Problems with the Bump-and-Revalue Method

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Delta and Common Random Numbers

• In estimating delta, it is natural to start with the finite-difference estimate

e−rτ E[ P (S + ) ] − E[ P (S − ) ]

2 .

– P (x) is the terminal payoff of the derivative security when the underlying asset’s initial price equals x.

• Use simulation to estimate E[ P (S + ) ] first.

• Use another simulation to estimate E[ P (S − ) ].

• Finally, apply the formula to approximate the delta.

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Delta and Common Random Numbers (concluded)

• This method is not recommended because of its high variance.

• A much better approach is to use common random numbers to lower the variance:

e−rτ E

 P (S + ) − P (S − ) 2

 .

• Here, the same random numbers are used for P (S + ) and P (S − ).

• This holds for gamma and cross gammas (for multivariate derivatives).

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Problems with the Bump-and-Revalue Method

• Consider the binary option with payoff

⎧⎨

1, if S(T ) > X, 0, otherwise.

• Then

P (S+)−P (S−) =

⎧⎨

1, if S +  > X and S −  < X, 0, otherwise.

• So the finite-difference estimate per run for the (undiscounted) delta is 0 or O(1/).

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Problems with the Bump-and-Revalue Method (concluded)

• The price of the binary option equals e−rτN (x − σ√

τ ).

• Its delta is

N(x − σ√

τ )/(Sσ√ τ ).

(5)

Gamma

• The finite-difference formula for gamma is e−rτ E

 P (S + ) − 2 × P (S) + P (S − )

2

 .

• For a correlation option with multiple underlying assets, the finite-difference formula for the cross gamma

2P (S1, S2, . . . )/(∂S1∂S2) is:

e−rτ E

 P (S1 + 1, S2 + 2) − P (S1 − 1, S2 + 2)

412

−P (S1 + 1, S2 − 2) +P (S1 − 1, S2 − 2)  .

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Gamma (continued)

• Choosing an  of the right magnitude can be challenging.

– If  is too large, inaccurate Greeks result.

– If  is too small, unstable Greeks result.

• This phenomenon is sometimes called the curse of differentiation.a

aA¨ıt-Sahalia & Lo (1998); Bondarenko (2003).

(7)

Gamma (continued)

• In general, suppose

i

∂θie−rτE[ P (S) ] = e−rτE

 iP (S)

∂θi



holds for all i > 0, where θ is a parameter of interest.

– A common requirement is Lipschitz continuity.a

• Then formulas for the Greeks become integrals.

• As a result, we avoid , finite differences, and resimulation.

aBroadie & Glasserman (1996).

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Gamma (continued)

• This is indeed possible for a broad class of payoff functions.a

– Roughly speaking, any payoff function that is equal to a sum of products of differentiable functions and indicator functions with the right kind of support.

– For example, the payoff of a call is

max(S(T ) − X, 0) = (S(T ) − X)I{ S(T )−X≥0 }. – The results are too technical to cover here (see next

page).

aTeng (R91723054) (2004); Lyuu & Teng (R91723054) (2011).

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Gamma (continued)

• Suppose h(θ, x) ∈ H with pdf f(x) for x and gj(θ, x) ∈ G for j ∈ B, a finite set of natural numbers.

• Then

∂θ



h(θ, x) 

j∈B1{gj (θ,x)>0}(x) f (x) dx

=



hθ (θ, x) 

j∈B1{gj (θ,x)>0}(x) f (x) dx

+ 

l∈ B

⎣h(θ, x)Jl(θ, x) 

j∈B\l1{gj (θ, x)>0}(x) f (x)

x=χl (θ) ,

where

Jl(θ, x) = sign

∂gl(θ, x)

∂xk

∂gl(θ, x)/∂θ

∂gl(θ, x)/∂x for l ∈ B.

