Topic 9 Binomial Option Pricing

1
Topic 9Binomial Option Pricing
2
Introduction

• In most textbooks, including Hull and McDonald,
  the first option pricing model that is discussed
  is the binomial model. There are a number of
  reasons for this
• It is a good way to introduce the basic ideas of
  option pricing.
• It is computationally simple.
• It is also a very useful tool for pricing complex
  options, such as American-style options, exotic
  options, really any path-dependent options.
• Binomial (and trinomial) models are particularly
  important in pricing interest-rate dependent
  derivatives.
• Hull and McDonald to take significantly different
  approaches in the way the introduce the binomial
  model.
• Hull quickly builds the Cox, Ross, and Rubinstein
  model.
• McDonald begins by working with forward
  contracts.
• We will follow Hull initially, but then will
  return to McDonald.

3
One Step Binomial Model

• Begin with a simple situation
• Assume that a stock is currently trading at 20
  and it is known that at the end of three months
  the stock price will either be 22 or 18. Further
  assume no dividends during the life of the
  option.
• Further assume that we want to know the value (or
  price) of a call option with strike of 21.
• Recall that the terminal value of a call option
  is max(0,ST-X).
• Thus, at the end of the three months, the option
  will take on one of two values. If the stock is
  worth 22, the option will be worth 1, otherwise
  it will be worth 0. What is this option worth
  today?

4
One-Step Binomial Model

• We can price this option rather easily if we make
  one very crucial assumption that there is no
  arbitrage in the market.
• To do this, we create a portfolio consisting of
  positions in the stock and the option that has
  absolutely no uncertainty. That is, it has the
  same value at the end of the three months
  regardless of what happens to the stock price.
  This lets us discount the cash flows at the
  risk-free rate (since there is no risk!).
• Since there are two securities and two possible
  outcomes, there will always be a solution.
• To see this consider portfolio which consists of
• ? shares of the stock.
• 1 short position in the option.

5
One Step Binomial Model

• Now consider the value of the portfolio in both
  states of the world
• When ST 22
• portfolio 22 ? -1
• When ST 18
• portfolio 18 ?.
• Therefore 22 ?-1 18 ?
• or 4 ?1
• so ? .25.
• Thus the portfolio should consist of the
  following
• Long .25 shares of the stock
• short 1 call option
• If the stock price moves up the portfolio is
  worth
• .25 22 -1 4.5
• If the stock price moves down the portfolio is
  worth
• 18.25 - 0 4.5

6
One Step Binomial Model

• So regardless of the final value of the stock,
  the portfolio is worth 4.5. Of course that is
  its value in three months. Since there is no
  risk in this portfolio, it must be discounted at
  the risk-free rate. Suppose that the risk free
  rate is 12, then the value of the portfolio
  today is
• 4.5e-.12.25 4.367.
• We know that the stock price today is 20.
  Denote the option price as f. The value of the
  portfolio must be, therefore
• 20.25 -f 5 - f,
• and since we know the portfolios value today is
  4.367, it follows that
• 5 - f 4.367,
• so,
• f 0.633.

7
Generalization

• This basic logic can be applied in a very general
  way, and not only to call options. To see this,
  consider the following.
• Assume a stock is currently priced at S, and
  there is a derivative on the stock with price f.
  
• The derivative pays a payout at time T. During
  the life of the derivative, the stock can either
  move up from S to a new level Su or down from s
  to Sd (ugt1, dlt1).
• The proportional increase when there is an up
  movement is u-1, when there is a down movement it
  is d-1. The derivative pays off fu and fd
  respectively.

8
Generalization

• Graphically, the situation is

S0u fu
S0 f0
S0d fd
9
Generalization

• As before imagine a portfolio consisting of a
  long position in ? shares and a short in one
  derivative. We determine the value of ? that
  makes the portfolio riskless. The ending value
  of the portfolio at the end of the derivative's
  life is given by
• S0u ? - fu
• and
• S0d ? - fd.
• The two are equal when
• S0u ? - fu Sdd ? - fd
• or when
• ? (fu - fd) / S0u S0d.
• In this case the portfolio is riskless and must
  earn the risk-free rate, r. The present value of
  the portfolio must be, therefore
• (S0u? - fu)e-rT.

10
Generalization

• Of course getting into this portfolio today
  costs
• S ? f
• and assuming no arbitrage, it must be that the
  cost to enter into the portfolio and the present
  value of the portfolio must be the same, i.e.
• (Su ? - fu)e-rT. S ? f
• Substituting in from our previous equation this
  reduces to
• f e-rT p fu (1-p) fd
• where
• p (erT - d) / (u-d)
• Note that in the above equation, d, u and r are
  not really related to probability explicitly,
  they are just prices and interest rates.

11
Generalization

• This allows for pricing of any derivative in a
  one period world. In the example given earlier,
  for example
• u 1.1 d .9
• r .12 T .25
• fu 1 fd 0
• And so we can determine P
• p (e.12.25 - .9) / (1.1 - .9) .65233
• We can then use p to determine the price of the
  option
• f e-.03 (.65233.1 .3477 0) 0.633.
• Note that this is the same price that we got
  earlier.

12
Generalization

• Let us work a second example, in this case a put
  with a strike price of 23.
• We will work with the same stock as before
• S0 20 u 1.1 d 0.9
• r .12 T 0.25
• Of course the payoff to a put is given by
• put payoff max(0,X-ST)
• Graphically, the situation looks like

S0u 22
fumax(0,23-22)1
20
Sd18
fd max(0,23-18)5
13
Generalization

• First, lets use the formulas that we just
  derived to determine the price of the option
• The price of the put option, therefore is given
  by

14
Generalization

• We can also get the same answer by forming our
  risk-free portfolio and backing out the price
  of the option.
• We again want to create a portfolio consisting of
  ? units of the stock and a long position in the
  option. This portfolio must have the same value
  in both the up and down state.

S0u 22
fumax(0,23-22)1 ?u?(22)1
20
Sd18
fd max(0,23-18)5 ?d?(18)5
15
Generalization

• So equating ?u and ?d and solving for ? yields
• Indeed, the portfolio will be worth exactly the
  same in both the up and down state.

16
Generalization
S0u 22
fumax(0,23-22)1 ?u?(22)122123
20
Sd18
fd max(0,23-18)5 ?d?(18)518523

• So the portfolio has no risk in it, and as a
  result it is appropriate to discount it at the
  risk-free rate. This means that the value of the
  portfolio is

17
Generalization

• Since we know that the portfolio consists of 1
  share of the stock and 1 put, we know
• Which is the same price that we obtained by the
  first method.

