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    Mathematical Statistics

    Stockholm University

    How to predict crashes in financial

    markets with the Log-Periodic Power

    Law

    Emilie Jacobsson

    Examensarbete 2009:7

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    Postal address:Mathematical StatisticsDept. of Mathematics

    Stockholm UniversitySE-106 91 StockholmSweden

    Internet:http://www.math.su.se/matstat

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    Mathematical StatisticsStockholm UniversityExamensarbete 2009:7,http://www.math.su.se/matstat

    How to predict crashes in financial markets

    with the Log-Periodic Power Law

    Emilie Jacobsson

    September 2009

    Abstract

    Speculative bubbles seen in financial markets, show similarities inthe way they evolve and grow. This particular oscillating movementcan be captured by an equation called Log-Periodic Power Law. Theending crash of a speculative bubble is the climax of this Log-Periodicoscillation. The most probable time of a crash is given by a parameterin the equation. By fitting the Log-Periodic Power Law equation to

    a financial time series, it is possible to predict the event of a crash.With a hybrid Genetic Algorithm it is possible to estimate the pa-rameters in the equation. Until now, the methodology of performingthese predictions has been vague. The ambition is to investigate if thefinancial crisis of 2008, which rapidly spread through the world, couldhave been predicted by the Log-Periodic Power Law. Analysis of theSP500 and the DJIA showed the signs of the Log-Periodic Power Lawprior to the financial crisis of 2008. Even though the analyzed indicesstarted to decline slowly at first and the severe drops came much fur-ther, the equation could predict a turning point of the downtrend.The opposite of a speculative bubble is called an anti-bubble, moving

    as a speculative bubble, but with a negative slope. This log-periodicoscillation has been detected in most of the speculative bubbles thatended in a crash during the Twentieth century and also for some anti-bubbles, that have been discovered. Is it possible to predict the courseof the downtrend during the financial crisis of 2008, by applying thisequation? The equation has been applied to the Swedish OMXS30index, during the current financial crisis of 2008, with the result of apredicted course of the index.

    Postal address: Mathematical Statistics, Stockholm University, SE-106 91, Sweden.E-mail: [email protected]. Supervisor: Ola Hammarlid.

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    109

    {Pt=1,T}

    Pmax= Pk Pmin = Pk+n k 2

    n 1 n

    Pk> Pk+1 > ... > Pk+n Pmax Pk Pk1 Pk > Pk+1 Pmin Pk+n < Pk+n1 Pk+n Pk+n+1

    D=Pmin Pmax

    Pmax

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    f(x) =

    zxz1

    z

    exp

    x

    zf or x 0

    0, f or x

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    =

    N1 ni=1

    (ri+1 E[r])2

    r

    ri+1 = logp(ti+1) logp(ti)

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    =3= 0.0456

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    D1 < D2 < ... < DN D1 n= 1, 2, , N Dn

    N(D) = A exp {b|D|z} b= z

    log N(D) = log A b|D|z

    3

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    3

    =

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    = 3

    =

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    =

    z

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    log( log(p)) = z log(x) z log()

    pi= 1i 0.5

    N i= 1,...,N

    N

    z

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    [0, 0.00025]

    [0.035, 0.31]

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    z

    c

    f(x) =

    z(x

    c)

    z1

    z

    exp(x

    c)z

    x c, z, >00, f or x < c

    c

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    c= 3= 0.0345

    z

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    i= 1,...,I N(i) N(i)

    i

    N(i)

    j = 1, , n

    selling = 1 buying = +1

    Ni=1 si

    s

    P P+ P= P+

    nj=1 sj

    si(t + 1) =signK n

    j=1

    sj(t) + i

    N(0, 1) K K

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    K

    K Kc K < Kc

    K

    Kc

    Kc K K

    A (Kc K)

    tc

    h(t) = B (tc t)

    t = ( 1)1 2< <

    1

    tc

    t0

    h(t)dt > 0

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    t

    Et[p(t)] = p(t) t > t

    p(t) =p(t0) = 0 t0

    (0, 1)

    dp= (t)p(t)dt p(t)dj

    j (t)

    Et[dp] = (t)p(t)dt p(t)h(t)dt= 0

    (t) = h(t)

    dp= h(t)p(t)dt h(t)dj

    p(t) = h(t)p(t)

    p(t)

    p(t) =h(t)

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    log p(t)

    p(t0)

    =

    t

    t0

    h(t

    )dt

    log(p(t)) = log(p(t0)) +

    t

    t0

    h(t)dt

    t

    t0

    h(t)dt =B

    t

    t0

    (tc t)

    dt= B

    (tc t

    )(+1)

    + 1

    t=tt=t

    =

    =

    B2tc t t0 + 1

    ={

    t0

    = tc

    = + 1

    (0, 1)}

    =

    B

    (t

    c t)

    log(p(t)) = log(p(to))B

    (tc t)

    A (Kc K)

    A (Kc K)

    Re

    A0(Kc K)

    + A1(Kc K)+i

    + ...

