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  • Solutions to Examples on

    Stochastic Differential Equations

    December 4, 2012

  • 2

    Q 1. LetW1(t) andW2(t) be two Wiener processes with correlated increments ∆W1 and ∆W2 such

    that E [∆W1∆W2] = ρ∆t. Prove that E [W1(t)W2(t)] = ρ t. What is the value of E [W1(t)W2(s)]?

    Answer

    Let 0 = t0 < t1 < t2 < · · · < tn = t be a dissection of the interval [0, t], then

    W (1) t =

    n∑ k=1

    ∆W (1) k , W

    (2) t =

    n∑ k=1

    ∆W (2) k .

    where W (1) k and W

    (2) k are zero mean Gaussian deviates satisfying E[W

    (1) k W

    (2) k ] = ρ(tk − tk−1). Thus

    E[W (1)t W (2) t ] = E

    [( n∑ k=1

    ∆W (1) k

    )( n∑ j=1

    ∆W (2) j

    )] =

    n∑ j,k=1

    E [ ∆W

    (1) k ∆W

    (2) j

    ]

    However, ∆W (1) k and ∆W

    (2) j are independent Gaussian deviates if k ̸= j and so E

    [ ∆W

    (1) k ∆W

    (2) j

    ] =

    ρ(tk − tk−1)δ(k − j). Thus

    E[W (1)t W (2) t ] =

    n∑ k=1

    ρ(tk − tk−1) = ρ(tn − t0) = ρ t .

    Q 2. Let W (t) be a Wiener process and let λ be a positive constant. Show that λ−1W (λ2t) and

    tW (1/t) are each Wiener processes.

    Answer

    This question concerns the properties of a random variable under changes of variable. We need to

    show that each random variable is Gaussian distributed with mean value zero and variance t.

    (a) Here t is a parameter and W (λ2t) is a Gaussian random variable with mean value zero and

    variance λ2t. Let Y = λ−1W (λ2t) then

    fY = fW dW

    dY = λ fW = λ

    1√ 2πλ2t

    exp ( − W

    2

    2λ2t

    ) =

    1√ 2πt

    exp ( − λ

    2Y 2

    2λ2t

    ) =

    1√ 2πt

    exp ( − Y

    2

    2t

    ) .

    Thus Y is a Gaussian deviate with mean value zero and variance t.

    (b) Here t is again a parameter and W (1/t) is a Gaussian random variable with mean value zero

    and variance 1/t. Let Y = tW (1/t) then

    fY = fW dW

    dY =

    1

    t fW =

    1

    t

    1√ 2π(1/t)

    exp ( − W

    2

    2(1/t)

    ) =

    1√ 2πt

    exp ( − (Y/t)

    2

    2λ2(1/t)

    ) =

    1√ 2πt

    exp ( − Y

    2

    2t

    ) .

    Thus Y is a Gaussian deviate with mean value zero and variance t.

  • 3

    Q 3. Suppose that (ε1 , ε2) is a pair of uncorrelated N (0, 1) deviates.

    (a) By recognising that ξx = σx ε1 has mean value zero and variance σ 2 x, construct a second deviate

    ξy with mean value zero such that the Gaussian deviate X = [ ξx , ξy ] T has mean value zero

    and correlation tensor

    Ω =

     σ2x ρ σxσy ρ σxσy σ

    2 y

     where σx > 0, σy > 0 and | ρ | < 1.

    (b) Another possible way to approach this problem is to recognise that every correlation tensor is

    similar to a diagonal matrix with positive entries. Let

    α = σ2x − σ2y

    2 , β =

    1

    2

    √ (σ2x + σ

    2 y)

    2 − 4(1− ρ2)σ2xσ2y = √ α2 − ρ2σ2xσ2y .

    Show that

    Q = 1√ 2β

     √ β + α −√ β − α√ β − α

    √ β + α

     is an orthogonal matrix which diagonalises Ω, and hence show how this idea may be used to

    find X = [ ξx , ξy ] T with the correlation tensor Ω.

    (c) Suppose that (ε1, . . . , εn) is a vector of n uncorrelated Gaussian deviates drawn from the dis-

    tribution N (0, 1). Use the previous idea to construct an n-dimensional random column vector

    X with correlation structure Ω where Ω is a positive definite n× n array.

    Answer

    (a) Let X = [ ξx , ξy ] T then ξx has variance σ

    2 x and so we may write ξx = σx ε1 without any loss in

    generality. The task is now to find ξy. The idea is to write ξy = α ε1+β ε2. It now follows that

    E [ ξx ξy ] = E [σx ε1 (α ε1 + β ε2) ] = ασx

    E [ ξy ξy ] = α 2 + β2

    Therefore, choose ασx = ρ σxσy and α 2 + β2 = σ2y . Thus α = ρ σy and β

    2 = σ2y(1 − ρ2). One possible vector deviate is

    X = [ σx ε1 , ρ σy ε1 +

    √ 1− ρ2 σy ε2

    ]T .

