Binomial Trees
Read Chapters 12 and 18 in Hull
Outline
- A Simple Binomial Model
- Generalizing the Binomial Tree
- Risk-Neutral Valuation
- Two-Step Binomial Tree Examples
- Binomial Trees in Practice
- An American Call Example
- Calculating the Greeks
A Simple Binomial Model
- A stock price is currently $25
- In three months, it will be either $28 or $22
- A 6-month call option on the stock has a strike price of 24
- Set up a riskless portfolio
- We create the portfolio
- Long Δ shares
- Short 1 call option
- Creates a delta neutral portfolio
- Refer to binomial tree below
- Portfolio is riskless when
-
FV = 28 Δ – 4 = 22 Δ – 0
-
Δ = 0.6667
- Both scenarios yield the same portfolio future value
- Portfolio value
- Risk-free interest rate is 10%
- The riskless portfolio is:
- long 0.6667 shares
- short 1 call option
- The future value of the portfolio in 6 months is
-
FV = 28 Δ – 4 = 28x0.6667 – 4 = $14.6676
- The value of the discount
-
discount = e –0.10x0.5 = 0.9512
- The value of the portfolio today
-
PV= $25 Δ – f = $25x0.6667 – f = $16.6675 – f
- The option's value equals
-
PV = FV x discount
-
$16.6675 – f = $14.6676 x 0.9512
-
f = 2.7157
- Note: A put portfolio has the following construction
- Short Δ shares of stock
- Long one put option
- Creates a delta neutral portfolio
Generalizing the Binomial Tree
- A derivative lasts for time T and depends on a stock price, S 0
- Percentage up, u
- Percentage down, d
- Portfolio value
- Long Δ shares
- Short 1 derivative
- The portfolio is riskless when
-
S 0 u Δ – ƒ u = S 0 d Δ – ƒ d
- Solve for Δ
- Future Value of the portfolio at time T
- Present Value of the portfolio today
-
PV = (S 0 u Δ – f u )e –rT
- Today's portfolio value
- Hence
-
PV = FV x discount
-
S 0 Δ - ƒ = (S 0 u Δ – ƒ u )e –rT
-
ƒ = S 0 Δ – (S 0 u Δ – ƒ u )e –rT
- Substituting for Δ to obtain
-
ƒ = [ p ƒ u + (1 – p) ƒ d ] e –rT
- where
- Refer to handout for derivation
- Define probability, p, as going up while 1 − p is probability of moving down
- Binomial distribution – two outcomes: Up or down
-
Average = p Outcome 1 + (1 – p) Outcome 2
- The derivative's value equals its expected payoff that is discounted at the risk-free rate
Risk-Neutral Valuation
- The probability of an up and down movements are p and 1-p
- The expected stock price at time T is S 0e rT
- Thus, the stock price earns the risk-free rate
- We calculate the average of the outcomes
- Binomial trees show how to value a derivative as the expected return of the underlying asset is assumed to earn the risk-free rate
- Thus, we can discount using the risk-free rate, which we call risk-neutral valuation
- Irrelevance of Stock's Expected Return
- When we value an option in terms of the underlying stock, the expected return on the stock is irrelevant
- Returning to the original example
- Calculate the parameters to the tree
-
u = 28 / 25 = 1.12
-
d = 22 / 25 = 0.88
-
p = ( e rt - d ) / ( u - d ) = ( e 0.1x0.5 - 0.88 ) / ( 1.12 - 0.88 ) = 0.7136
- Since p is a risk-neutral probability, then
- You can solve for p using the formula
-
25e 0.10x0.5 = 28 p + 22 ( 1 – p )
-
p = 0.7136
- The value of the option is
-
f = [ 0.7136x4 + 0.2864x0 ] e –0.10x0.5
-
f = 0.2.7152
Two-Step Binomial Tree Examples
- Refer to the tree below
- The tree represents a European call option
- Each time step is 3 months
- Discount rate, r =10%
- Probability, p = 0.7091
- u = 1.0583
- d = 0.9449
- Calculate the payoff for each node
- Determine whether you can exercise or not
- Node D:
-
payoff = max ( 28 - 26, 0 ) = 2.0000
- Node E:
-
payoff = max ( 25 - 26, 0 ) = 0.0000
- Node F:
-
payoff = max ( 22.3214 - 26, 0 ) = 0.0000
- The other nodes, we use the discounted binomial average
- You cannot exercise an European option at these nodes
- You are calculating the intrinsic value
- Node B
-
f = ( 0.7091 x 2.0000 + 0.2909 x 0.00) e –0.10x0.25 = 1.3832
- Node C
-
