M5A42 APPLIED STOCHASTIC PROCESSES PROBLEM SHEET 1 SOLUTIONS Term

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M5A42 APPLIED STOCHASTIC PROCESSES

PROBLEM SHEET 1 SOLUTIONS

Term 1 2010-2011

1. Calculate the mean, variance and characteristic function of the following probability density functions. (a) The exponential distribution with density

f(x) =

λe−λx _{x >}₀_,

0 x <0, withλ >0.

(b) The uniform distribution with density f(x) =

1

b−a a < x < b,

0 x /∈(a, b), witha < b.

(c) The Gamma distribution with density f(x) = ( λ Γ(α)(λx) α−1_e−λx _{x >}₀_, 0 x <0, withλ >0, α >0andΓ(α)is the Gamma function

Γ(α) = Z ∞ 0 ξα−1e−ξdξ, α >0. SOLUTION (a) E(X) = Z +∞ −∞ xf(x)dx=λ Z +∞ 0 xe−λxdx = 1 λ. E(X2) = Z +∞ −∞ x2f(x)dx=λ Z +∞ 0 x2e−λxdx = 2 λ2.

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Consequently,

var(X) =E(X2)−(EX)2= 1 λ2.

The characteristic function is

φ(t) =E(eitx) =λ Z ∞ 0 eitxe−λxdt= λ λ−it. (b) E(X) = Z +∞ −∞ xf(x)dx= Z b a x b−adx = a+b 2 . E(X2) = Z +∞ −∞ x2f(x)dx=λ Z b a x2 x 2 b−adx = b 2₊_ab₊_a2 3 . Consequently, var(X) =E(X2)−(EX)2 = (b−a) 2 12 .

φ(t) =E(eitx) =λ Z b a eitx 1 b−adt= eitb−eita it(b−a). (c) E(X) = λ α Γ(α)λ Z +∞ 0 xαe−λxdx = Γ(α+ 1) λΓ(α) = α λ. E(X2) = λ Z +∞ 0 x1+αe−λxdx = Γ(α+ 2) λ2_Γ(_α₎ = α(α+ 1) λ2 . Consequently, var(X) =E(X2)−(EX)2= α λ2.

φ(t) = E(eitx) = λ α Γ(α) Z ∞ 0 eitxxα−1dt = λ α Γ(α) 1 (λ−it)α Z ∞ 0 e−yyα−1dy = λ α (λ−it)α.

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2. (a) LetXbe a continuous random variable with characteristic functionφ(t). Show that EXk= 1

ikφ

(k)₍₀₎_,

whereφ(k)(t)denotes thek-th derivative ofφevaluated att.

(b) LetXbe a nonnegative random variable with distribution functionF(x). Show that E(X) =

Z +∞

0

(1−F(x))dx.

(c) LetXbe a continuous random variable with probability density functionf(x)and characteristic functionφ(t). Find the probability density and characteristic function of the random variable Y =aX+bwitha, b∈_R.

(d) LetXbe a random variable with uniform distribution on[0,2π]. Find the probability density of the random variableY = sin(X).

SOLUTION (a) We have φ(t) =E(eitx) = Z R eitxf(x)dx. Consequently φ(k)(t) = Z R (ix)keitxf(x)dx. Thus: φ(k)(0) = Z R (ix)kf(x)dx=ikEXk, andEXk = _i1kφ(k)(0). (b) LetR >0and consider

P(X < R) = Z R 0 xf(x)dx = Z R 0 xdF dx dx = xF(x)|R₀ − Z R 0 F(x)dx = Z R 0 (F(R)−F(x))dx. Thus, EX = lim R→∞P(X < R) = Z ∞ 0 (1−F(x))dx,

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where the factlimx→∞F(x) = 1was used. Alternatively: Z ∞ 0 (1−F(x))dx = Z ∞ 0 Z ∞ x f(y)dydx = Z ∞ 0 Z y 0 f(y)dxdy = Z ∞ 0 yf(y)dx=EX. (c) We have: P(Y ≤y) = P(aX +b≤y) = P(X ≤ y−b a ) = Z y−_ab −∞ f(x)dx. Consequently, fY(y) = ∂ ∂yP(Y ≤y) = 1 af y−b a . Similarly, φY(t) = EeitY =Eeit(aX+b)

= eitbEeitaX =eitbφ(at).

(d) The density of the random variableXis fX(x) =

n 1

2π, x∈[0,2π],

0, x /∈[0,2π]. The distribution function is

FX(x) =

n 0 x <0,

x

2π, x∈[0,2π],

1, x >2π.

The random variable Y takes values on [−1,1]. Hence, P(Y ≤ y) = 0 for y ≤ −1 and P(Y ≤y) = 1fory ≥1. Let nowy∈(−1,1). We have

FY(y) = P(Y ≤y) =P(sin(X)≤y).

The equationsin(x) =yhas two solutions in the interval[0,2π]:x= arcsin(y), π−arcsin(y)

fory >0andx=π−arcsin(y),2π+ arcsin(y)fory <0. Hence, FY(y) = π+ 2 arcsin(y)

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The distribution function ofY is FY(y) = n 0 y ≤0, π+2 arcsin(y) 2π , y∈(−1,1), 1, y≥1.