18
Generalization

• Although a simple example, this really
  illustrates the most fundamental concept behind
  the pricing of options (really all derivatives)
  written on traded securities
• They are priced by forming a portfolio that is
  riskless over the next period, discounting that
  portfolio back at the risk-free rate,
  substituting in the market value(s) of the
  underlying instruments, and then solving for the
  derivatives price.
• In the binomial model, the portfolio is riskless
  over the next time step. In a continuous-time
  model (such as Black-Scholes), the portfolio need
  only be instantaneously riskless.
• Indeed, it is this property that leads us to the
  entire idea of risk-neutral valuation.

19
Irrelevance of the Stocks Expected Return

• Notice that none of the equations we have used
  explicitly states the probability of the stocks
  moving up or down. This is surprising, but it is
  true. The probability of moving up or down is
  essentially already in the price of the stock -
  we dont have to take them into account a second
  time.
• Another way of saying this is that the option
  prices are calculated relative to the prices of
  the underlying assets.

20
Puts, Calls, and Risk-Neutrality

• Risk neutral valuation is perhaps the most
  important and fundamental idea in the valuation
  of derivatives.
• The following extended example is designed to
  demonstrate the basics of what risk-neutrality
  is, why we can price derivatives with it, and why
  risk-neutral prices give the same prices for
  derivatives that we would get in a risky world.
• We will illustrate these examples via call and
  put options, simply because they are really
  simple instruments with which to start, but the
  basic idea extends to any instrument which is
  publicly traded.
• For ease of exposition we will return to the
  first example we worked.

21
Risk Neutrality

• Lets return to our first example. A stock is
  currently worth 20. In three months it will be
  worth either 22 or 18.
• The risk free rate is currently 12.
• You hold a three month call option with a strike
  (X) of 21, so it will be worth either 1, if
  ST22, or 0 if ST18 (since the payoff to a call
  is given by Payoff max(0,ST-X), where ST is Su
  or Sd
• Our goal is to determine the price of the call
  option today.

Su22S0u cu max(0,22-21)1
S020c0 ??
Sd18 S0dcd max(0,18-21)0
22
Risk Neutrality

• Now, our first thought is that since we know the
  cash flows, that we can multiply them by the
  probability of receiving them and then discount
  them at the appropriate rate.
• This would be the classical discounted cash
  flow method for determining value.
• The difficulty, of course, is that we dont know
  either the correct discount rate or the
  probabilities associated with the cash flows
  (okay we know the collective probability of
  achieving either the 22 or 18 state is 100,
  but we dont know the individual probabilities,
  which is what matters.)
• This means that we have to come up with some
  other way of determining the price of the option.

23
Risk Neutrality

• Fortunately, there is a way we can do this. It
  involves creating a portfolio that is worth the
  same in both potential future states of the world
  (i.e. that has the same value regardless of
  whether the stock is worth 18 or 22).
• This portfolio must contain the option and
  another asset (or assets) for which we know the
  current (time 0) price.
• Since this portfolio is riskless, it is
  appropriate to discount it at the risk-free.
• Since we know the price today of everything in
  the portfolio except the option, we can back
  out the price of the option.

24
Risk Neutrality

• The key, of course, was to create a riskless
  portfolio.
• We did this by creating a portfolio consisting of
  1 short position in the call option and ? (.25)
  shares of the stock.
• At the end of the period, this portfolio value is
  given by

Su22cu max(0,22-21)1 Pu?Su-cu ?22-1
S020c0 ?? P0 ?S0-c0
Sd18c0 max(0,18-21)0 Pd?Sd-cd ?18-0
25
Risk Neutrality

• Obviously we found the price to be 0.63.
• Notice that we were able to find the value of
  this option without explicitly stating the
  probability of the up or down moves.
• Theoretically, the reason we can do this is
  because we are pricing the option, conditional on
  the value of the underlying stock.
• Financial economic theory says that a stock price
  is a martingale, that is, its price contains all
  publicly available information about the stock.
• This includes the probability of the stocks up
  and down moves.
• By pricing the option conditional on the stocks
  price, we implicitly incorporate those
  probabilities into the options price as well.

26
Risk Neutrality

• Now, if the expected return on the stock is 15,
  this means that its expected value at the end of
  the 3 months must be
• EST20e.15(.25)20.76
• Recalling the possible ending values we can see
  that the up and down probabilities are the ones
  that make this statement true20.76
  (p)22(q)18
• Where p is the probability of an up jump and q
  is the probability of a down jump.

Su22
S020
Sd18
27
Risk Neutrality

• Since q(1-p), we can rewrite this as
• 20.76p(22)(1-p)(18)
• Or
• 20.7622p18-18p
• Or
• 2.7622p-18p
• Or
• 2.764p
• Or
• .69p, so q(1-p).31

28
Risk Neutrality

• Indeed, we can verify that this must be the case
• 20.76.69(22)(.31)18
• Or, written another way
• S0ereturn(dt)p(Su)(1-p)Sd
• Which is
• S0ereturn(dt)p(S0u)(1-p)(S0d)
• Or, simplifying
• ereturn(dt)pu (1-p)d
• Or,
• ereturn(dt)pu d-pd
• So that
• p (ereturn(dt)-d)/(u-d)
• Is the general formula.

S0u 22
p.69
S020
S0d18
(1-p).31
Note that in this example, u1.1 (201.122) and
d.9 (20.9)18
29
Risk Neutrality

• The equation establishes a
  very important principal
• The probability of the stock going up and down is
  fundamentally related to the expected return on
  the stock.
• For example, if instead of having a 15 expected
  return, the stock had an 18 expected return, the
  probabilities would be
• p (e.18(.25)-.9)/(1.1-.9) (1.046-.9)/(1.1-.9)
  .7301
• What is interesting is the effect that this would
  have on the option.

30
Risk Neutrality

• Lets say that the return on the stock is 15,
  such that p.69. What would be the return on the
  option?
• Recall that we know that the option is worth
  .6329 today, and will be worth 1 if the stock
  ends up at 22, and 0 if it the stock ends up at
  18.
• So what is the expected return to buying the
  call?
• Based on these probabilities, the expected payoff
  to the call will be 0.69 EcT.69(1).31(0)

S0u 22 cu1
p.69
S020 c0.6329
S0d18 cd0
(1-p).31
31
Risk Neutrality

• The expected return to the call, is thus given
  by
• c0eEreturn(.25)EcT
• Or
• eEreturn(.25)EcT/c0
• Or
• Ereturn(.25)ln(ECT/c0)
• Or
• Ereturnln(EcT/c0)/.25
• Which is
• Ereturnln(.69/.6329)/.25
• 34.55 (!)