f = ( 0.7091 x 0.00 + 0.2909 x 0.00) e –0.10x0.25 = 0.00
- Node A
-
f = ( 0.7091 x 1.3832 + 0.2909 x 0.00) e –0.10x0.25 = 0.9566
- Refer to the tree below
- The tree represents a European put option
- K = 160
- Each time step is 3 months
- Discount rate, r =10%
- Probability, p = 0.7270
- u = 1.0541
- d = 0.9487
- Calculate the payoff for each node
- Determine whether you can exercise or not
- Node D:
-
payoff = max ( 160 - 166.6667, 0 ) = 0.0000
- Node E:
-
payoff = max ( 160 - 150, 0 ) = 10.0000
- Node F:
-
payoff = max ( 160 - 135, 0 ) = 25.0000
- The other nodes, we use the discounted binomial average
- You cannot exercise an European option at these nodes
- You are calculating the intrinsic value
- Node B
-
f = ( 0.7270 x 0.00 + 0.2730 x 10.00) e –0.10x0.25 = 2.6627
- Node C
-
f = ( 0.7270 x 10.00 + 0.2730 x 25.00) e –0.10x0.25 = 13.7471
- Node A
-
f = ( 0.7270 x 2.6627 + 0.2730 x 13.7471) e –0.10x0.25 = 5.5483
- How to use binomial tree to calculate the price of an American option
- Using the put example from the last example
- A holder can exercise the American put at maturity just like the European
- So Nodes D, E, and F stay the same
- A holder can exercise an American option anywhere on the tree
- Node B
- Intrinsic value: f = ( 0.7270 x 0.00 + 0.2730 x 10.00) e –0.10x0.25 = 2.6627
- Option value if holder does not exercise
- Exercise early: payoff = max ( 160 - 158.1139, 0 ) = 1.8861
- Holder goes for the higher value, which means he does not exercise
- Node C
- Intrinsic value: f = ( 0.7270 x 10.00 + 0.2730 x 25.00) e –0.10x0.25 = 13.7471
- Exercise early: payoff = max ( 160 - 142.3025, 0 ) = 17.6975
- Holder earns greater value by exercising early
- Node A
- Intrinsic value: f = ( 0.7270 x 2.6627 + 0.2730 x 17.6975) e –0.10x0.25 = 5.5483
- Note, Node C caused a number change in the payoff
- Exercise early: payoff = max ( 160 - 150.0000, 0 ) = 10.0000
- Holder exercises immediately earns a payout of 10
- If he or she waits, he or she only earns 5.5483
Binomial Trees in Practice
- Analysts frequently use binomial trees to approximate the price movements of a stock or other asset
- In each small interval of time, the stock price is assumed to
- move up by a proportional amount u
- or to move down by a proportional amount d
- Movements in Time Δt
- Risk-Neutral Valuation
- We choose the tree parameters p, u, and d so the tree gives correct values for the mean and standard deviation of the stock price changes in a risk-neutral world
- Two conditions are
-
e rΔt = pu + (1 – p) d
- or S 0 e rΔt = p S 0 u + (1 – p) S 0 d
-
σ 2Δt = p u 2 + (1 – p)d 2 – [p u + (1 – p) d ] 2
- where σ is the volatility
- The following condition is often imposed
-
u = 1/ d
- Cox, Ross, and Rubinstein (1979) used this approach
- Be careful on exams whether this condition holds
- For a small time step, Δt, a solution to the equations is
- Backwards Induction
- We know the value of the option at the final nodes
- We work back through the tree using risk-neutral valuation to calculate the value of the option at each node
- For European options, keep working backwards
- For American options, you must check whether you can exercise at each node
- Changing the binomial tree for different cases
- Constructing the tree remains the same
- Only the probability changes
- Four cases
- For plain vanilla (ice-cream) options, q = 0
- If a stock price pays continuous dividends at rate q
- For options on stock indices, q equals the dividend yield on the index
-
a = e (r - q)Δt
- For options on a foreign currency, q equals the foreign risk-free rate, r f
- For options on futures contracts q = r
An American Call Example
- Parameters for example
-
S 0 = 75
-
K = 75
-
r =10%
-
σ = 15%
-
T = 3 months or 0.25 year
- Time step: Δt = 1 month or 1 /12 = 0.0833 year
- Calculate the parameters
-
u = 1.0443
-
d = 0.9576
-
p = 0.5858
-
1 – p = 0.4142
-
