on at each node before expiry, and take the larger one as the value at that node⁹. The put from the previous section is valued again, this time as an American option. The values at the three nodes at expiry are the same.

The value of exercising now is the intrinsic value at that node (for a put, the strike price minus the underlying price, or 0 if that is negative). The value of holding on is found from the two nodes of the next step, as in the previous section. At the 90 node at step 1, exercising gives you 100 − 90 = 10 now. If you hold on, the payoff at the next step is 1 or 19, and weighting by $p$ and discounting gives only 8.04, so exercising is the better choice at this node. At time 0, the value of holding on becomes (0.6 × 0.3922 + 0.4 × 10) ÷ 1.02 = 4.15, with the value at the 90 node replaced by 10.

The American put's price of 4.15 is 0.77 higher than the 3.38 of a European put with the same terms, and this difference is the value of being able to exercise before expiry.

For an American call on an underlying that pays no dividends, no node appears at which exercising before expiry is the better choice, so its price is the same as that of a European call⁹. If you value a call with a strike price of 100 in the same two-step tree, the value of holding on at the 110 node at step 1 is 0.6 × 21 ÷ 1.02 = 12.35, which is above the value of exercising, 10. So the price is 7.27 whether the call is American or European.

Setting the tree factors from volatility

In practice, the up factor $u$ and the down factor $d$ are set from the volatility of the underlying and the step length $\Delta t$. In a Cox-Ross-Rubinstein (CRR) tree, the down factor is the reciprocal of the up factor (1 ÷ $u$), and the up factor is larger when the volatility is higher and when the step length is longer¹⁰. The name comes from Cox, Ross, and Rubinstein, who published a paper on this binomial model in 1979¹¹.

Consider a hypothetical example in which the underlying price is 100, the volatility is 20%, and the time to expiry is 1 year, divided into 3 steps. The step length is 1/3 of a year, so the up factor is 1.1224 and the down factor is 1 ÷ 1.1224 = 0.8909. The interest rate is 5% a year (continuously compounded), and the interest factor for one step is 1.0168. $p$ is (1.0168 − 0.8909) ÷ (1.1224 − 0.8909) = 0.5438¹². After one step, the price is 112.24 if it rises by a factor of 1.1224 and 89.09 if it falls by a factor of 0.8909. The third step has four nodes: 141.40, 112.24, 89.09, and 70.72.

Deep diveFormulas for the factors and the probability in a CRR tree

With the volatility $\sigma$ (annualized) and the step length $\Delta t$ (in years), the up and down factors of a CRR tree are as follows¹⁰.

$$u = e^{\sigma \sqrt{\Delta t}}$$ $$d = \frac{1}{u}$$

With a continuously compounded interest rate $r$, the interest factor for one step is $e^{r \Delta t}$, and $p$ is given by the following formula. It is the formula from the earlier section with $1 + r$ replaced by $e^{r \Delta t}$.

$$p = \frac{e^{r \Delta t} - d}{u - d}$$

In the example above, $\sigma = 0.2$ and $\Delta t = 1/3$, so $\sigma \sqrt{\Delta t} = 0.11547$ and $u = e^{0.11547} = 1.1224$. The interest factor for one step is $e^{0.05/3} = 1.0168$.

How the tree price approaches the Black-Scholes price as steps increase

As the number of steps grows, the tree price of a European option approaches the price given by the formula of the Black-Scholes model¹⁰. Using the terms from the previous section (an underlying price of 100, a time to expiry of 1 year, an interest rate of 5%, and a volatility of 20%), the table compares the tree price of a European put with a strike price of 100 for different numbers of steps. The Black-Scholes price is 5.57 (5.5735).

In this example, the tree price is below the Black-Scholes price when the number of steps is even and above it when the number is odd, so it swings up and down as it converges. This is because the position of the nodes at expiry relative to the strike price differs between even and odd numbers of steps¹³. As the number of steps goes from 10 to 20, 50, and 100, the size of the difference shrinks from 0.197 to 0.099, 0.040, and 0.020, and doubling the number of steps roughly halves the difference¹⁴.