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Gamma (concluded)

• Similar results have been derived for Levy processes.a

• Formulas are also recently obtained for credit derivatives.b

• In queueing networks, this is called infinitesimal perturbation analysis (IPA).c

aLyuu, Teng (R91723054), & S. Wang (2013).

bLyuu, Teng (R91723054), & Tzeng (2014).

cCao (1985); Ho & Cao (1985).

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Biases in Pricing Continuously Monitored Options with Monte Carlo

• We are asked to price a continuously monitored up-and-out call with barrier H.

• The Monte Carlo method samples the stock price at n discrete time points t1, t2, . . . , tn.

• A sample path

S(t0), S(t1), . . . , S(tn) is produced.

– Here, t0 = 0 is the current time, and tn = T is the

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Biases in Pricing Continuously Monitored Options with Monte Carlo (continued)

• If all of the sampled prices are below the barrier, this sample path pays max(S(tn) − X, 0).

• Repeating these steps and averaging the payoffs yield a Monte Carlo estimate.

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1: C := 0;

2: for i = 1, 2, 3, . . . , N do

3: P := S; hit := 0;

4: for j = 1, 2, 3, . . . , n do

5: P := P × e(r−σ2/2) (T /n)+σ

(T /n) ξ;

6: if P ≥ H then

7: hit := 1;

8: break;

9: end if

10: end for

11: if hit = 0 then

12: C := C + max(P − X, 0);

13: end if

14: end for

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Biases in Pricing Continuously Monitored Options with Monte Carlo (continued)

• This estimate is biased.a

– Suppose none of the sampled prices on a sample path equals or exceeds the barrier H.

– It remains possible for the continuous sample path that passes through them to hit the barrier between sampled time points (see plot on next page).

aShevchenko (2003).

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H

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Biases in Pricing Continuously Monitored Options with Monte Carlo (concluded)

• The bias can certainly be lowered by increasing the number of observations along the sample path.

• However, even daily sampling may not suffice.

• The computational cost also rises as a result.

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Brownian Bridge Approach to Pricing Barrier Options

• We desire an unbiased estimate which can be calculated efficiently.

• The above-mentioned payoff should be multiplied by the probability p that a continuous sample path does not hit the barrier conditional on the sampled prices.

• This methodology is called the Brownian bridge approach.

• Formally, we have

p = Prob[ S(t) < H, 0Δ ≤ t ≤ T | S(t0), S(t1), . . . , S(tn) ].

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Brownian Bridge Approach to Pricing Barrier Options (continued)

• As a barrier is hit over a time interval if and only if the maximum stock price over that period is at least H,

p = Prob



0≤t≤Tmax S(t) < H | S(t0), S(t1), . . . , S(tn)

 .

• Luckily, the conditional distribution of the maximum over a time interval given the beginning and ending stock prices is known.

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Brownian Bridge Approach to Pricing Barrier Options (continued)

Lemma 23 Assume S follows dS/S = μ dt + σ dW and define ζ(x) = expΔ



2 ln(x/S(t)) ln(x/S(t + Δt)) σ2Δt

 . (1) If H > max(S(t), S(t + Δt)), then

Prob



t≤u≤t+Δtmax S(u) < H 

 S(t),S(t + Δt)

= 1 − ζ(H).

(2) If h < min(S(t), S(t + Δt)), then Prob



t≤u≤t+Δtmin S(u) > h

 S(t),S(t + Δt)

= 1 − ζ(h).

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Brownian Bridge Approach to Pricing Barrier Options (continued)

• Lemma 23 gives the probability that the barrier is not hit in a time interval, given the starting and ending stock prices.

• For our up-and-out call,a choose n = 1.

• As a result,

p =

1 − exp

2 ln(H/S(0)) ln(H/S(T )) σ2T

, if H > max(S(0), S(T )),

0, otherwise.

aSo S(0) < H.

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Brownian Bridge Approach to Pricing Barrier Options (continued)

The following algorithms works for up-and-out and down-and-out calls.

