Final exam - exemplary problems on probability theory with hints and sketches of solutions of selected problems.

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Final exam - exemplary problems on probability

theory with hints and sketches of solutions of

selected problems.

Probability Theory and Stochastic Processes, summer semester 2017/2018

Probability spaces. One-dimensional random variables.

1. Find the smallestσ-fieldFonΩ ={1,2,3,4}and all probability measures P: F →[0,1]such thatA ={2,4} ∈ F, B ={3} ∈ F and P(A) = 0.5, P(B) = 0.3.

2. Verify if (Ω,F,_P)is a probability triple if: Ω = [0,1],F is the family of subsets of Ωsuch that A is finite or its complementA0 = Ω\Ais finite and_Pis defined in the following way:

P(A) = ( 0 ifAis finite; 1 ifA0 is finite. 3. * Verify thatΩ,F˜,P

is a probability triple if: Ω = [0,1],F˜is the family of subsets ofΩsuch that Ais countable or its complementA0= Ω\Ais countable and_Pis defined in the following way:

P(A) =

(

0 ifAis countable; 1 ifA0 is countable.

Hint. Union (sum) of countable sets is also countable.

4. Let Ω,F˜,_P be as in the previous problem. Verify if X : Ω → R,

X(ω) =ω is a random variable.

5. Verify that (Ω,F,P) is a probability triple if: Ω = (1,2,3,4), F is the family of all subsets of Ω and for A ∈ F we define P(A) = 101

P

a∈Aa.

Verify ifX : Ω→R, X(ω) = 2ωis a random variable. If so, find preimages

X−1([0,1]), X−1({1,1.2,1.4,1.6,1.8,2}), then find the probability mass function ofX and calculateEX.

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Solutions of selected problems

1.

2. LetAn =

1

n forn= 2,3,4, . . .. We see that forn= 2,3,4, . . .,An ∈ F

since An is countable. Let B = ∪∞n=2An and B0 = Ω\B = [0,1]\B.

We have B0 = (1₂,1]∪(₃1,1₂)∪. . .. If F was a σ-algebra then we would haveB0∈ F butB0 is not finite, neither its complement (B) is finite, thus

B0 ∈ F/ andF can not be aσ-algebra and the triple (Ω,F,P)can not be a probability triple.

3. 4. 5.

Probability laws of one-dimensional random variables and

their characteristics. Characteristic and moment

generat-ing functions.

1. LetX be a random variable with the following probability mass function

x= 1 2 3 4

P(X =x) = 0.2 0.3 0.3 0.2 .

Findm1(X) =EX, m2(X) =EX2,σ2X=µ2(X)andµ3(X).

2. A random variableXhasBin(3,0.4)distribution. FindP(X= 0),P(X= 2) and_P(X = 10). Calculate the standard deviation ofX.

3. LetX ∼P oi(4). Which valuek= 0,1, . . .doesX attain with the greatest probability? Estimate_P(X ≤3) and_P(X ≥5).

4. For whichcthe functionf :R→R,

fX(x) =

(

0 ifx <1;

c

x2 ifx≥1

is a probability density function of a continuous random variableX? Cal-culate _P(X ≤2), _P(X = 2), _P(X∈[2,3]). Find _EX2 _and

E √

X if they are well defined.

5. Find the moment generating functionMX(t)of (a random variable with)

the Erlang distribution Erlang(5,5). For whicht it is finite? For which

k= 1,2, . . .,kth moment of this distribution is well defined?

6. Using the moment generating function of a standard normal random vari-able calculate the fourth moment of this varivari-able.