discount = e –rt = 0.9917
- Holders can exercise American and European at Time T, maturity
- Node G, Payoff = 10.4039
- Node H, Payoff = 3.3189
- Node I, Payoff = 0.0000
- Node J, Payoff = 0.0000
- Since a holder can exercise an American option early, we must check each node
- Node D
- Intrinsic value = 7.4071
- Can holder exercise American early, payoff = 81.7847 - 75 = 6.7847
- Thus, holder waits since payoff is higher for intrinsic value
- Node E
- Intrinsic value = 1.9280
- Can holder exercise American early, payoff = 75 - 75 = 0
- Thus, holder waits since payoff is higher for intrinsic value
- Node F
- Intrinsic value = 0
- Can holder exercise American early, payoff = 0
- Thus, holder does not exercise early
- Node B
- Intrinsic value = 5.0949
- Can holder exercise American early, payoff = 78.3189 -75 = 3.3189
- Thus, holder does not exercise early since waiting has higher payoff
- Node C
- Intrinsic value = 1.1200
- Can holder exercise American early, payoff = 0
- Thus, holder does not exercise early
- Node A
- Intrinsic value = 3.4198
- Can holder exercise American early, payoff = 0
- Thus, holder does not exercise early
- The top number at each node is the stock price
- The value of the option is below the stock price
- The holder does not exercise the American option early at any nodes
- So, in this case, the American call option equals the European call, or C = c
- Increasing the Time Steps
- A tree should have 30 time steps or more to give good option values
- DerivaGem calculates up to 500 time steps
- The Black-Scholes-Merton Model
- We can derive the Black-Scholes-Merton model by looking at what happens to the price of a European call option as the time step tends to zero
- Means the number of branches increases to infinity
- The binomial tree for European options converges to Black Scholes
Calculating the Greeks
- Calculating delta
- Delta means how much the value of the option changes if the spot price increases by $1
- Delta (Δ) is the ratio of the change in the price of a stock option to a one-unit change in the price of the underlying stock
- Once can calculate delta at any adjacent nodes
- The value of Δ varies from node to node
- At node B, stock price = 78.3189 and call price = 5.0949
- At node C, stock price = 71.8217 and call price = 1.1200
- delta = ( 5.0949 – 1.1200 ) / ( 78.3189 – 71.8217 ) = 0.6118
- Delta hedging – used delta to make our portfolio riskless
- Calculating Gamma
- Gamma – how delta changes to a one $1 increase in the spot price
- Gamma requires two deltas from the same time point
- Calculate the following from Nodes D, E, and F
- delta 1 = 0.8076
- delta 2 = 0.3099
- denominator = average ( 81.7847 – 75.0000, 75.0000 – 68.7781 ) = 6.5033
- gamma = ( 0.8076 – 0.3099 ) / 6.5033 = 0.0765
- If the stock prices rises by $1, then delta increases by 0.0765
- Calculating Theta
- Theta is how the option's value changes if you increase the time by one unit
- Theta is calculated from the central nodes at 0 months and 2 months
- That way, the stock price remains constant
- Only the time and option prices differ
- Denominator starts at 0 and ends at 2 months (or 0.1667 year)
- theta = ( 1.9280 – 3.4198 ) / ( 0.1667 – 0 ) = –8.9506
- If time increases by one year, then the option value decreases by 8.9506
- Calculating Vega
- Vega is is how the option's value change if volatility increases by one unit
- We use the following procedure
- From the original tree, volatility equals 15% and the option price equals 3.4198
- Construct a new tree with a volatility of σ = 16%
- Option price equals 3.5744
- vega = ( 3.5744 – 3.4198 ) / 1 = 0.1546
- A 1% increase in volatility raises the option's value by approximately 0.1516
- More volatility means the holder will more likely exercise the option
- Note: Delta, Gamma, Theta, and Vega apply to Black Scholes as well
- Same ideas but calculated differently
|