1: C := 0;

2: for i = 1, 2, 3, . . . , N do

3: P := S × e(r−q−σ2/2) T +σ

T ξ( )

;

4: if (S < H and P < H) or (S > H and P > H) then

5: C := C+max(P −X, 0)×



1 − exp

2 ln(H/S)×ln(H/P ) σ2T



;

6: end if

7: end for

8: return Ce−rT/N ;

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Brownian Bridge Approach to Pricing Barrier Options (concluded)

• The idea can be generalized.

• For example, we can handle more complex barrier options.

• Consider an up-and-out call with barrier Hi for the time interval (ti, ti+1 ], 0 ≤ i < n.

• This option thus contains n barriers.

• Multiply the probabilities for the n time intervals to obtain the desired probability adjustment term.

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Variance Reduction

• The statistical efficiency of Monte Carlo simulation can be measured by the variance of its output.

• If this variance can be lowered without changing the expected value, fewer replications are needed.

• Methods that improve efficiency in this manner are called variance-reduction techniques.

• Such techniques become practical when the added costs are outweighed by the reduction in sampling.

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Variance Reduction: Antithetic Variates

• We are interested in estimating E[ g(X1, X2, . . . , Xn) ].

• Let Y1 and Y2 be random variables with the same distribution as g(X1, X2, . . . , Xn).

• Then

Var

 Y1 + Y2 2



= Var[ Y1 ]

2 + Cov[ Y1, Y2 ]

2 .

– Var[ Y1 ]/2 is the variance of the Monte Carlo method with two independent replications.

• The variance Var[ (Y1 + Y2)/2 ] is smaller than

Var[ Y1 ]/2 when Y1 and Y2 are negatively correlated.

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Variance Reduction: Antithetic Variates (continued)

• For each simulated sample path X, a second one is obtained by reusing the random numbers on which the first path is based.

• This yields a second sample path Y .

• Two estimates are then obtained: One based on X and the other on Y .

• If N independent sample paths are generated, the antithetic-variates estimator averages over 2N

estimates.

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Variance Reduction: Antithetic Variates (continued)

• Consider process dX = at dt + bt

dt ξ.

• Let g be a function of n samples X1, X2, . . . , Xn on the sample path.

• We are interested in E[ g(X1, X2, . . . , Xn) ].

• Suppose one simulation run has realizations

ξ1, ξ2, . . . , ξn for the normally distributed fluctuation term ξ.

• This generates samples x1, x2, . . . , xn.

• The estimate is then g(x), where x = (xΔ 1, x2 . . . , xn).

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Variance Reduction: Antithetic Variates (concluded)

• The antithetic-variates method does not sample n more numbers from ξ for the second estimate g(x).

• Instead, generate the sample path x Δ= (x1, x2 . . . , xn) from −ξ1,−ξ2, . . . ,−ξn.

• Compute g(x).

• Output (g(x) + g(x))/2.

• Repeat the above steps for as many times as required by accuracy.

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Variance Reduction: Conditioning

• We are interested in estimating E[ X ].

• Suppose here is a random variable Z such that

E[ X | Z = z ] can be efficiently and precisely computed.

• E[ X ] = E[ E[ X | Z ] ] by the law of iterated conditional expectations.

• Hence the random variable E[ X | Z ] is also an unbiased estimator of E[ X ].

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Variance Reduction: Conditioning (concluded)

• As

Var[ E[ X | Z ] ] ≤ Var[ X ],

E[ X | Z ] has a smaller variance than observing X directly.

• First obtain a random observation z on Z.

• Then calculate E[ X | Z = z ] as our estimate.

– There is no need to resort to simulation in computing E[ X | Z = z ].

• The procedure can be repeated a few times to reduce

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Control Variates

• Use the analytic solution of a similar yet simpler problem to improve the solution.

• Suppose we want to estimate E[ X ] and there exists a random variable Y with a known mean μ = E[ Y ].Δ

• Then W = X + β(YΔ − μ) can serve as a “controlled”

estimator of E[ X ] for any constant β.