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Solutions of selected problems 1. 2. 3. Fork= 0,1, . . .we haveP(X=k+1) P(X=k) = e−4₄k+1_/₍_k_+1)! e−4₄k_/k_! = 4 k+1. Thus P(X=k+1) P(X=k) > 1 for k = 0,1,2,P(X=k+1) P(X=k) = 1 for k = 3 and P(X=k+1) P(X=k) < 1 for k = 4,5, . . .and we have _P(X = 0)<_P(X = 1)<_P(X = 2)<_P(X= 3) =

P(X = 4) > P(X = 5) > . . .. X attains with the greatest probability values3and4. 4. We calculate 1 = Rx=+∞ x=−∞ fX(x)dx = Rx=+∞ x=1 c x2dx = Rx=+∞ x=1 cx −2_d_x ₌ cx₋−₂₊₁2+1|x=+∞ x=1 =c − 1 x |x=+∞ x=1 =c − 1 +∞− − 1 1 =c 0 + 1₁ =c. Thus we see thatc= 1. We have: P(X≤2) =Rxx==2−∞fX(x)dx=

Rx=2 x=1 1 x2dx= −1 x |x=2 x=1=−12− − 1 1 = ₂1, P(X = 2) =Rxx=2=2fX(x)dx= Rx=2 x=2 1 x2dx= −1 x |x=2 x=2=−12− − 1 2 = 0,P(X ∈[2,3]) =Rxx=2=3fX(x)dx= Rx=3 x=2 1 x2dx= −1 x |x=3 x=2=− 1 3− − 1 2 =1₆. Next, we calculate_EX2₌Rx=+∞ x=−∞ x 2_f X(x)dx= Rx=+∞ x=1 x 2 1 x2dx = Rx=+∞ x=1 1dx = x| x=+∞ x=1 = +∞ −1 = +∞, thus EX2

is not finite (it is not well defined). E √ X = R_xx₌=+_−∞∞x1/2fX(x)dx = Rx=+∞ x=1 x 1/2 1 x2dx = Rx=+∞ x=1 x −3/2_d_x ₌ x−3/2+1 −3/2+1| x=+∞ x=1 = 0− 1 −1/2 = 2, thusE √

X is finite (it is well defined).

5. By the definition of the moment generating function forX ∼Erlang(5,5)

we haveMX(t) =Eexp (tX) = 5 5 (5−1)! R+∞ 0 e tx_x5−1_e−5x_d_x₌ 55 (5−1)! R+∞ 0 x 5−1_e−(5−t)x_d_x₌ 55 (5−1)! (5−1)! (5−t)5 = 5 5−t 5

fort <5(this follows from the fact that(5₍₅−₋t_1)!)5x5−1e−(5−t)x

is the density ofErlang(5,5−t)distribution thus(5₍₅−₋t_1)!)5R+∞

0 x

5−1_e−(5−t)x_d_x₌

1). Otherwise we have that MX(t) = +∞. For any k = 1,2, . . . we

have _EXk ₌ 55 (5−1)! R+∞ 0 x k_x5−1_e−5x_d_x ₌ 55 (5−1)! R+∞ 0 x k+5−1_e−5x_d_x ₌ 55 (5−1)! (k+5−1)! 5k+5 = (k+4)!

5k_4! (this follows from the fact that

5k+5 (k+5−1)!x

k+5−1_e−5x

is the density ofErlang(k+5,5)distribution thus₍_k5₊₅k+5₋_1)!R+∞

0 x

k+5−1_e−5x_d_x₌

1). EXk is well defined for anyk= 1,2, . . .. 6. ForX ∼ N(0,1) we have MX(t) = exp

t2

2

. Using differentiation rules we calculate M_X0 (t) = t·expt₂2, M_X00(t) = expt₂2+t2·expt₂2,

M_X000(t) =t·expt₂2+2t·expt₂2+t3_·_expt2 2 = 3t+t3 expt₂2and MIV X (t) = 3 + 3t 2 expt2 2 + 3t+t3 texpt2 2 . Substitutingt= 0we get_EX4₌_MIV X (0) = 3.

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Tails, cumulative distribution functions and joint

distribu-tion of two real random variables.

1. Find the formula for the tails of_P(X > t)ofX and of the cumulative dis-tribution function ofX whenX is the random variable with the following cumulative distribution function:

fX(x) =

(

0 ifx <1;

2

x3 ifx≥1

2. Let X be a random variable with the Erlang distribution Erlang(2,5). Using integration by parts find the formula for the tails ofP(X > t)ofX and of the cumulative distribution function ofX.