  • 4

    (b) To check that Q is an orthogonal matrix, it is enough to observe that the two columns of Q are

    orthogonal to each other, and that each column of Q is a unit vector. Thus

    QTΩQ = 1

     √ β + α √ β − α − √ β − α

    √ β + α

     σ2x ρ σxσy ρ σxσy σ

    2 y

     √ β + α −√ β − α√ β − α

    √ β + α

     Without substituting for α and β, the computation of QTΩQ simplifies to

    1

     (β + α)σ2x + (β − α)σ2y + 2ρσxσy√ β2 − α2 (σ2y − σ2x)√ β2 − α2 + 2αρσxσy (σ2y − σ2x)

    √ β2 − α2 + 2αρσxσy (β + α)σ2y + (β − α)σ2x − 2ρσxσy

    √ β2 − α2

     . We need to demonstrate that this is a diagonal matrix. Consider therefore

    (σ2y − σ2x) √ β2 − α2 + 2αρσxσy = −2α

    √ β2 − α2 + 2αρσxσy

    = 2α(ρ σxσy − √ β2 − α2) .

    It follows directly from the definition of β that this entry is zero. Consequently, QTΩQ is a

    diagonal matrix. By noting that, first that σ2x−σ2y = 2α and, second that ρ σxσy = √ β2 − α2,

    the array QTΩQ now becomes

    1

     β(σ2x + σ2y) + 2α2 + 2ρσxσy√ β2 − α2 0 0 β(σ2y + σ

    2 x)− 2α2 − 2ρσxσy

    √ β2 − α2

     =

    1

     β(σ2x + σ2y) + 2β2 0 0 β(σ2y + σ

    2 x)− 2β2

     =

    1

    2

     σ2x + σ2y + 2β 0 0 σ2y + σ

    2 x − 2β

     It is again obvious from the definition of β that σ2y+σ

    2 x ≥ 2β and so the entries of this diagonal

    matrix are non-negative. Let

    Y = 1√ 2

    [ √ σ2x + σ

    2 y + 2β ε1 ,

    √ σ2x + σ

    2 y − 2β ε2

    ]T where ε1 ∼ N(0, 1) and ε2 ∼ N(0, 1), then the random variable X = QY has mean value zero and covariance tensor QTΩQ.

    (c) Since Ω is a symmetric positive definite n × n array, it is possible to find an n × n orthog- onal matrix Q such that Ω = QDQT in which D is a diagonal matrix whose entries are the

    eigenvalues of Ω in some order. Since Ω is positive definite, then each entry of D is positive.

    Let Y = [ √ λ1 ε1, · · · ,

    √ λn εn ]

    T where λk > 0 is the k-th diagonal entry of D. Consider the

    properties of X = QY . Clearly

    E [X ] = E [QY ] = QE [Y ] = 0 ,

    E [XXT ] = E [QY Y TQT ] = QE [Y Y T ]QT = QDQT = Ω .

  • 5

    Thus X has the required properties.

    Q 4. Let X be normally distributed with mean zero and unit standard deviation, then Y = X2 is

    said to be χ2 distributed with one degree of freedom.

    (a) Show that Y is Gamma distribution with λ = ρ = 1/2.

    (b) What is now the distribution of Z = aX2 if a > 0 and X ∼ N(0, σ2).

    (b) If X1, · · · , Xn are n independent Gaussian distributed random variables with mean zero and unit standard deviation, what is the distribution of Y = XTX where X is the n dimensional

    column vector with k-th entry Xk.

    Answer

    (a) Since Y = X2 then clearly Y ≥ 0 and so every non-zero value of Y arises from either X or −X. Therefore, the density of Y is

    fY (y) = 2fX(x) dX

    dY =

    2√ 2π

    e−x 2/2 1

    2 y−1/2 =

    1√ 2π

    y−1/2 e−y/2 .

    Evidently the distribution function for Y is the special case of the Gamma distribution in which

    λ = ρ = 1/2.

    (b) Since Z = aX2 with a > 0 then clearly Z ≥ 0 and so once again every non-zero value of Z arises from either X or −X. The density of Z is now

    fZ(z) = 2fX(x) dX

    dZ =

    2√ 2π σ

    e−x 2/2σ2 1

    2 √ a z−1/2 =

    1√ 2π

    √ a σ

    z−1/2 e−z/2aσ 2 .

    The distribution function for Z is now the Gamma distribution with ρ = 1/2 and λ = (2aσ2)−1.

    (c) If X1, · · · , Xn are each normally distributed with mean zero and unit standard deviation, then Y = X21 +X

    2 2 + · · ·+X2n is likewise Gamma distributed with parameters λ = 1/2 and ρ = n/2.

    This result follows from the previous example. In this case we say that Y is χ2 distributed with

    n degrees of freedom.

    The chi-squared distribution with n degrees of freedom plays an important role in statistical

    hypothesis testing in which deviates are assigned to “bins”.

    Q 5. Calculate the bounded variation (when it exists) for the functions

    (a) f(x) = |x| x ∈ [−1, 2] (b) g(x) = log x x ∈ (0, 1]

    (c) h(x) = 2x3 + 3x2 − 12x+ 5 x ∈ [−3, 2] (d) k(x) = 2 sin 2x x ∈ [π/12, 5π/4]

    (e) n(x) = H(x)−H(x− 1) x ∈ R (f) m(x) = (sin 3x)/x x ∈ [−π/3, π/3] .

  • 6

    Answer

    The simplest approach here is to draw each function.

    (a) The bounded variation of f(x) = |x| for x ∈ [−1, 2] is 1 + 2 = 3.

    (b) The function g(x) = log x→ −∞ as x→ 0+.

    (c) Here we need to find the stationary values of h(x