– However β is chosen, W remains an unbiased estimator of E[ X ] as

E[ W ] = E[ X ] + βE[ Y − μ ] = E[ X ].

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Control Variates (continued)

• Note that

Var[ W ] = Var[ X ] + β2 Var[ Y ] + 2β Cov[ X, Y ],

(110)

• Hence W is less variable than X if and only if

β2 Var[ Y ] + 2β Cov[ X, Y ] < 0. (111)

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Control Variates (concluded)

• The success of the scheme clearly depends on both β and the choice of Y .

– For example, arithmetic average-rate options can be priced by choosing Y to be the otherwise identical geometric average-rate option’s price and β = −1.

• This approach is much more effective than the antithetic-variates method.

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Choice of Y

• In general, the choice of Y is ad hoc,a and experiments must be performed to confirm the wisdom of the choice.

• Try to match calls with calls and puts with puts.b

• On many occasions, Y is a discretized version of the derivative that gives μ.

– Discretely monitored geometric average-rate option vs. the continuously monitored geometric

average-rate option given by formulas (50) on p. 401.

aBut see Dai (B82506025, R86526008, D8852600), Chiu (R94922072),

& Lyuu (2015).

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Optimal Choice of β

• For some choices, the discrepancy can be significant, such as the lookback option.a

• Equation (110) on p. 826 is minimized when β = −Cov[ X, Y ]/Var[ Y ].

– It is called beta in the book.

• For this specific β,

Var[ W ] = Var[ X ] Cov[ X, Y ]2

Var[ Y ] = 

1 − ρ2X,Y 

Var[ X ], where ρX,Y is the correlation between X and Y .

aContributed by Mr. Tsai, Hwai (R92723049) on May 12, 2004.

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Optimal Choice of β (continued)

• Note that the variance can never be increased with the optimal choice.

• Furthermore, the stronger X and Y are correlated, the greater the reduction in variance.

• For example, if this correlation is nearly perfect (±1), we could control X almost exactly.

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Optimal Choice of β (continued)

• Typically, neither Var[ Y ] nor Cov[ X, Y ] is known.

• Therefore, we cannot obtain the maximum reduction in variance.

• We can guess these values and hope that the resulting W does indeed have a smaller variance than X.

• A second possibility is to use the simulated data to estimate these quantities.

– How to do it efficiently in terms of time and space?

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Optimal Choice of β (concluded)

• Observe that −β has the same sign as the correlation between X and Y .

• Hence, if X and Y are positively correlated, β < 0, then X is adjusted downward whenever Y > μ and upward otherwise.

• The opposite is true when X and Y are negatively correlated, in which case β > 0.

• Suppose a suboptimal β +  is used instead.

• The variance increases by only 2Var[ Y ].a

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A Pitfall

• A potential pitfall is to sample X and Y independently.

• In this case, Cov[ X, Y ] = 0.

• Equation (110) on p. 826 becomes

Var[ W ] = Var[ X ] + β2 Var[ Y ].

• So whatever Y is, the variance is increased!

• Lesson: X and Y must be correlated.

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Problems with the Monte Carlo Method

• The error bound is only probabilistic.

• The probabilistic error bound of

N does not benefit from regularity of the integrand function.

• The requirement that the points be independent random samples are wasteful because of clustering.

• In reality, pseudorandom numbers generated by completely deterministic means are used.

• Monte Carlo simulation exhibits a great sensitivity on the seed of the pseudorandom-number generator.

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Matrix Computation

(41)

To set up a philosophy against physics is rash;

philosophers who have done so have always ended in disaster.

— Bertrand Russell

(42)

Definitions and Basic Results

• Let A = [ aΔ ij ]1≤i≤m,1≤j≤n, or simply A ∈ Rm×n, denote an m × n matrix.

• It can also be represented as [ a1, a2, . . . , an ] where ai ∈ Rm are vectors.

– Vectors are column vectors unless stated otherwise.

• A is a square matrix when m = n.