3. LetX be a random variable with the following probability mass function: fork= 1,2, . . .,

P(X=k) = 1

k4 −

1 (k+ 1)4.

Find the cumulative distribution function (CDF) ofX. Fork= 0,1,2, . . . ,

calculateP(X≥k).

4. The joint density function of the vector(X, Y)is given by

fX,Y(x, y) =

1

21[0,1](x)1[0,2](y).

Find the CDF of the vector(X, Y).

5. The joint probability mass function of the vector (X, Y) is given in the

table:

X=\Y = 1 2 3

0 0.2 0.1 0

1 0.1 0.3 0

2 0 0 0.3

. Find the marginal probability

mass functions ofXandY, and conditional probabilitiesP(X= 0|Y = 1),

P(X = 1|Y = 1),P(X = 2|Y = 1),P(X= 0|Y = 2),P(X = 1|Y = 2),P(X= 2|Y = 2). 6. The random vector (X, Y) is uniformly distributed on the circular disc

D=

(x, y)∈R2:x2+y2≤2 which means that the joint density func-tion of(X, Y)is given by fX,Y(x, y) = 1 2π1D(x, y) = ( ₁ 2π ifx 2₊_y2_≤_2; 0 ifx2₊_y2_>₂_.

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1. To calculate the CDF (cumulative distribution function) of X we write

FX(t) = P(X ≤t) = Rxx==−∞t fX(x)dx. We notice that fort <1 we have

FX(t) = P(X≤t) = Rxx==−∞t 0dx = 0 and for t ≥ 1 we have FX(t) =

P(X ≤t) =Rxx=1=t 2 x3dx= 2 x−3+1 −3+1| x=t x=1 = 2 t−2 −2 − 1−2 −2 = 1− 1 t2. Finally,

we calculate the tails of X to be equal P(X > t) = 1−P(X≤t) =

1−FX(t)which ie equal1fort <1 and _t12 fort≥1.

2. TheErlang(2,5)distribution has the following density52xe−5x. Using in-tegration by partsRu(x)v0(x)dx=u(x)v(x)−R u0(x)v(x)dxwe calculate R xe−5x_d_x₌R x −1 5e −5x0_d_x₌₋_x1 5e −5x₊R x01 5e −5x_d_x₌₋1 5xe −5x₊ 1 5 R e−5x_d_x₌₋1 5xe −5x₋ 1 52e− 5x ₌₋1 25e −5x₍₅_x_{+ 1)}_{. Now we calculate}

the tails: for t > 0, P(X > t) = Rt+∞25xe

−5x_d_x _{= 25}R+∞

t xe

−5x_d_x ₌

−e−5x(5x+ 1)|xx=+=t∞= 0+e−5t(5t+ 1)while fort≤0we haveP(X > t) =

1. The CDF ofX is equalFX(t) = 1−P(X > t)thusFX(t) = 1fort <0

andFX(t) = 1−e−5t(5t+ 1)fort≥0. 3. Fork= 1,2, . . .we haveFX(k) =P(X≤k) =Pki=1P(X =i) = Pk i=1 1 i4 − 1 (i+1)4 = 1 14 − 1 24 + 1 24 − 1 34 +. . .+ 1 k4 − 1 (k+1)4 = 1− 1

(k+1)4. Now, for anyt ∈R

we have FX(t) = P(X ≤t) = 0 for t < 1 and FX(t) = P(X ≤t) = P(X ≤ btc) = 1− (btc1+1)4 for t ≥ 1. To calculate P(X ≥k) for k =

1,2, . . . we write P(X≥k) = P+i=∞kP(X=i) = P+∞ i=k 1 i4 − 1 (i+1)4 = 1 k4 − 1 (k+1)4 + 1 (k+1)4 − 1 (k+2)4 +. . .= 1 k4. 4. 5.