S0u 22 cu1
p.69
S020 c0.6329
S0d18 cd0
(1-p).31
32
Risk Neutrality

• So why is the expected rate of return on the
  option so much higher than that of the stock?
  Simply because the option is riskier (i.e. it has
  more systematic risk), and as such, investors
  demand a higher rate of return to compensate for
  the risk.
• What would happen to the expected return on the
  option if the expected return on the stock were
  to be higher?
• Recall that if the expected return on the stock
  were 18, we demonstrated that the probability of
  an up jump would be 73.01

33
Risk Neutrality

• With p.7301, the expected call value at time T
  is .7301
• EcT.7301(1).2699(0)0.7301
• So the expected return is
• c0eEreturn(.25)EcT
• Or
• eEreturn(.25)EcT/c0
• Or
• Ereturn(.25)ln(ECT/c0)
• Or
• Ereturnln(EcT/c0)/.25
• Which is
• Ereturnln(.7301/.6329)/.25
• 57.14

S0u 22 cu1
p.7301
S020 c0.6329
S0d18 cd0
(1-p).2699
34
Risk Neutrality

• Notice that even though the expected return
  changed we held the prices constant we simply
  changed the probabilities.
• Also notice that when the expected return on the
  stock increased, the expected return on the
  option increased by a lot more (in both absolute
  and proportionate terms).
• This is because the option is riskier than the
  stock. When the stocks expected rate of return
  increases, it means investors are demanding a
  higher rate of return for bearing risk, and so
  they demand an even higher rate of return for the
  riskier investment the option.
• The chart on the following page lists the
  probability of an up-jump and the expected return
  on the option for various expected returns on the
  stock

35
Risk Neutrality
36
Risk Neutrality

• There are three major points to notice from this
  chart
• The option price is exactly the same in each of
  these cases the change in the expected option
  payoff caused by the change in the probabilities
  is exactly offset by the change in the options
  implicit rate of return.
• As the expected return on the stock increases,
  the expected return on the option increases by a
  larger amount. One way of thinking of this is
  that if the stocks expected return increases, it
  means that investors are demanding a higher rate
  of return for bearing risk. The option is riskier
  than the stock, so when the premium that
  investors demand for bearing risk increases, it
  goes up even faster for riskier investments.
• When the stocks expected return is set to the
  risk-free rate (12), the options return is also
  the risk-free rate!
• The third point is actually very, very useful,
  and we will examine it in more detail.

37
Risk Neutrality

• The third point demonstrates that when the
  stocks expected return is the same as the risk
  free return, then the options expected return is
  also the risk free return.
• Theoretically, this would only happen in a world
  in which all investors were indifferent (neutral)
  toward risk.
• In this world, all assets, including options,
  must be discounted at the risk free rate!
• Notice that in this risk-neutral world, our stock
  has the same initial price (20), and the same
  two potential ending prices (18 or 22), the
  only thing that is different is the relative
  probabilities of reaching the two ending points.

38
Risk Neutrality

• Perhaps most importantly, however, the option
  price is the same (its still 0.63).
• This gives us a really neat trick for determining
  option values
• 1. First, assume a risk-neutral world, i.e. one
  in which the stock will grow at the risk-free
  rate.
• 2. Determine the probabilities of an up and down
  jump in that risk-neutral world (the up jump
  probability was 65.23 in our example)
• 3. Determine the expected payoff to the option
  using those risk-neutral probabilities.
• 4. Discount the expected payoff by the risk-free
  rate to get the price of the option.
• Even though we have priced the option in a
  risk-neutral world, the option will have the
  same value in the risky world just as we
  demonstrated in the chart above.

39
Risk Neutrality

• This is actually the secret behind all
  derivatives pricing derivatives will have the
  same value in a risky or a risk-neutral world.
• How would this work for a put?
• Well, assume for a moment that a put had a strike
  of 21 as well. We can calculate its price noting
  that it will pay 0 in the up state and 3 in the
  down state.
• We can get its price by discounting the expected
  payout (with the risk-neutral probabilities) at
  the risk-free rate
• Put (.65230.34773)e-.12.25 1.012

40
Risk Neutrality

• Note, however, that we are not saying the
  probability of the up jump is 0.6523 in reality.
  We are saying that in a risk-neutral world, to
  get the same price for the stock and the option
  that we observe in the real world, the up jump
  probability would be 0.6523.
• Another way of seeing this is to realize that the
  real expected return to the option is not the
  risk-free rate of 12. To see this, lets assume
  that the stock is really returning 15.
• This means that the real probability of an
  up-jump is 0.6911.

41
Risk Neutrality

• The real expected payoff to the put, therefore
  would be
• put (.690.313)e-return(.25)
• Or
• put.93e-return(.25)
• But we know the price of the option is 1.012, so
  the real return to the put is
• -ln(1.012/.93)/.25 -33.80
• The rate is negative because the option is
  negatively correlated with the market. In
  essence, it is an insurance contract and those
  typically have negative real returns to the
  purchaser of the contract.

42
Risk Neutrality

• So what should you take away from this?
• Options are priced relative to the underlying
  instrument, and as a result they implicitly
  incorporate the probability distribution of the
  underlying into the options price.
• Given a probability distribution (and investor
  risk-preferences), there is a precise
  relationship between the stocks price (or
  expected return) and the options price (or
  expected return.) This relationship guarantees
  that the option will have the same price
  relative to the stock regardless of the
  expected returns on the stock.
• If we assume that the stock and the option will
  each only earn the risk-free rate, we can solve
  for one set of probabilities. While these
  risk-neutral probabilities are not the real
  probabilities, if we use them to determine the
  expected cash flows from the option and then
  discount those cash flows at the risk-free rate,
  we will calculate the correct options prices.
• In order for all of the above to work, we have to
  be able to form the riskless portfolio involving
  the underlying and the option.

43
Two Step Binomial Trees

• We can extend the analysis to two-period binomial
  trees (and both authors later extend them to
  multi-period binomial trees).
• Here we assume the stock price starts at 20 and
  that u is 1.1 and d is 0.9 in each of the two
  time steps.
• Suppose that each time step is three months in
  length and the risk-free rate is 12. Again
  consider an option with strike of 21.

24.2
22
19.8
20
18
16.20
44
Two Step Binomial Trees

• We want to know the value of the option at time
  0.
• We begin at the terminal time step of the
  lattice. Note that at the bottom and middle
  nodes, the value of the option is 0, while at the
  top node it is worth 3.2 dollars.

24.2 cuumax(0,24.2-21)3.2
22
19.8 cudmax(0,19.80-21)0
20
18
16.20 cddmax(0,16.20-21)0
45
Two Step Binomial Trees

• At the second time step we have two nodes to
  consider. At the bottom node it must be worth
  zero, since it will be worth zero at time 2 with
  certainty.
• What about at the upper node? To do this we need
  to calculate some numbers. First, lets notice a
  few facts
• u 1.1 d .9 r .12 T .12
• Therefore
• p (e.03 - .9)/ (1.1 - .9) .6523 (same as
  before)
• and so we can use our standard formula to get the
  price
• e-.12.25(.65233.2 .34770) 2.0257.
• So now we go back to time 0 and do the same
  thing. P works out the same, so therefore the
  price is
• e-.12.25(.65232.0257 .34770) 1.2823

46
General Result

• This can be seen in the lattice at time 1,

24.2 cuumax(0,24.2-21)3.2
22 cu2.20257
19.8 cudmax(0,19.80-21)0
20
18 cd0
16.20 cddmax(0,16.20-21)0
47
General Result

• As well as at time 0

24.2 cuumax(0,24.2-21)3.2
22 cu2.20257
19.8 cudmax(0,19.80-21)0
20 c01.2823
18 cd0
16.20 cddmax(0,16.20-21)0
48
Two-Step Binomial Trees

• Its neat to see the full lattice, but sometimes
  its more convenient to work with simply the
  option values.
• We see that after each time step the stock is
  worth either Su or Sd where S is its initial
  value. We then calculate the terminal values of
  fuu and fdd and fud. Then we need to know the
  values of fu and fd
• fu e-rTpfuu (1-p) fud
• fd e-rTpfud (1-p) fdd
• and ultimately
• f e-rTpfu (1-p) fd
• substituting in gives
• f e-r2Tp2fuu 2p(1-p)fud (1-p)2fdd.