    A0(Kc K)

    + A1(Kc K)cos( log(Kc K) + )...

    A0 A

    1

    = /2

    h(t) Re

    B0(tc t)

    + B1(tc t)+i

    + ...

    B0(tc t)

    + B1(tc t)cos( log(tc t) + )

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    log(p(t)) = log(p(to))

    B0(tc t)

    + B1(tc t)

    cos( log(tc t) + )

    y= A + B (tc t)z

    + C(tc t)z cos ( log(tc t) + )

    y(t)

    A p(tc) B B

    C= 0

    tc z [0, 1]

    z = 0.33 0.18 [5, 15]

    = 6.36 1.56

    [0, 2]A B C

    B (tc t)z

    cos ( log(tc t) + )

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    y(t) A + Bf(t) + Cg(t)

    f(t) =

    (tc t)

    z

    (t tc)z

    g(t) =

    cos ( log(tc t) + )

    cos ( log(t tc) + )

    Ni y(ti)

    Ni y(ti)f(ti)Ni y(ti)g(ti)

    =

    NN

    i f(ti) N

    i g(ti)

    Ni f(ti)

    Ni f(ti)

    2

    Ni f(ti)g(ti)N

    i g(ti)N

    i f(ti)g(ti) N

    i g(ti)2

    AB

    C

    Xy= (X

    X)b

    X=

    1 f(t1) g(t1)

    1 f(tn) g(tn)

    b=

    AB

    C

    b= (X

    X)1Xy

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    min

    F() =Ni

    (y(ti) y(ti))2

    , = (tc, , , z)

    F()

    P Rn F

    F =F(P) F(P), P

    F :Rn R

    [2, 2]

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    tc

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    tc

    k k = 2

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    tc

    min

    F() =N

    i (y(ti) y(ti))2

    , = (tc, )

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    tc

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    tc (tc t) tc (tc t)

    tc tstart tc

    tc< tstart

    tstart

    tc< tstart |tct|

    tstart tc

    tc

    tc tstart tc tstart

    tstart

    [2, 2] R2 tc

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    tc R2

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    tc

    R2

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    tc R2

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    tc

    R2

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    tc R2

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    tc

    R2

    tc

    R2

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    tc

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    tc R2

    tc R2

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    tc R2

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    t tc

    tc

    tc tc

    tc

    tc

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    tc

    tc z

    tc

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    z z = 1

    Xi ] , 0] DN=X1+ X2+ ... + XN Xi

    F s(p+) p(n) = pn

    1 p+ p+ p p+ = 1p

    X(k) = E[ekX ] =

    0

    ekxf(x)dx

    Xi

    D(k) =n=1

    E[eD|N=n]P(N=n) =n=1

    E[eD]pn1

    p+ =

    =p+

    n=1

    E[ek(X1+...+Xn)]pn1

    =p+

    n=1

    E[ekX1 ...ekXn]pn1

    =

    =

    n=0

    xn = 1

    1 x

    n=1

    1

    1 x 1 =

    x

    1 x

    =

    =p+

    p

    n=1

    (X(k)p)n

    =

    =

    P(k) = pX(k) = p

    0

    ekxf(x)dx P(0) =p

    =

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    = p+

    p

    pX(k)

    1pX(k)

    = p

    +

    p

    P(k)

    1 P(k)

    = 1

    1 1p+

    P(k)pP(k)

    =

    =

    P(0) =p

    =

    1

    1 1p+

    P(k)P(0)

    P(k)

    P(k)P(0)

    P(k)

    P(k) P(0)

    P(k)

    kP(0)

    P(0)

    + O(k2)

    kP(0)

    P(0)=

    pX(k)

    p=

    0

    xf(x)dx= E[X]

    = E[DN] = E[Nj=1

    Xj |N=n] = E[N E[X]] = E[N]E[X] = 1

    p+E[X]

    P(k) = 1

    1 1p+

    P(k)P(0)

    P(k)

    = 11 + k

    p+E[X]

    = 1

    1 + k

    f(d) =e

    |d|

    p+ = p= 1/2