6. We apply the formulafX(x) =R y=+∞ y=−∞ fX,Y(x, y)dy= 1 2π Ry=+∞ y=−∞ 1D(x, y)dy.

Let us notice that for x <−√2 and x >√2 we have x2₊_y2 _>₂ _thus

1D(x, y) = 0andfX(x) = 0. Now let us assume thatx∈[−

√ 2,√2]. We have 1D(x, y) = 1 iffy ∈ −√2−x2_,√₂₋_x2 and otherwise 1D(x, y) =

0. From this we calculatefX(x) =₂1_πR y=+∞ y=−∞ 1D(x, y)dy= 1 2π Ry=+ √ 2−x2 y=−√2−x2 1dy= 1 π √ 2−x2_{. Similarly,} _f Y(y) = 0 fory <− √ 2 andy >√2, andfY(y) = 1 π p 2−y2_for_y_∈_[₋√₂_,√_2]_.

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Independence of random variables.

Moments and other

characteristics of two-dimensional random variables.

Co-variance, correlation.

1. Verify, whether the random variablesXandY whose joint density function

fX,Y is given by fX,Y(x, y) = 1 2π1D(x, y) = ( ₁ 2π ifx 2₊_y2_≤_2; 0 ifx2₊_y2_>₂_.

are independent. (*) Calculate the covariance betweenX andY.

2. Calculate the covariance and correlation matrix of the vector(X, Y)whose

joint probability mass function is given in the table:

X=\Y = 1 2 3

0 0.2 0.1 0

1 0.1 0.3 0

2 0 0 0.3

.

3. Verify, whether the random variablesXandY whose joint density function

fX,Y is given by

fX,Y(x, y) =

(₁

2 if0≤x≤y≤2;

0 otherwise.

are independent. Calculate the correlation matrix of the vector(X, Y).

1. Using the marginal densities calculated in the solution of Problem 6. from the previous section we see thatfX,Y(x, y)6=fX(x)fY(y) thusX andY

are dependent. 2.

3. Using the joint density ofXandY we calculate_EXY =Rx=+∞

x=−∞ Ry=+∞ y=−∞ xyfX,Y(x, y)dy dx= Rx=2 x=0 _R_y₌₂ y=xxy 1 2dy dx= 1₂R_xx₌₀=2xR_yy₌=2_xydydx= 1₂R_xx₌₀=2x1₂y2|y=2 y=xdx= 1 4 Rx=2 x=0 x 4−x 2 dx= 1₄Rx=2 x=0 4x−x 3_d_x₌ 1 4 4 1 2x 2₋1 4x 4 |x=2 x=0= 1 4 2·2 2₋1 42 4 = 1. Next, we calculateEX =Rxx==+−∞∞x _R_y₌₊_∞ y=−∞ fX,Y(x, y)dy dx=R_xx₌₀=2xR_yy₌=2_x 1₂dydx= 1 2 Rx=2 x=0 x Ry=2 y=x1dy dx = 1₂Rx=2 x=0 x·y| y=2 y=xdx = 1 2 Rx=2 x=0 x(2−x)dx = 1 2 Rx=2 x=0 2x−x 2_d_x₌1 2 2 1 2x 2₋1 3x 3 |x=2 x=0= 12 2 2₋1 32 3 =2 3 andEX2=Rxx==+−∞∞x 2Ry=+∞ y=−∞ fX,Y(x, y)dy dx=Rx=2 x=0 x 2Ry=2 y=x 1 2dy dx= 1 2 Rx=2 x=0 x 2Ry=2 y=x 1dy dx= 1 2 Rx=2 x=0 x 2_·_y_|y=2 y=xdx= 12 Rx=2 x=0 x 2₍₂₋_x₎_d_x₌ 1 2 Rx=2 x=0 2x 2₋_x3_d_x₌ 1 2 2 1 3x 3₋1 4x 4 |x=2 x=0= 1 2 2 1 32 3₋1 42 4 = 2₃.