• The rank of a matrix is the largest number of linearly independent columns.

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Definitions and Basic Results (continued)

• A square matrix A is said to be symmetric if AT = A.

• A real n × n matrix

A = [ aΔ ij ]i,j is diagonally dominant if | aii | >

j=i | aij | for 1 ≤ i ≤ n.

– Such matrices are nonsingular.

• The identity matrix is the square matrix I = diag[ 1, 1, . . . , 1 ].Δ

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Definitions and Basic Results (concluded)

• A matrix has full column rank if its columns are linearly independent.

• A real symmetric matrix A is positive definite if xTAx =

i,j

aijxixj > 0 for any nonzero vector x.

• A matrix A is positive definite if and only if there exists a matrix W such that A = WTW and W has full

column rank.

(45)

Cholesky Decomposition

• Positive definite matrices can be factored as A = LLT,

called the Cholesky decomposition.

– Above, L is a lower triangular matrix.

(46)

Generation of Multivariate Distribution

• Let x = [ xΔ 1, x2, . . . , xn ]T be a vector random variable with a positive definite covariance matrix C.

• As usual, assume E[ x ] = 0.

• This covariance structure can be matched by P y.

– C = P PT is the Cholesky decomposition of C.a – y = [ yΔ 1, y2, . . . , yn ]T is a vector random variable

with a covariance matrix equal to the identity matrix.

aWhat if C is not positive definite? See Lai (R93942114) & Lyuu (2007).

(47)

Generation of Multivariate Normal Distribution

• Suppose we want to generate the multivariate normal distribution with a covariance matrix C = P PT.

– First, generate independent standard normal distributions y1, y2, . . . , yn.

– Then

P [ y1, y2, . . . , yn ]T has the desired distribution.

– These steps can then be repeated.

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Multivariate Derivatives Pricing

• Generating the multivariate normal distribution is essential for the Monte Carlo pricing of multivariate derivatives (pp. 748ff).

• For example, the rainbow option on k assets has payoff max(max(S1, S2, . . . , Sk) − X, 0)

at maturity.

• The closed-form formula is a multi-dimensional integral.a

aJohnson (1987); Chen (D95723006) & Lyuu (2009).

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Multivariate Derivatives Pricing (concluded)

• Suppose dSj/Sj = r dt + σj dWj, 1 ≤ j ≤ k, where C is the correlation matrix for dW1, dW2, . . . , dWk.

• Let C = P PT.

• Let ξ consist of k independent random variables from N (0, 1).

• Let ξ = P ξ.

• Similar to Eq. (109) on p. 791, Si+1 = Sie(r−σ2j/2) Δt+σj

Δt ξj, 1 ≤ j ≤ k.

(50)

Least-Squares Problems

• The least-squares (LS) problem is concerned with

x∈Rminn  Ax − b , where A ∈ Rm×n, b ∈ Rm, and m ≥ n.

• The LS problem is called regression analysis in statistics and is equivalent to minimizing the mean-square error.

• Often written as

Ax = b.

(51)

Polynomial Regression

• In polynomial regression, x0 + x1x + · · · + xnxn is used to fit the data { (a1, b1), (a2, b2), . . . , (am, bm)}.

• This leads to the LS problem,

⎢⎢

⎢⎢

⎢⎢

1 a1 a21 · · · an1 1 a2 a22 · · · an2 ... ... ... . .. ... 1 am a2m · · · anm

⎥⎥

⎥⎥

⎥⎥

⎢⎢

⎢⎢

⎢⎢

x0 x1 ... xn

⎥⎥

⎥⎥

⎥⎥

=

⎢⎢

⎢⎢

⎢⎢

b1 b2 ... bm

⎥⎥

⎥⎥

⎥⎥

.

• Consult p. 273 of the textbook for solutions.

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American Option Pricing by Simulation

• The continuation value of an American option is the conditional expectation of the payoff from keeping the option alive now.