49
Cox, Ross, Rubinstein Model

• Now ultimately, what determines our model is how
  we calculate u and d. The most famous model is
  the one published by Cox, Ross and Rubinstein. In
  that model they make the following definitions
• Where s is the volatility of the stocks returns.
  Using the same value for p as before, then
• p (er ?t d)/(u-d)

50
Cox, Ross, Rubinstein Model

• Now notice also that we have the parameter ?t.
• This is the amount of time that passes between
  levels of the lattice.
• Some books will refer to this quantity as dt,
  others as h.
• You need to be aware of the fact that there are 3
  parameters that held define time in any binomial
  lattice.
• T the time to maturity (in years) of the option
  you are pricing.
• ?t the time step, or amount of time between
  levels of the lattice.
• N the total number of levels (i.e. time
  steps) in the lattice.
• Realize that ?tT/N.
• Frequently you are given two of these and have to
  determine the third parameter.

?
?
?
?
?
?
N4
?
?
?
?t
?
T
51
Cox, Ross and Rubinstein Model

• Normally if you say you are working with the
  binomial model, this is what people assume that
  you mean. The only additional complication is
  that now we have a parameter, s, the volatility
  of the expected return on the stock.
• The major advantage of the CRR model is that as
  you reduce the time between levels on the
  lattice, the price the CRR model generates
  converges to that given by Black-Scholes.
• McDonald points out that there can be some
  problems with CRR if your time step is too
  large, but frankly those are never issues in
  practice.

52
American Options

• American options make it somewhat more difficult,
  although not much so.
• All you have to do is realize that at each node,
  the option is worth the maximum of its immediate
  exercise price or its discounted value.
• Well work an extended example with an American
  put option to illustrate.

53
An Extended Example

• Assume that company T had stock selling at
  31/share, and that the volatility of the stock
  price was 25 (annually), and that the risk-free
  rate of return was 1.
• How would you use the binomial options pricing
  model to value a call option on that stock if the
  strike price of the option were 30, and the
  option expired in exactly one month?

54
Parameters

• First, you should put all of the parameters into
  order
• S0 31
• K 30
• T 1/12 0.083333
• s 0.25
• r .01
• You still have to make a crucial decision, which
  how many steps (N) you want to have in your
  lattice. Lets use N4 time steps, thus, dt
  (1/12)/4 0.083333/4 .0208333

55
Parameters

• We now have enough information to begin setting
  up our lattice. First, we need to calculate u and
  d.
• First, u
• Now, d
• Notice that d 1/u!

56
Parameters

• We can also calculate the other major parameter,
  p, the risk-neutral probability of an up jump.
• Now, I can begin to lay out the lattice that I
  will use.
• The first point (or node) on the lattice is the
  current stock price, which is 31 dollars.
• The nodes at the next time step are determined by
  multiplying the original stock price by u and d.

57
The Lattice Layout

• Thus,
• Su S0u 31 1.036743 31.10229, and
• Sd S0d 31 0.964559 29.90133.
• Thus, my lattice would appear be

58
The Lattice Layout
32.14
31
29.90
4
0
2
3
1
59
The Lattice Layout

• As we add additional nodes to our lattice, we are
  going to want to be able to rapidly refer to a
  given node. It is normal to refer to these nodes
  by (t,j) notation, where t is the time step the
  node is on, and j is the number of nodes below
  it on that time step.
• Thus, the first node in the lattice is node
  (0,0), meaning the node on time step 0, with 0
  other nodes below it.
• Sd is node (1,0), meaning the bottom-most node on
  time step 1 and Su is (1,1) meaning the node on
  time step 1 with 1 node below it.

60
The Lattice Layout

• We can refer to the stock price at a given node
  as S(t,j). Thus, we can say that S(0,0) 31,
  S(1,0) 29.90 and S(1,1) 32.14.
• Indeed, our goal is to fill out the S(t,j) values
  for the entire lattice. When complete it will
  look like

61
The Lattice Layout
S(4,4)
S(3,3)
S(4,3)
S(2,2)
S(3,2)
S(1,1)32.14
S(4,2)
S(2,1)
S(0,0)31
S(3,1)
S(1,0) 29.90
S(4,1)
S(2,0)
S(3,0)
S(4,0)
4
0
2
3
1
62
The Lattice Layout

• One nice thing is that I can start to create some
  formulas that will help me easily calculate the
  S(t,j) values.
• To see this, consider S(2,0). Now, to get to
  S(2,0), one starts at S(0,0) and makes two down
  jumps, that is S(2,0) S(0,0)dd
• Of course, S(0,0) d S(1,0), so we can rewrite
  this as
• S(2,0) S(1,0) d.
• In this case, we can see that since S(1,0)
  29.90, that S(2,0) 29.90 0.96455 28.84

63
The Lattice Layout
S(4,4)
S(3,3)
S(4,3)
S(2,2)
S(3,2)
S(1,1)32.14
S(4,2)
S(2,1)
S(0,0)31
S(3,1)
S(1,0) 29.90
S(4,1)
S(2,0)S(1,0)d 29.90 0.96455 28.84
S(3,0)
S(4,0)
4
0
2
3
1
64
The Lattice Layout

• Similarly, one can figure out the value of S(2,1)
  by multiplying S(1,0) by u S(2,1) S(1,0) u
  29.90 1.0367 31.00
• Indeed, from any node S(t,j) one can easily
  determine S(t1,j) and S(t1,j1), by these
  simple formulas
• S(t1,j) S(t,j) d
• and
• S(t1,j1) S(t,j)u

65
The Lattice Layout

• Indeed, lets use these formulas from node (1,1).
  Recall that S(1,1) 32.14, so
• S(2,1) S(1,1)d 32.14 .96455 31.00
• S(2,2) S(1,1)u 32.14 1.0367 33.32
• Notice that you can get S(2,1) either by S(2,1)
  S(1,1) d
• Or by
• S(2,1) S(1,0) u
• So now we can lay out the second full time step