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We also haveEY =Ryy==+−∞∞y Rx=+∞ x=−∞ fX,Y(x, y)dx dy=Ry=2 y=0 y Rx=y x=0 1 2dx dy= 1 2 Ry=2 y=0 y Rx=y x=0 1dx dy=1 2 Ry=2 y=0 y·x| x=y x=0dy= 12 Ry=2 y=0 y 2_d_y₌1 2 1 3y 3_|y=2 y=0= 1 62 3₌4 3 andEY2=Ryy==+−∞∞y 2Rx=+∞ x=−∞ fX,Y(x, y)dx dy=R_yy₌₀=2y2 R_xx₌₀=y1₂dxdy= 1 2 Ry=2 y=0 y 2 Rx=y x=0 1dx dy= 1₂Ry=2 y=0 y 2_·_x_|x=y x=0dy= 1 2 Ry=2 y=0 y 3_d_y₌ 1 2 1 4y 4_|y=2 y=0= 1 82 4_{= 2}_.

Now we have all ingredients needed to calculate ρ(X, Y) = cov(_σ X,Y) XσY = EXY−EXEY √ EX2−(EX)2 √ EY2−(EY)2 = 1−23 4 3 q 2 3−( 2 3) 2q 2−(4 3) 2 = 1 9 √ 2 9 √ 2 9 = 1

2 and the

cor-relation matrix of (X, Y) reads as

1 1₂ 1 2 1 . Since ρ(X, Y) 6= 0 the variablesX andY are dependent.

Covariance and correlation matrix. Sums of random

vari-ables, their distributions, characteristics and characteristic

functions.

1. The joint density function of the random variablesX andY is given by

f(x, y) = (

2 if0≤x≤y≤1; 0 otherwise.

The variableZis independet fromXandY, and has the same distribution as the variableX. (a) Find the covariance matrix of the vector(X, Y, Z). (b) Calculate first two moments of the variableX+Y +Z. (c) Compute cov(X, Y+Z). Hint: cov(X, Y+Z) =cov(X, Y)+cov(X, Z). (c) Compute the covariance matrix of the vector(X, X+Z, Y +Z).

2. The covariance matrixΣof the vector(X, Y, Z)reads as

Σ =  

cov(X, X) cov(X, Y) cov(X, Z)

cov(Y, X) cov(Y, Y) cov(Y, Z)

cov(Z, X) cov(Z, Y) cov(Z, Z)  =   2 0 1 0 4 −1 1 −1 4  .

(a) Calculate the variance of the variable X +Y +Z. (b) Compute cov(X, Y+Z). Hint: cov(X, Y+Z) =cov(X, Y)+cov(X, Z). (c) Compute the covariance matrix of the vector(X, X+Z, Y +Z).

3. X and Y are two independent random variables, X ∼P oi(2) and Y ∼

P oi(3). (a) Find the expectation and the variance of the variableX+Y. (b) Find the characteristic function of the variableX +Y. (c) Find the probability mass function of the variableX+Y.

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4. X1, X2, . . . , Xn are independent random variables such thatXj∼P oi(2)

forj= 1,2, . . . , n. (a) Find the expectation, the variance and the standard deviation of the variable

¯

X =X1+X2+. . .+Xn

n .

(b) Find the characteristic function of the variable

S =nX¯ =X1+X2+. . .+Xn

and identify its distribution.

5. X ∼ N(0,1)andY ∼ N(0,4)are two independent random variables. (a) Find the expectation and the variance of the variable X+Y. (b) Find the characteristic function of the variableX+Y. (c) Find the probability density function of the variableX+Y.

6. X1, X2, . . . , Xn are independent random variables such thatXj ∼ N(0,4)

for j = 1,2, . . . , n. (a) Find the characteristic function of the variable

S = X1+X2+. . .+Xn and identify its distribution. (b) Find the

ex-pectation, the variance and the standard deviation of the variable

¯

X = 1

nS=

X1+X2+. . .+Xn

n

and identify its distribution.

7. X ∼Erlang(3,0.2) andY ∼Erlang(2,0.2) are two independent random variables. (a) Find the characteristic function of the variableX+Y. (b) Identify the distribution ofX+Y. (c) Calculate_E(X+Y)3.

8. Given are twodependentdiscrete random variablesX andY,X with the following joint probability mass function:

P(X =n, Y =k) =e−3 1 n! _n k 2n−k n= 0,1,2, . . . , k= 0,1, . . . , n.