• The option holder must compare the immediate exercise value and the continuation value.

• In standard Monte Carlo simulation, each path is treated independently of other paths.

• But the decision to exercise the option cannot be reached by looking at one path alone.

(53)

The Least-Squares Monte Carlo Approach

• The continuation value can be estimated from the cross-sectional information in the simulation by using least squares.a

• The result is a function (of the state) for estimating the continuation values.

• Use the function to estimate the continuation value for each path to determine its cash flow.

• This is called the least-squares Monte Carlo (LSM) approach.

a

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The Least-Squares Monte Carlo Approach (concluded)

• The LSM is provably convergent.a

• The LSM can be easily parallelized.b

– Partition the paths into subproblems and perform LSM on each of them independently.

– The speedup is close to linear (i.e., proportional to the number of cores).

• Surprisingly, accuracy is not affected.

aCl´ement, Lamberton, & Protter (2002); Stentoft (2004).

bHuang (B96902079, R00922018) (2013); Chen (B97902046, R01922005) (2014); Chen (B97902046, R01922005), Huang (B96902079, R00922018) & Lyuu (2015).

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A Numerical Example

• Consider a 3-year American put on a non-dividend-paying stock.

• The put is exercisable at years 0, 1, 2, and 3.

• The strike price X = 105.

• The annualized riskless rate is r = 5%.

• The current stock price is 101.

– The annual discount factor hence equals 0.951229.

• We use only 8 price paths to illustrate the algorithm.

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A Numerical Example (continued)

Stock price paths

Path Year 0 Year 1 Year 2 Year 3

1 101 97.6424 92.5815 107.5178

2 101 101.2103 105.1763 102.4524 3 101 105.7802 103.6010 124.5115

4 101 96.4411 98.7120 108.3600

5 101 124.2345 101.0564 104.5315

6 101 95.8375 93.7270 99.3788

7 101 108.9554 102.4177 100.9225 8 101 104.1475 113.2516 115.0994

(57)

0 0.5 1 1.5 2 2.5 3 95

100 105 110 115 120 125

1

2

3

4 5

6 7

8

(58)

A Numerical Example (continued)

• We use the basis functions 1, x, x2. – Other basis functions are possible.a

• The plot next page shows the final estimated optimal exercise strategy given by LSM.

• We now proceed to tackle our problem.

• The idea is to calculate the cash flow along each path, using information from all paths.

aLaguerre polynomials, Hermite polynomials, Legendre polynomials, Chebyshev polynomials, Gedenbauer polynomials, and Jacobi polynomi- als.

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0 0.5 1 1.5 2 2.5 3 95

100 105 110 115 120 125

1

2 3

4 5

6 7 8

(60)

A Numerical Example (continued)

Cash flows at year 3

Path Year 0 Year 1 Year 2 Year 3

1 — — — 0

2 — — — 2.5476

3 — — — 0

4 — — — 0

5 — — — 0.4685

6 — — — 5.6212

7 — — — 4.0775

8 — — — 0

(61)

A Numerical Example (continued)

• The cash flows at year 3 are the exercise value if the put is in the money.

• Only 4 paths are in the money: 2, 5, 6, 7.

• Some of the cash flows may not occur if the put is exercised earlier, which we will find out step by step.

• Incidentally, the European counterpart has a value of

0.9512293 × 2.5476 + 0.4685 + 5.6212 + 4.0775

8 = 1.3680.

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A Numerical Example (continued)

• We move on to year 2.

• For each state that is in the money at year 2, we must decide whether to exercise it.

• There are 6 paths for which the put is in the money: 1, 3, 4, 5, 6, 7 (p. 851).

• Only in-the-money paths will be used in the regression because they are where early exercise is relevant.

– If there were none, we would move on to year 1.

(63)

A Numerical Example (continued)

• Let x denote the stock prices at year 2 for those 6 paths.

• Let y denote the corresponding discounted future cash flows (at year 3) if the put is not exercised at year 2.