66
The Lattice Layout
S(4,4)
S(3,3)
S(4,3)
S(2,2)33.32
S(3,2)
S(1,1)32.14
S(4,2)
S(2,1)31.00
S(0,0)31
S(3,1)
S(1,0) 29.90
S(4,1)
S(2,0) 28.84
S(3,0)
S(4,0)
4
0
2
3
1
67
The Lattice Layout

• Indeed, laying out the next time step becomes
  relatively simple
• S(3,0) S(2,0)d 28.84 0.9646 27.82
• S(3,1) S(2,0)U 28.84 1.0367 29.90
• S(3,2) S(2,1)U 31.00 1.0367 32.14
• S(3,3) S(2,2)U 33.32 1.0367 34.54

68
The Lattice Layout
S(4,4)
S(3,3)34.54
S(4,3)
S(2,2)33.32
S(3,2)32.14
S(1,1)32.14
S(4,2)
S(2,1)31.00
S(0,0)31
S(3,1)29.09
S(1,0) 29.90
S(4,1)
S(2,0) 28.84
S(3,0)27.82
S(4,0)
4
0
2
3
1
69
The Lattice Layout

• And we can then follow the same procedure for the
  last time step
• S(4,0) S(3,0)d 27.82 0.9646 26.83
• S(4,1) S(3,0)u 27.82 1.0367 28.84
• S(4,2) S(3,1)u 29.90 1.0367 31.00
• S(4,3) S(3,2)u 32.14 1.0367 33.32
• S(4,4) S(3,3)u 34.54 1.0367 35.81

70
The Lattice Layout
S(4,4)35.81
S(3,3)34.54
S(4,3)33.32
S(2,2)33.32
S(3,2)32.14
S(1,1)32.14
S(4,2)31.00
S(2,1)31.00
S(0,0)31
S(3,1)29.09
S(1,0) 29.90
S(4,1)28.84
S(2,0) 28.84
S(3,0)27.82
S(4,0)26.83
4
0
2
3
1
71
Valuation at Terminal Step

• Now that we have laid out our lattice, we can
  begin to consider pricing the option.
• We do this through a process known as backwards
  induction. This is just a fancy way of saying
  that we begin at the end (right side) of the
  lattice and work our way toward the beginning
  (the left side.)
• Why do it this way? Because, we can easily
  identify the options value at each of the
  terminal nodes (nodes (4,0) through (4,4). We
  know this from the basic (call) option payoff
  formula
• c max(0,ST-K)

72
Valuation at Terminal Step

• So we can now discuss the value of the options at
  each of the terminal nodes. Let C(t,j) denote the
  value of the option at node (t,j), and realize
  that at the terminal node time step
• c(t,j) max(0,S(t,j)-K) (for t4)
• So, lets examine node (4,0). Since
• S(4,0) 26.83, it must be the case that
• c(4,0) max(0,26.84-30) 0
• At node (4,4), however, we see that
• S(4,4) 35.81, and so
• c(4,4) max(0,35.81 30)
• Indeed, we can value the option at each node

73
Terminal Option values
S(4,4)35.81
c(4,4)max(0,35.81-30)5.81
S(3,3)34.54
S(4,3)33.32
S(2,2)33.32
c(4,3)max(0,33.32-30)3.32
S(3,2)32.14
S(1,1)32.14
S(4,2)31.00
S(2,1)31.00
S(0,0)31
c(4,2)max(0,31.00-30) 1.00
S(3,1)29.09
S(1,0) 29.90
S(4,1)28.84
S(2,0) 28.84
c(4,1)max(0,28.84-30) 0
S(3,0)27.82
S(4,0)26.83
c(4,0)max(0,26.83-30) 0
4
0
2
3
1
74
Interior Option Values

• So how do we proceed once we have the terminal
  option values? Assuming that it is an
  European-style option, meaning that you can only
  exercise the option at its expiration date, its
  really simple.
• All you have to do at a given node (t,j), is to
  discount the expected value of the option at the
  two adjacent nodes (t1,j) and (t1,j1) at the
  risk free rate.
• How do you get the expected value at the next
  time step? Since p is the probability of an up
  jump, just multiply c(t1,j1) by p, and c(t1,j)
  by (1-p).
• Thus, the actual formula to use is

75
Interior Option Values

• Thus, we can calculate the options value at each
  node on time step 3 (remember that dt .0208333
  and p 0.493866)
• This lets us fill the lattice for level 3

76
Call Option values
S(4,4)35.81
c(4,4)5.81
S(3,3)34.54c(3,3)4.550
S(4,3)33.32
S(2,2)33.32
c(4,3)3.32
S(3,2)32.14 c(3,2)2.145
S(1,1)32.14
S(4,2)31.00
S(2,1)31.00
S(0,0)31
c(4,2)1.00
S(3,1)29.09 c(3,1).49376
S(1,0) 29.90
S(4,1)28.84
S(2,0) 28.84
c(4,1)0
S(3,0)27.82 c(3,0)0
S(4,0)26.83
c(4,0)0
4
0
2
3
1
77
Interior Option Values

• We can then repeat this process for time step 2,
• And fill in our lattice

78
Call Option values
S(4,4)35.81
c(4,4)5.81
S(3,3)34.54c(3,3)4.550
S(4,3)33.32
S(2,2)33.32 c(2,2)3.33
c(4,3)3.32
S(3,2)32.14 c(3,2)2.145
S(1,1)32.14
S(4,2)31.00
S(2,1)31.00 c(2,1)1.309
S(0,0)31
c(4,2)1.00
S(3,1)29.09 c(3,1).49376
S(1,0) 29.90
S(4,1)28.84
S(2,0) 28.84 c(2,0) 0.244
c(4,1)0
S(3,0)27.82 c(3,0)0
S(4,0)26.83
c(4,0)0
4
0
2
3
1
79
Interior Option Values

• We can then repeat this process for time step 2,
• now solving for time step 1,
• and filling in the lattice

80
Call Option values
S(4,4)35.81
c(4,4)5.81
S(3,3)34.54c(3,3)4.550
S(4,3)33.32
S(2,2)33.32 c(2,2)3.33
c(4,3)3.32
S(3,2)32.14 c(3,2)2.145
S(1,1)32.14 c(1,0)2.308
S(4,2)31.00
S(2,1)31.00 c(2,1)1.309
S(0,0)31
c(4,2)1.00
S(3,1)29.09 c(3,1).49376
S(1,0) 29.90 c(1,0)0.7698
S(4,1)28.84
S(2,0) 28.84 c(2,0) 0.244
c(4,1)0
S(3,0)27.82 c(3,0)0
S(4,0)26.83
c(4,0)0
4
0
2
3
1
81
Interior Option Values

• We can then repeat this process for time step 2,
• Followed by time step 1,
• And finally, we solve for time step 0 to
  determine the current option value!