An American put has no pricing formula like the Black-Scholes formula for a European option, so it is valued by increasing the number of steps in the tree¹⁵. With the same terms, the American put is worth 6.50 with a 3-step tree and 6.08 with a 100-step tree. With a 100-step tree, the European put is worth 5.55, so the value of being able to exercise before expiry is 0.53.

Notes

1. 重川一郎, "数理ファイナンス入門" (Introduction to mathematical finance), Kyoto University (intensive lectures at the Faculty of Science, Ehime University, 2009), "ポートフォリオ" (Portfolio) and "単期間のポートフォリオ" (Single-period portfolio), pages 5–6 (in Japanese). The source uses an example in which a share price of 10 becomes 20 or 7.5, and finds the price as the replication cost of 1 (the source calls replication "duplication"). ↩

2. The same source, "単期間のポートフォリオ" (Single-period portfolio), pages 6–7 (in Japanese). It shows, for both the seller and the buyer of the option, that if the price differs from the replication cost, one of them can make a profit with no chance of a loss. ↩

3. Markus K. Brunnermeier, "Lecture 10: Multi-period Model, Options – Black-Scholes-Merton model," lecture slides, Princeton University, slide 6, "Single-period replication." ↩

4. 重川一郎, "数理ファイナンス入門," "リスク中立測度" (Risk-neutral measure), pages 8–9 (in Japanese). The source states that the real-world probability (written p in the source) is "not directly related to pricing" (our translation), and notes that the probability q in equation (1.2) "is not related to H" (our translation), where H is the payoff of the option. The source's q corresponds to $p$ on this page. ↩ ↩²

5. Brunnermeier's lecture slides, slides 5–7 ("Single-period replication" and "Risk-neutral probabilities"). In the slides, $R$ is the interest factor for one period ($1 + r$ on this page), and $C_u$ and $C_d$ are the payoffs of the option when the price goes up and when it goes down. ↩ ↩²

6. Brunnermeier's lecture slides, slide 3, "Assumptions," and slide 8, "Risk-neutral probabilities." The slides state "No arbitrage: u > R > d." ↩

7. 重川一郎, "数理ファイナンス入門," "多期間 2 項モデル=CRR モデル" (Multi-period binomial model = CRR model), pages 10–12 (in Japanese). ↩

8. Brunnermeier's lecture slides, slides 12–13, "Two-period binomial tree." The slides state "Synthetic option requires dynamic hedging." ↩

9. New York University Courant Institute, "American Options" (Lecture 8), lecture notes, section 1.4, pages 10–11. The notes explain the comparison of values at each node and why an American call with no dividends is priced like a European call. ↩ ↩²

10. Brunnermeier's lecture slides, slides 16–17, "Towards Black-Scholes." The slides set $u = e^{\sigma \sqrt{T/n}}$ and $d = 1/u$, and state that the Black-Scholes formula is obtained as the number of steps $n$ becomes large. The slides' $T/n$ ($T$ is the time to expiry) corresponds to $\Delta t$ on this page. ↩ ↩² ↩³

11. ESMA Working Paper No. 1 (2015), page 7 and page 28 (references). ↩

12. The factors and $p$ are rounded to four decimal places. Calculating with the unrounded values also gives $p$ = 0.5438. ↩

13. Paul Johnson, "Lecture 5: Analysis of binomial option pricing," lecture notes for MATH60082, The University of Manchester, section 5.2, "Errors," pages 51–52. The notes call this swing the "odd-even effect." ↩

14. The same source, section 5.1, "Convergence," pages 49–50. The notes state that the prices of binomial models such as CRR approach the Black-Scholes price at a rate of 1/n, and that halving the error requires doubling the number of steps. ↩

15. Paul Johnson's lecture notes, section 5.4, "Exotic options," page 55, and "Summary," page 57. ↩

16. Paul Johnson's lecture notes, section 5.5, "Computational Effort," page 56. ↩ ↩²