(64)

A Numerical Example (continued)

Regression at year 2

Path x y

1 92.5815 0 × 0.951229

2 — —

3 103.6010 0 × 0.951229 4 98.7120 0 × 0.951229 5 101.0564 0.4685 × 0.951229 6 93.7270 5.6212 × 0.951229 7 102.4177 4.0775 × 0.951229

8 — —

(65)

A Numerical Example (continued)

• We regress y on 1, x, and x2.

• The result is

f (x) = 22.08 − 0.313114 × x + 0.00106918 × x2.

• f(x) estimates the continuation value conditional on the stock price at year 2.

• We next compare the immediate exercise value and the continuation value.a

aThe f(102.4177) entry on the next page was corrected by Mr. Du, Yung-Szu (B79503054, R83503086) on May 25, 2017.

(66)

A Numerical Example (continued)

Optimal early exercise decision at year 2 Path Exercise Continuation 1 12.4185 f (92.5815) = 2.2558

2 — —

3 1.3990 f (103.6010) = 1.1168 4 6.2880 f (98.7120) = 1.5901 5 3.9436 f (101.0564) = 1.3568 6 11.2730 f (93.7270) = 2.1253 7 2.5823 f (102.4177) = 1.2266

8 — —

(67)

A Numerical Example (continued)

• Amazingly, the put should be exercised in all 6 paths: 1, 3, 4, 5, 6, 7.

• Now, any positive cash flow at year 3 should be set to zero or overridden for these paths as the put is exercised before year 3 (p. 851).

– They are paths 5, 6, 7.

• The cash flows on p. 855 become the ones on next slide.

(68)

A Numerical Example (continued)

Cash flows at years 2 & 3

Path Year 0 Year 1 Year 2 Year 3

1 — — 12.4185 0

2 — — 0 2.5476

3 — — 1.3990 0

4 — — 6.2880 0

5 — — 3.9436 0

6 — — 11.2730 0

7 — — 2.5823 0

8 — — 0 0

(69)

A Numerical Example (continued)

• We move on to year 1.

• For each state that is in the money at year 1, we must decide whether to exercise it.

• There are 5 paths for which the put is in the money: 1, 2, 4, 6, 8 (p. 851).

• Only in-the-money paths will be used in the regression because they are where early exercise is relevant.

– If there were none, we would move on to year 0.

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A Numerical Example (continued)

• Let x denote the stock prices at year 1 for those 5 paths.

• Let y denote the corresponding discounted future cash flows if the put is not exercised at year 1.

• From p. 863, we have the following table.

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A Numerical Example (continued)

Regression at year 1

Path x y

1 97.6424 12.4185 × 0.951229 2 101.2103 2.5476 × 0.9512292

3 — —

4 96.4411 6.2880 × 0.951229

5 — —

6 95.8375 11.2730 × 0.951229

7 — —

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A Numerical Example (continued)

• We regress y on 1, x, and x2.

• The result is

f (x) = −420.964 + 9.78113 × x − 0.0551567 × x2.

• f(x) estimates the continuation value conditional on the stock price at year 1.

• We next compare the immediate exercise value and the continuation value.

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A Numerical Example (continued)

Optimal early exercise decision at year 1

Path Exercise Continuation

1 7.3576 f (97.6424) = 8.2230 2 3.7897 f (101.2103) = 3.9882

3 — —

4 8.5589 f (96.4411) = 9.3329

5 — —

6 9.1625 f (95.8375) = 9.83042

7 — —

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A Numerical Example (continued)

• The put should be exercised for 1 path only: 8.

– Note that f(104.1475) < 0.

• Now, any positive future cash flow should be set to zero or overridden for this path.

– But there is none.

• The cash flows on p. 863 become the ones on next slide.

• They also confirm the plot on p. 854.