Form= 0,2,4 calculate (a)_P(X+Y =m). Remark: 0₀ = 1.

1. 1(a) Hint: ifX andZ are independent then cov(X, Z) = 0(provided that it is well defined).

2.

3. (a) Since forZ ∼P oi(λ), _EZ =σ2

Z =λandX ∼P oi(2), Y ∼P oi(3)we

have _EX = 2 _EY = 3 and _E(X+Y) = _EX +_EY = 2 + 3 = 5. Next, we have σ2

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σ2_X₊_Y =σ2_X+σ_Y2 = 2 + 3 = 5. (b) The characteristic function of a ran-dom variable Z such that Z ∼ P oi(λ) reads as ϕZ(t) = Eexp (itZ) =

e−λP+∞ k=0e itk λk k! = exp λ e it₋₁ . Hence ϕX(t) = exp 2 eit−1

and ϕY(t) = exp 3 eit−1. SinceX and Y are independent, we have

ϕX+Y(t) = ϕX(t)ϕY(t) = exp 5 eit−1

. (c) Since the characteristic function uniquely determines the distribution, we identify the distribution ofX+Y asP oi(5)and_P(X+Y =k) =e−5₅k_/k_!_for_k_{= 0}_,₁_,₂_{, . . .}_.

An-other method to obtain this distribution is to apply convolution formula P(X+Y =k) =Pkl=0P(X=l)P(Y =k−l).

4. (a) Since forZ ∼P oi(λ),EZ =σZ2 =λandXj∼P oi(2),j= 1,2, . . . , n

we haveEX¯ =En1 Pn j=1Xj = Pn j=1 1 nEXj = 1 nn·2 = 2. SinceX1, X2, . . . , Xn

are independent, we haveσX2¯ = 1 n2σ 2 X1+X2+...+Xn = 1 n2 σ 2 X1+σ 2 X2+. . .+σ 2 Xn = 1 n2n·2 = 2 n and σX¯ = q 2

n. (b) Similarly as in the previous problem we

see that ϕS(t) = (ϕX1(t))

n

= exp 2 eit₋₁n

= exp 2n eit₋₁

and since the characteristic function uniquely determines the distribution, we identify the distribution ofS asP oi(2n).

5. (a) Since for Z ∼ N(µ, σ2₎_,

EZ = µ, σ2Z = σ2 and X ∼ N(0,1),

Y ∼ N(0,4) we haveEX = 0, EY = 0 and E(X+Y) =EX +EY =

0 + 0 = 0. Next, we have σ2

X = 1, σ2Y = 4 and since X and Y are

independent, we have σ_X2₊_Y =σ2_X+σ_Y2 = 1 + 4 = 5. (b) The charac-teristic function of a random variable Z such that Z ∼ N(µ, σ2) reads as ϕZ(t) = Eexp (itZ) = exp itµ−12t

2_σ2

. Hence ϕX(t) = exp −1₂t2

and ϕY(t) = exp −1₂4t2. Since X and Y are independent, we have

ϕX+Y(t) =ϕX(t)ϕY(t) = exp −1₂5t2

. (c) Since the characteristic func-tion uniquely determines the distribufunc-tion, we identify the distribufunc-tion of

X +Y as N(0,5) and fX+Y(z) = √₁₀1_πexp −₁₀1z2

. Another method to obtain this distribution is to apply convolution formula fX+Y(z) =

R+∞

x=−∞fX(x)fY(z−x)dx.

6. Similarly as in the previous problem we see that ϕS(t) = (ϕX1(t))

n = e−2t2 n =e−2nt2 = e−124nt 2

and we identify S ∼ N(0,4n). (b) Since

¯ X = 1 nS we have EX¯ = 1 nES = 1 n0 = 0 and σ 2 ¯ X = 1 n2σ 2 S = 1 n24n = 4 n.

Finally, we notice that ϕ_X¯(t) =_Eexp itX¯=_Eexp it1_nS=ϕS nt

= e−124n( t n) 2 =e−12 4 nt 2 and we identifyX¯ ∼ N(0,_n4).

