• 沒有找到結果。

# Problems with the Bump-and-Revalue Method

N/A
N/A
Protected

Share "Problems with the Bump-and-Revalue Method"

Copied!
86
0
0

(1)

### 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.

(2)

### 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).

(3)

### 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/).

(4)

### 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)

412

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

(6)

### 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.

(8)

### 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).

(9)

### 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.

(10)

### 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).

(11)

### 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

(12)

### 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.

(13)

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

(14)

### 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).

(15)

H

(16)

### 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.

(17)

### 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) ].

(18)

### 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.

(19)

### 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).

(20)

### 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.

(21)

### 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 ;

(22)

### 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.

(23)

### 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.

(24)

### 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.

(25)

### 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.

(26)

### 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).

(27)

### 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= (x1, x2 . . . , xn) 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.

(28)

### 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 ].

(29)

### 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

(30)

### 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 ].

(31)

### 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)

(32)

### 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.

(33)

### 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).

(34)

### 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.

(35)

### 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.

(36)

### 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?

(37)

### 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

(38)

### 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.

(39)

### 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.

(40)

## 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.

(43)

### 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

(44)

### 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.

(48)

### 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).

(49)

### 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.

(52)

### 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

(54)

### 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).

(55)

### 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.

(56)

### 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.

(59)

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.

(62)

### 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.

(70)

### 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.

(71)

### 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 — —

(72)

### 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.

(73)

### 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 — —

(74)

### 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.

(75)

### 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

(76)

### 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.

(77)

### 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).

(78)

## 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.

(81)

### GARCH Option Pricing (continued)

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

ln St+1

St = r h2t

2 + htt+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.

(82)

### 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.

(83)

### 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

(84)

### GARCH Option Pricing (concluded)

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

yt+1 = yt + r h2t

2 + htt+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)

(85)

### 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.

(86)

### 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).

An n×n square is called an m–binary latin square if each row and column of it filled with exactly m “1”s and (n–m) “0”s. We are going to study the following question: Find

6 《中論·觀因緣品》，《佛藏要籍選刊》第 9 冊，上海古籍出版社 1994 年版，第 1

Reading Task 6: Genre Structure and Language Features. • Now let’s look at how language features (e.g. sentence patterns) are connected to the structure

All steps, except Step 3 below for computing the residual vector r (k) , of Iterative Refinement are performed in the t-digit arithmetic... of precision t.. OUTPUT approx. exceeded’

Optim. Humes, The symmetric eigenvalue complementarity problem, Math. Rohn, An algorithm for solving the absolute value equation, Eletron. Seeger and Torki, On eigenvalues induced by

Light rays start from pixels B(s, t) in the background image, interact with the foreground object and finally reach pixel C(x, y) in the recorded image plane. The goal of environment

• But Monte Carlo simulation can be modiﬁed to price American options with small biases..

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(t n ) −