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A Numerical Example (continued)

Cash flows at years 1, 2, & 3

Path Year 0 Year 1 Year 2 Year 3

1 — 0 12.4185 0

2 — 0 0 2.5476

3 — 0 1.3990 0

4 — 0 6.2880 0

5 — 0 3.9436 0

6 — 0 11.2730 0

7 — 0 2.5823 0

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A Numerical Example (continued)

• We move on to year 0.

• The continuation value is, from p 870,

(12.4185 × 0.9512292 + 2.5476 × 0.9512293 +1.3990 × 0.9512292 + 6.2880 × 0.9512292 +3.9436 × 0.9512292 + 11.2730 × 0.9512292 +2.5823 × 0.9512292 + 0.8525 × 0.951229)/8

= 4.66263.

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A Numerical Example (concluded)

• As this is larger than the immediate exercise value of 105 − 101 = 4,

the put should not be exercised at year 0.

• Hence the put’s value is estimated to be 4.66263.

• Compare this with the European put’s value of 1.3680 (p. 856).

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Time Series Analysis

(79)

The historian is a prophet in reverse.

— Friedrich von Schlegel (1772–1829)

(80)

GARCH Option Pricing

a

• Options can be priced when the underlying asset’s return follows a GARCH process.

• Let St denote the asset price at date t.

• Let h2t be the conditional variance of the return over the period [ t, t + 1 ] given the information at date t.

– “One day” is merely a convenient term for any elapsed time Δt.

aARCH (autoregressive conditional heteroskedastic) is due to Engle (1982), co-winner of the 2003 Nobel Prize in Economic Sciences. GARCH (generalized ARCH) is due to Bollerslev (1986) and Taylor (1986). A Bloomberg quant said to me on Feb 29, 2008, that GARCH is seldom used in trading.

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GARCH Option Pricing (continued)

• Adopt the following risk-neutral process for the price dynamics:a

ln St+1

St = r h2t

2 + htt+1, (112) where

h2t+1 = β0 + β1h2t + β2h2t(t+1 − c)2, (113)

t+1 ∼ N(0, 1) given information at date t, r = daily riskless return,

c ≥ 0.

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GARCH Option Pricing (continued)

• The five unknown parameters of the model are c, h0, β0, β1, and β2.

• It is postulated that β0, β1, β2 ≥ 0 to make the conditional variance positive.

• There are other inequalities to satisfy (see text).

• The above process is called the nonlinear asymmetric GARCH (or NGARCH) model.

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GARCH Option Pricing (continued)

• It captures the volatility clustering in asset returns first noted by Mandelbrot (1963).a

– When c = 0, a large t+1 results in a large ht+1,

which in turns tends to yield a large ht+2, and so on.

• It also captures the negative correlation between the asset return and changes in its (conditional) volatility.b

– For c > 0, a positive t+1 (good news) tends to decrease ht+1, whereas a negative t+1 (bad news) tends to do the opposite.

a. . . large changes tend to be followed by large changes—of either sign—and small changes tend to be followed by small changes . . . ”

b

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GARCH Option Pricing (concluded)

• With yt = ln SΔ t denoting the logarithmic price, the model becomes

yt+1 = yt + r h2t

2 + htt+1. (114)

• The pair (yt, h2t) completely describes the current state.

• The conditional mean and variance of yt+1 are clearly E[ yt+1 | yt, h2t ] = yt + r h2t

2 , (115) Var[ yt+1 | yt, h2t ] = h2t. (116)

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GARCH Model: Inferences

• Suppose the parameters c, h0, β0, β1, and β2 are given.

• Then we can recover h1, h2, . . . , hn and 1, 2, . . . , n from the prices

S0, S1, . . . , Sn

under the GARCH model (112) on p. 876.

• This property is useful in statistical inferences.

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The Ritchken-Trevor (RT) Algorithm

a

• The GARCH model is a continuous-state model.

• To approximate it, we turn to trees with discrete states.

• Path dependence in GARCH makes the tree for asset prices explode exponentially (why?).

• We need to mitigate this combinatorial explosion.

aRitchken & Trevor (1999).

Figure

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