82
Call Option values
S(4,4)35.81
c(4,4)5.81
S(3,3)34.54c(3,3)4.550
S(4,3)33.32
S(2,2)33.32 c(2,2)3.33
c(4,3)3.32
S(3,2)32.14 c(3,2)2.145
S(1,1)32.14 c(1,0)2.308
S(4,2)31.00
S(2,1)31.00 c(2,1)1.309
S(0,0)31 c(0,0)1.529
c(4,2)1.00
S(3,1)29.09 c(3,1).49376
S(1,0) 29.90 c(1,0)0.7698
S(4,1)28.84
S(2,0) 28.84 c(2,0) 0.244
c(4,1)0
S(3,0)27.82 c(3,0)0
S(4,0)26.83
c(4,0)0
4
0
2
3
1
83
Terminal Option Value

• Thus we say that the option value is 1.529.
• As mentioned earlier, the above process prices a
  European Call option. What would change if it
  were a European put option?
• The only significant change would be that your
  terminal boundary condition would become p(t,j)
  max(0,K-S(t,j))
• The following chart demonstrates this using the
  same lattice and a put with a strike of 35.

84
Put Option values
S(4,4)35.81
p(4,4)0
S(3,3)34.54p(3,3)0.85
S(4,3)33.32
S(2,2)33.32 p(2,2)1.86
p(4,3)1.68
S(3,2)32.14 p(3,2)2.85
S(1,1)32.14 p(1,0)2.94
S(4,2)31.00
S(2,1)31.00 p(2,1)3.99
S(0,0)31 p(0,0)4.02
p(4,2)4.00
S(3,1)29.09 p(3,1)5.09
S(1,0) 29.90 p(1,0)5.08
S(4,1)28.84
S(2,0) 28.84 p(2,0) 6.14
p(4,1)6.16
S(3,0)27.82 p(3,0)7.17
S(4,0)26.83
p(4,0)8.17
4
0
2
3
1
85
Terminal Option Value

• So this option has a value of 4.02.
• What if the option were an American-style option,
  what would change?
• You would have to determine at each node whether
  or not it is optimal for the option-holder to
  exercise their option early. You do this by
  comparing the options value against its
  intrinsic value and taking the higher of the two.
  In essence your formula would become

86
Put Option Values

• Now it turns out that you will never exercise a
  call option early on a non-dividend paying stock,
  but you might exercise early a put option on the
  same stock.
• Lets redo the same put option as the last time,
  but now allowing early exercise.
• The terminal points are exactly the same, so it
  is only the interior points that matter.
• Lets begin by looking at time step 3, then.
• Here is the lattice right before we begin our
  analysis.

87
Put Option values
S(4,4)35.81
p(4,4)0
S(3,3)34.54
S(4,3)33.32
S(2,2)33.32
p(4,3)1.68
S(3,2)32.14
S(1,1)32.14
S(4,2)31.00
S(2,1)31.00
S(0,0)31
p(4,2)4.00
S(3,1)29.09
S(1,0) 29.90
S(4,1)28.84
S(2,0) 28.84
p(4,1)6.16
S(3,0)27.82
S(4,0)26.83
p(4,0)8.17
4
0
2
3
1
88
Put Option Values

• And so we can now begin to analyze put options at
  each node on level 3.
• Just to illustrate one such node, consider p(3,0)
• Which means that it is exercised early.

89
Put Option values
S(4,4)35.81
p(4,4)0
S(3,3)34.54p(3,3)0.85
S(4,3)33.32
S(2,2)33.32 p(2,2)1.87
p(4,3)1.68
S(3,2)32.14 p(3,2)2.86
S(1,1)32.14 p(1,0)2.95
S(4,2)31.00
S(2,1)31.00 p(2,1)4.00
S(0,0)31 p(0,0)4.04
p(4,2)4.00
S(3,1)29.09 p(3,1)5.10
S(1,0) 29.90 p(1,0)5.10
S(4,1)28.84
S(2,0) 28.84 p(2,0) 6.16
p(4,1)6.16
S(3,0)27.82 p(3,0)7.18
S(4,0)26.83
p(4,0)8.17
4
0
2
3
1
90
A Shorthand for the Stock Price

• So far we have manually built our trees, that is
  we start at time 0 with S0 and then at each level
  we multiply the stock price at the previous node
  by u and d to get the current level.
• Its actually easier to realize that the value of
  S at a particular node (t,j) the value of the
  stock is given by
• Just remember that on any level t, the
  bottom-most node is node (t,0), and the top-most
  node is (t,t).

91
A Shorthand for the Stock Price

• The next page shows the first three levels from
  the example that we have been using.
• Recall that S031, and that u 1.0367, and that
  d0.96455.

92
A Shorthand for the Stock Price
S2,2S0u2d(2-2) S2,2 31(1.0367)2(0.96455)0 S2,2
33.32
S1,1S0u1d(1-1) S1,1 31(1.0367)1(0.96455)0 S1,1
32.14
S2,1S0u1d(2-1) S2,1 31(1.0367)1(0.96455)1 S2,1
31
S0,031
S1,0S0u0d(1-0) S1,0 31(1.0367)0(0.96455)1 S1,0
29.90
S2,0S0u0d(2-0) S2,0 31(1.0367)0(0.96455)2 S2,0
28.84
0
2
1
93
McDonalds Forward Lattice

• McDonald (in Chapter 10) uses a different way of
  building the lattice. Instead of working with the
  stock price directly, he works with the forward
  price of the stock.
• He begins with the question of what would happen
  if there were no uncertainty regarding the future
  stock price.
• The stock would have to have a total return equal
  to the risk free rate, so the growth rate in the
  stock price would be equal to the risk-free rate,
  less any dividend yield the stock paid.

94
McDonalds Forward Lattice

• He then modifies this to permit uncertainty. He
  defines s to be the annualized standard deviation
  the continuously compounded return.
• This allows him to model the stock price
  evolution as
• Which can be simplified to

95
McDonalds Forward Lattice

• Aside from changing the manner in which you
  calculate u and d, there is no change in the
  process for calculating the option price.
• An interesting question is, how much difference
  is there in prices between these two lattices?
• Answer not much, really.
• Example 3 month European call option with a
  K80, S080, s25, r4.
• With 1 step per month CRR 4.712 McDonald
  4.702
• With 2 steps per month CRR 4.216 McDonald
  4.328
• With 3 steps per month CRR 4.488 McDonald
  4.476

96
McDonalds Forward Lattice

• Indeed, this raises an interesting point, what
  happens to these two models as you increase the
  number of steps per month, i.e. as you decrease
  dt?
• They both converge toward a limit (which is the
  Black-Scholes price.)
• The following graphs illustrate this.