7. Let Z ∼Erlang(n, λ). By the definition of a characteristic function we haveϕZ(t) =Eexp (itZ) = λ n (n−1)! R+∞ 0 e itz_zn−1_e−λz_d_z₌ λn (n−1)! R+∞ 0 z n−1_e−(λ−it)z_d_z₌ λn (n−1)! (n−1)! (λ−it)n = λ λ−it n

. (This formula is analogous to the formula for

MZ(t) =

λ λ−t

n

.) Now, since X ∼Erlang(3,0.2), Y ∼Erlang(2,0.2)

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indepen-dent, we haveϕX+Y(t) =ϕX(t)ϕY(t) =

0.2 0.2−it

5

. (b) From the form of the characteristic function ofX+Y we see thatX+Y ∼Erlang(5,0.2). (c) For Z ∼ Erlang(n, λ) we have mk(Z) = EZk = (n(+n−k−1)!1)!

1 λk thus E(X+Y)3=(5+3(5−−1)!1)! 1 0.23 = 7! 4!5 3_{= 5}_·₆_·₇_·₅3_{= 26250}_.

Conditional distributions. Conditional expectation and

con-ditional variance with respect to a variable.

1. A hen laysN eggs, whereN has geometric distribution with the following probability mass function: forn= 0,1,2, . . .

P(N =n) =q·pn,

where q = 1−p ∈ (0,1). Each egg hatches with probability r ∈ (0,1)

independently of the other eggs. LetKbe the number of chicks. (a) Find the joint probability mass function of N and K. (b) Find the marginal probability mass function ofK.

2. X ∼ N(0,1) and Y ∼ N(0,4) are two independent random variables. Find the conditional probability density function of Z = X +Y given

X = 3.

3. X ∼ N(0,1) and Y ∼ N(0,4) are two independent random variables. Find the conditional probability density function ofY givenZ=X+Y = 3.

4. X1, X2, . . . , X10are independent random variables such thatXj ∼ N(0,4)

for j = 1,2, . . . ,10. Find the conditional probability density function of

X1 givenX1+X2+. . .+X10= 3.

5. A hen laysN eggs, whereN has geometric distribution with the following probability mass function: forn= 0,1,2, . . .

P(N =n) =q·pn,

where q = 1−p ∈ (0,1). Each egg hatches with probability r ∈ (0,1)

independently of the other eggs. Let K be the number of chicks. (a) CalculateE(K|N= 3). (b) Calculate E(K|N). (c) Calculate EK. 6. X ∼ N(0,1)andY ∼ N(0,4) are two independent random variables and

Z=X+Y. (a) CalculateE(Z|X = 3). (b) CalculateE(Z|X).

7. X1, X2, . . . , X10are independent random variables such thatXj ∼ N(0,4)

forj= 1,2, . . . ,10andS=X1+X2+. . .+X10. (a) CalculateE(X1|S= 3).

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8. Each driver in a small town participates in a random number N of ac-cidents during a year. The number of acac-cidents N depends on driver’s parameterP which describes her/his driving skills and the number of ac-cidents givenP =phasBin(4, p)distribution,N|P =p∼Bin(4, p). The parameter P in the population of drivers of the town has uniform distri-bution on the interval[0,1]. (a) Find the marginal distribution ofN. (b) FindE(N|P). (c) Find_EN.

1.

2. Since Z = X +Y we have that Z given X = 3 has the same distri-bution as 3 +Y given X = 3. Since Y is independent from X and

3 +Y ∼ N(3,4)this yields that Z|X = 3∼ N (3,4). Thus fZ|X=3(z) = 1 2√2πexp −1 8(z−3) 2 .

3. We will use the following formula fY|Z=3(y) =

fY,Z(y,3)

fZ(3) =

fY,X(y,3−y)

fZ(3) .