We differentiate the above expression to obtain the probability density: fY(y) =

n _π1√1

1−y2, y∈(−1,1),

0, y /∈(−1,1).

3. LetX be a discrete random variable taking vales on the set of nonnegative integers with probability mass functionpk=P(X =k)withpk≥0,P+k=0∞pk= 1. Thegenerating functionis defined as

g(s) =E(sX) =

+∞ X

k=0

pksk. (a) Show that

EX=g0(1) and EX2 =g00(1) +g0(1), where the prime denotes differentiation.

(b) Calculate the generating function of the Poisson random variable with pk=P(X=k) = e

−λ_λk

k! , k= 0,1,2, . . . and λ >0.

(c) Prove that the generating function of a sum of independent nonnegative integer valued random variables is the product of their generating functions.

(a) We have g0(s) = +∞ X k=0 kpksk−1 and g00(s) = +∞ X k=0 k(k−1)pksk−2. Hence g0(1) = +∞ X k=0 kpk =EX and g00(1) = +∞ X k=0 k2pk− +∞ X k=0 kpk=EX2−g0(1)

from which it follows

EX2 =g00(1) +g0(1). (b) We calculate g(s) = +∞ X k=0 e−λλk k! s k = eλ(s−1).

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(c) Consider the independent nonnegative integer valued random variablesXi, i= 1, . . . d. Since theXi’s are independent, so are the random variableseXi_{, i}_{= 1}_{, . . .}_{. Consequently,}

gPd i=1Xi(s) =E(e Pd i=1Xi_{) = Π}d i=1E(eXi) = Πdi=1gXi(s). 4. Let b ∈ _Rn _and _Σ _∈

Rn×n a symmetric and positive definite matrix. Let X be the multivariate Gaussian random variable with probability density function

γ(x) = 1 (2π)n/2 1 p det(Σ)exp −1 2hΣ −1_(x₋_b)_,_x₋_b_i . (a) Show that

Z

Rd

γ(x)dx= 1.

(b) Calculate the mean and the covariance matrix ofX. (c) Calculate the characteristic function ofX.

(a) From the spectral theorem for symmetric positive definite matrices we have that there exists a diagonal matrixΛwith positive entries and an orthogonal matrixBsuch that

Σ−1=BTΛB. Letz=x−bandy=Bz. We have

hΣ−1z,zi = hBTΛBz,zi = hΛBz, Bzi=hΛy,yi = d X i=1 λiyi2.

Furthermore, we have that det(Σ−1) = Πd_i₌₁λi, that det(Σ) = Πdi=1λ

−1

i and that the Jacobian of an orthogonal transformation isJ =det(B) = 1. Hence,

Z Rd exp −1 2hΣ −1_(x₋_b)_,_x₋_b_i dx = Z Rd exp −1 2hΣ −1_z_,_z_i dz = Z Rd exp −1 2 d X i=1 λiyi2 ! |J|dy = Πn_i₌₁ Z R exp −1 2λiy 2 i dyi = (2π)n/2Πn_i₌₁λ_i−1/2= (2π)n/2pdet(Σ), from which we get that

Z

Rd

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(b) From the above calculation we have that γ(x)dx = γ(BTy+b)dy = 1 (2π)n/2p det(Σ)Π n i=1exp −1 2λiy 2 i dyi. Consequently EX = Z Rd xγ(x)dx = Z Rd (BTy+b)γ(BTy+b)dy = b Z Rd γ(BTy+b)dy=b.

We note that, sinceΣ−1 ₌_BT_Λ_{B, we have that}_{Σ =}_BT_Λ−1_B_{. Furthermore,}_z ₌_BT_y_{. We} calculate E((Xi−bi)(Xj−bj)) = Z Rd zizjγ(z+b)dz = 1 (2π)n/2p det(Σ) Z Rd X k Bkiyk X m Bmiymexp − 1 2 X ` λ`y2` ! dy = 1 (2π)n/2p det(Σ) X k,m BkiBmj Z Rd ykymexp − 1 2 X ` λ`y2` ! dy = X k,m BkiBmjλ−_k1δkm = Σij.

(c) Let y be a multivariate Gaussian random variable with mean 0 and covariance I. Let also C=B√Λ. We have thatΣ =CCT =CTC. We have that

X=CY+b.

To see this, we first note thatXis Gaussian since it is given through a linear transformation of a Gaussian random variable. Furthermore,

EX=b and E((Xi−bi)(Xj−bj)) = Σij. Now we have: φ(t) = EeihX,ti=eihb,tiEeihCY,ti = eihb,tiEeihY,C T_t_i = eihb,tiEei P j( P kCjktk)yj = eihb,tie−12 P j| P kCjktk| 2 = eihb,tie−12hCt,Cti = eihb,tie−12ht,C T_C_t_i = eihb,tie−12ht,Σti.

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Consequently,