97
McDonalds Forward Lattice
98
McDonalds Forward Lattice
99
McDonalds Forward Lattice
100
McDonalds Forward Lattice
101
Dividend Yields

• A nice feature of the way in which McDonald
  develops the forward tree, is that he already has
  built into it how to handle dividend yields. In
  the formulae for d and u we have
• where d is the dividend yield on the stock.
• The logic of this is exactly the same as in the
  case of a forward contract in a risk-neutral
  world, all assets return the risk-free rate.
  Return is comprised of two components price
  appreciation and the dividend (if any). If the
  asset pays a dividend yield, its price
  appreciation must be reduced by that amount.

102
Dividend Yields

• It is just as simple to adjust the CRR model for
  dividend yields, simply define u and d as usual,
  but define p slightly differently (Hull presents
  this on page 269-270 in Chapter 13.)

103
Dividend Yields

• Example Consider a 2 month European call option
  on a stock index currently at 50. The strike is
  at 51, the volatility is 30, the risk-free rate
  is 5, and the dividend yield is 2. Assuming a
  monthly time step, what is the price of the
  option?
• First, calculate u, d, and p.

104
Dividend Yield

• The next step, of course, is to build the
  lattice, and to price the option backwards
  through the lattice

59.91 cuumax(0,59.91-51)8.91
54.73 cu(0.481238.910.5180)e-.05(1/12)4.27
50.0 cudmax(0,50-51)0
50 c0(0.481234.270.5180)e-.05(1/12)2.046
45.68 cd 0
41.73 cddmax(0,41.73)0
105
Dividend Yield

• The dividend yield-based model is particularly
  useful for options on stock indices.
• It is also used for foreign currency options
  treat the current exchange rate of the foreign
  currency as a stock with a dividend yield equal
  to the risk-free rate of the foreign currency,
  and the domestic risk-free rate has the same role
  as the risk-free rate in the standard binomial
  model.

106
Discrete Dividends

• In some sense, however, these are really special
  cases. Normally we are concerned with the way
  individual stocks pay dividends, which is
  discretely. We will examine two methods for
  valuing options on stocks that pay a discrete
  dividend
• Assume that the stock will pay a discrete
  dividend in the future that is proportional to
  the stocks price. This is, in essence, a
  discrete dividend yield model.
• Assume that the stock will pay a fixed dollar
  amount in dividends, that is independent of the
  price of the stock on the ex-dividend date.
• Of the two, the dollar dividend is the more
  difficult case.

107
Discrete Dividends

• Hull discusses how to deal with the discrete
  dividend yield case on pages 402-403 in chapter
  18. You are responsible for this material.
• The dividend yield model is really pretty simple
  to handle in the general method we currently use.
  
• Assume that on the ex-dividend date (d) the
  company will pay a dividend equal to dSd.
• In the period prior to the dividend, simply use
  the standard binomial model (i.e. CRR) to
  determine the stock price, i.e. S0ujdt-j.
• In the period after the dividend, modify the
  formula such that the stock price is given by
  S0(1-d)ujdt-j.

108
Discrete Dividends

• So lets work an example of this. Assume that we
  have a stock with a price of 45, that will pay a
  dividend of 2 between time 1 and 2 months. How
  much is a 3 month American call option with K51
  worth today, given r.05 and s.30, and dt1/12?
• First, calculate u, d, and p as normal

109
Discrete Dividends

• Next, calculate the stock prices. For nodes
  before time step 2, use the formula
• And for time step 2 and above, use the formula
• We can calculate the value of the stock at each
  node

110
Discrete Dividends
111
Discrete Dividends
S3,3 63.54
S1,154.52
S2,1 58.27
S3,2 53.43
S0,050
S2,1 49.00
S1,0 45.85
S3,1 44.94
S2,0 41.21
S3,0 37.79
0
2
3
1
112
Discrete Dividends

• Using backwards induction

S3,3 63.54 C3,3max(0,63.54-45)18.54
S1,154.52
S2,1 58.27 C2,2e-.05/12(.502418.54.49768.43)
C2,213.45
S3,2 53.43 C3,2max(0,53.43-45)8.43
S0,050
S2,1 49.00 C2,1e-.05/12(.50248.43) C2,14.22
S1,0 45.85
S3,1 44.94 C3,1max(0,44.94-45)0
S2,0 41.21 C2,00
S3,0 37.79 C3,0max(0,37.79-55)0
0
2
3
1
113
Discrete Dividends

• At time 1 we must also check for optimal early
  exercise!

S1,154.52 C1,1max(54.52-45,e-.05/12(.502413.45
.49764.22)) C1,1max(9.52,8.85)9.52 C1,19.52
S3,3 63.54 C3,318.54
S2,1 58.27 C2,213.45
S3,2 53.43 C3,28.43
S0,050
S2,1 49.00 C2,14.22
S1,0 45.85 C1,0max(45.85-45,e-.05/12(.50244
.22.49760)) C1,0max(0.85,2.11) C1,02.11
S3,1 44.94 C3,10
S2,0 41.21 C2,00
S3,0 37.79 C3,00
0
2
3
1
114
Discrete Dividends

• We also must check it at time 0.

S3,3 63.54 C3,312.54
S1,154.52 C1,19.52
S2,1 58.27 C2,27.46
S3,2 53.43 C3,22.43
S0,050 C0,05.81 C0,0max(50-45,e-.05/12(.5
0249.52.49762.11)) C0,0max(5,5.81)
S2,1 49.00 C2,11.22
S1,0 45.85 C1,02.11
S3,1 44.94 C3,10
S2,0 41.21 C2,00
S3,0 37.79 C3,00
0
2
3
1
115
Discrete Dividends

• So we can see that if we have a discrete dividend
  that is proportional to the stock price, that its
  relatively straightforward to model it in a
  binomial.
• Things are not quite a simple if we are modeling
  a discrete dividend that is a dollar amount, and
  not a proportional amount of the price.
• One approach is to basically carry on as we did
  in the proportional case, that is, at the first
  time-step after the ex-dividend date reduce the
  stock price by the amount of the dividend, in
  this case called D, and then let the new stock
  price rise or fall by u or d each period
  thereafter.

116
Discrete Dividends

• The difficulty is that in this case the tree no
  longer will recombine this is a down jump
  followed by an up jump will no longer get you to
  the same point as an up jump followed by a down
  jump.
• This is not really a problem from a conceptual
  point of view, but it can be a problem from a
  practical implementation, as the number of nodes
  you have analyze will jump very quickly.
• Example Use the same basic setup as in the last
  example, but now assume the dividend is equal to
  2 instead of 2, although it still occurs
  between time steps 1 and 2.