SinceXandY are independent, we havefY,X(y,3−y) =fX(3−y)fY(y) =

1 √ 2πexp − 1 2(3−y) 2 1 2√2πexp − 1 8y 2 and Z = X +Y ∼ N(0,5) thus fZ(3) = √₁₀1_πexp −₁₀132

. Finally we find thatfY|Z=3(y) = √ 1 2π·4/5exp − 1 2·4/5(y−12/5) 2 . Another solution. We will apply the fact that if(X1, X2, . . . , Xn) =X∼

N(m,Σ) and B is m×n matrix then BX ∼ N Bm, BΣBT. Since

(X, Y)∼ N 0 0 , 1 0 0 4 and(Y, Z) = (Y, X+Y) = Y X+Y = 0 1 1 1 X Y we have that(Y, Z)∼ N 0 1 1 1 0 0 , 0 1 1 1 1 0 0 4 0 1 1 1 = N 0 0 , 0 4 1 4 0 1 1 1 = N 0 0 , 4 4 4 5 . We see that ρ(Y, Z) = cov(_σ Y,Z) YσZ = 4 √ 4√5 = 2 √

5. By the formula for the conditional

dis-tribution of the bivariate normal disdis-tribution we getY|Z=z∼ NµY +ρσ_σY_Z (z−µZ), 1−ρ2

σ2 Y = N0 +√2 5 2 √ 5(z−0), 1− 4 5 4=N(4z/5,4/5)thusY|Z= 3∼ N(12/5,4/5). 4. We have that Y = X2 +X3 +. . .+X10 ∼ N(0,9·4) = N(0,36).

Now, similarly as in the previous problem, we write fX1|X1+Y=3(x1) =

fX₁,X_{1 +}Y(x1,3)

fX1 +Y(3) =

fX₁,Y(x1,3−x1)

fX1 +Y(3) . SinceX1andY are independent, we have

fX1,Y(x1,3−x1) =fX1(x1)fY(3−x1) = 1 2√2πexp − 1 8(x1) 2 1 6√2πexp −1 72(3−x1) 2 . We also have X1+Y ∼ N(0,10·4)and fX1+Y(3) =

1 √ 80πexp − 1 803 2 . Finally we find thatfX1|X1+Y=3(x1) =

1 √ 2π·18/5exp − 1 2·18/5(x1−0.3) 2 . 5.

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6. (a) Using the conditional distribution obtained in Problem 2. we get thatE(Z|X = 3) = 3. (b) Using properties of conditional expectation we calculateE(Z|X) =E(X+Y|X) =E(X|X) +E(Y|X) =X+EY =X. 7. (a) Using the conditional distribution obtained in Problem 4. we get that E(X1|S= 3) = 0.3. Another solution. By symmetry, for j =

sym-metry, we getE(X1|S) =S/10.

8. (a) The marginal distribution ofNreads asP(N =n) =R01P(N=n|P =p)fP(p)dp=

R1

0 4

n

pn₍₁₋_p₎4−n_d_p_for_n_{= 0}_,₁_,₂_,₃_,₄_{(this may be calulated explicitly}

Multidimensional normal random variables. Linear

regres-sion. Central Limit Theorem.

1. Let(X1, X2, X3) =X∼ N     1.2 2.3 5  ,   2 0 1 0 2 2 1 2 4   

. (a) Find the

dis-tribution of2X1+X2−3X3. Hint: If(X1, X2, . . . , Xn) =X∼ N(m,Σ)

andB ism×nmatrix thenBX∼ N Bm, BΣBT

. (b) Find the condi-tional distribution of(X2, X3)givenX1= 0.2.

2. There are 250 dogs at a dog show who weigh an average of 12 pounds, with a standard deviation of 8 pounds. If 4 dogs are chosen at random, what is the probability they have an average weight of greater than 8 pounds and less than 25 pounds?

1. (a) We notice thatY = 2X1+X2−3X3= [2,1,−3]

  X1 X2 X3  thus (applying

the hint) we getY ∼ N

 [2,1,−3]   1.2 2.3 5  ,[2,1,−3]   2 0 1 0 2 2 1 2 4     2 1 −3    = N  −10.3,[1,−4,−8]   2 1 −3    =N(−10.3,22). 2.