117
Discrete Dividends
The first two time steps are the same, but there
is a difference at step 2.
S1,154.52
S2,1 57.45
S0,050
S2,1 48.00
S1,0 45.85
S2,0 40.05
0
2
3
1
118
Discrete Dividends

• At time step 3, things become much more
  difficult. Consider that if you were at node 2,2,
  where S2,257.45 that at the next time step the
  up and down values would be
• S2,2 S2,2u 62.65, and S-2,2S2,2d52.68
• Next, consider node 2,1, where S2,148.00, the
  next period up and down values would be
• S2,1 S2,1u 52.34, and S-2,1S2,1d44.02
• Notice that S2,1 is not the same as S-2,1.
• In fact, the entire lattice would appear as

119
Discrete Dividends
The first two time steps are the same, but there
is a difference at steps 2 3.
S2,2 62.65
S1,154.52
S2,2 57.45
S-2,2 52.69
S2,1 52.34
S0,050
S2,1 48.00
S1,0 45.85
S-2,1 44.02
S2,0 43.67
S2,0 40.05
S-2,0 36.73
0
2
3
1
120
Discrete Dividends

• Once you get the lattice set up, however,
  determining the options price is relatively
  straightforward.
• You have to make sure that you price the option
  using the correct branches of each part of the
  tree.

121
Discrete Dividends
As usual, start at the terminal time step and
work backwards
S2,2 62.65 C2,2max(0,62.65-45)17.65
S-2,2 52.69 C-2,2max(0,52.69-45)7.69
S1,154.52
S2,2 57.45
S2,1 52.34 C2,1max(0,52.34-45)7.34
S0,050
S2,1 48.00
S-2,1 44.02 C-2,1max(0,44.02-45)0
S1,0 45.85
S2,0 43.67 C2,0max(0,43.67-45)0
S2,0 40.05
S-2,0 36.73 C-2,0max(0,36.73-45)0
0
2
3
1
122
Discrete Dividends
Proceed backwards as we normally to step 2
S2,2 62.65 C2,2 17.65
S-2,2 52.69 C-2,27.69
S1,154.52
S2,2 57.45 C2,1e-.05/12(.502417.65.49767.69)
C2,112.64
S2,1 52.34 C2,17.34
S0,050
S2,1 48.00 C2,1e-.05/12(.50247.34) C2,13.67
S-2,1 44.02 C-2,10
S1,0 45.85
S2,0 43.67 C2,00
S2,0 40.05 C2,00
S-2,0 36.73 C-2,0 0
0
2
3
1
123
Discrete Dividends
At time 1 we once again have to check for early
exercise
S1,154.52 C1,1max(54.52-45,e-.05/12(.502412.64
.49763.67)) C1,1max(9.52,8.15)9.52 C1,19.52
S2,2 62.65 C2,2 17.65
S-2,2 52.69 C-2,27.69
S2,2 57.45 C2,112.64
S2,1 52.34 C2,17.34
S0,050
S2,1 48.00 C2,13.67
S-2,1 44.02 C-2,10
S1,0 45.85 C1,0max(45.85-45,e-.05/12(.50243
.67.49760)) C1,0max(0.85,1.84) C1,01.84
S2,0 43.67 C2,00
S2,0 40.05 C2,00
S-2,0 36.73 C-2,0 0
0
2
3
1
124
Discrete Dividends
We again check for early exercise at time 0, but
find that it is not optimal.
S2,2 62.65 C2,2 17.65
S1,154.52 C1,19.52
S-2,2 52.69 C-2,27.69
S2,2 57.45 C2,112.64
S2,1 52.34 C2,17.34
S0,050 C0,05.68 C0,0max(50-45,e-.05/12(.5
0249.52.49761.84)) C0,0max(5,5.68)
S2,1 48.00 C2,13.67
S-2,1 44.02 C-2,10
S1,0 45.85 C1,01.84
S2,0 43.67 C2,00
S2,0 40.05 C2,00
S-2,0 36.73 C-2,0 0
0
2
3
1
125
Discrete Dividends

• You can see the problem that you quickly run
  into, however the number of nodes per time step
  doubles for each time step after the dividend.
• You have 3 nodes at time step 2, 6 at time 4, and
  you would have 12 at time step 5, 24 at time step
  6, etc.
• Note that if the tree recombines, the number of
  nodes per time step increases by 1 for each new
  time step 3 nodes for a 2 step tree, 6 nodes for
  a 3 step tree, 10 nodes for a 4 step tree, etc.
• If you were to have a 25 step tree a not
  unusual tree - and if the dividend occurred at
  step 3, then, if the tree did not recombine you
  would have to analyze 50,331,648 nodes on level
  25 alone. If the tree did recombine, you would
  have to analyze 26 nodes on level 25.

126
Discrete Dividends

• So, not surprisingly, researchers seek to develop
  methods for using a recombining tree.
• For the case of discrete dividends, such a method
  does exist.
• The key is to model the dividend as a certain
  cash flow, and the dividend-adjusted stock price
  movements as uncertain.
• This means, in practice, defining S0S0-D0 where
  D0 is the present value of the future dividends.
• You also have to determine a new s, although in
  practice this would normally be given to you.

127
Discrete Dividends

• So, lets go back to our previous example, except
  now we will assume that s.30, and that the
  dividend of 2.00 will be paid in exactly 6
  weeks, so that the present value of the dividend
  is 2.00e-.05(6/52)1.988 at time 0, and
  2.00e-.05(2/62)1.996 at time 1.
• Thus, S050-1.988 48.01. We then model S
  through the lattice (notice that d, u, and p will
  remain the same as in the last example.)
• We can then return to our normal way of
  calculating the lattice of St,j values
  St,jS0ujdt-j, giving this lattice

128
Discrete Dividends
S3,3 62.26
S2,1 57.09
S1,1 48.011.0905 52.53
S3,2 52.35
S2,1 48.01
S0,048.01
S1,0 48.01(.9170) 44.03
S3,1 44.03
S2,0 40.38
S3,0 37.03
0
2
3
1
129
Discrete Dividends

• Of course our option prices are based not on
  St,j, but rather on St,j. To recover St,j we
  must add back the present value of the dividends
  in the time steps before the dividend is paid in
  this case that is when tlt2.
• Recall that at time 0 the present value of the
  dividend is 1.988 and at time 1 the present value
  of the dividend is 1.996.
• We can now model S in a recombining lattice

130
Discrete Dividends
S3,3 S3,3 62.26
S2,2 S2,2 57.09
S1,1S1,1D1 52.351.996 54.35
S3,2 S3,2 52.35
S2,1 S2,1 48.01
S0,0 S0,0D0 48.01 1.988 50.00
S1,0 S1,0D1 44.031.996 46.02
S3,1 S3,1 44.03
S2,0 S2,0 40.38
S3,0 S3,0 37.03
0
2
3
1
131
Discrete Dividends

• We can then just use our normal backwards pricing
  algorithm to determine the price of the call
  but note that once again we must determine if
  there are any optimal early exercises.

132
Discrete Dividends
S3,3 62.26 C3,3max(0,62.26-45)17.36
S2,2 57.09
S1,154.35
S3,2 52.35 C3,3max(0,52.35-45)7.35
S2,1 48.01
S0,0 50.00
S1,046.02
S3,1 44.03 C3,10
S2,0 40.38
S3,0 37.03 C3,00
0
2
3
1
133
Discrete Dividends