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UNIVERSITY OF CALIFORNIA, DAVISDepartment of Electrical and Computer Engineering

EEC161 Probabilistic Analysis of Electrical and Computer Systems Spring 2012

Midterm #2 Solutions

Problem 1

a) In order to ensure that fX Y(x, y) is a valid PDF, we must have

1 =

fX Y(x, y)dxdy

= C

10

x[

x0

y2dy]dx

= C3

1

0x4dx = C

15,

so C = 15.

b) The marginal probability density fX(x) of X is given by

fX(x) =

x0

fX Y(x, y)dy = 15x4

3= 5x4

for 0 x 1 and fX(x) = 0 otherwise. Similarly, the marginal PDF of Y is given by

fY(y) =1

yfX Y(x, y)dx =

15

2 (1 y2)y2

for 0 y 1 and fY(y) = 0 otherwise.c) Since

fX(x)fY(y) =75

2x4(1 y2)y2

for 0 x, y 1, we conclude that

fX(x)fY(y) = fX Y(x, y)

so X and Y are not independent.

d) We have

fY|X(y|x) =fX Y(x, y)

fX(x)= 3

y2

x3

for 0 y x and fY|X(y|x) = 0 otherwise.

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EEC161 Probabilistic Analysis of Electrical and Computer Systems Spring 2012

e) The conditional expectation

E[Y|X = x] =

yfY|X(y|x)dy

=3

x3

x

0

y3dy =3

4x .

Problem 2:

a) For n 1, N = n visits to replacement parts stores are needed if the first n1 stores donot carry the desired replacement part (each with probability 1p), and the n-th store hasthe needed part (with probability p). Thus N is a type-1 geometric random variable with

P[N = n] = p(1 p)n1

for n 1. The moment generating function is

GN(z) = pz

N=1

((1 p)z)n1

=pz

1 (1 p)z ,

for (1 p)|z| < 1. Its first and second derivatives are given by

GN(z) =p(1 (1 p)z) +pz(1 p)

(1 (1 p)z)2=

p

(1 (1 p)z)2

d

2

GNdz2 = 2p(1 p)(1 (1 p)z)3 ,

so

mN =dGN

dz(1) =

1

p

E[N2] =d2GN

dz2(1) +

dGNdz

(1) =2(1 p)

p2+

1

p,

and

KN = E[N2] m2N =

1 pp2

.

b) The generating function of X is

MX(s) =

esxfX(x)dx =

0

e(s)xdx

=

s

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EEC161 Probabilistic Analysis of Electrical and Computer Systems Spring 2012

for (s) < . The first and second derivatives of MX(s) are given bydMX

ds=

( s)2d2MX

ds2=

2

( s)3,

so

mX =dMX

ds(0) =

1

E[X2] =d2MX

ds2(0) =

2

2,

and

KX = E[X2] m2X =

1

2.

c) If Charles visits N = n stores until he finds the desired part, the time spent is

Y =n

i=1

Xi

and since the random variables Xi are independent

MY|N(s|n) = E[esY|N = n] =n

i=1

E[esXi ] = (MX(s))n .

d) By using the principle of total probability

MY(s) =

n=1

E[esY|N = n]P[N = n]

= pMX(s)

n=1

(MX(s)(1 p))n1 = pMX(s)1 (1 p)MX(s)

for values of s such that (1 p)MX(s) < 1. This gives

MY(s) =p/( s)

1 (1 p)/( s)=

p

p s.

e) Comparing MY(s) with MX(s), we conclude that Y is therefore an exponential randomvariable with parameter p. Its mean mY and variance KY are therefore

mY =1

p, KY =

1

(p)2.

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EEC161 Probabilistic Analysis of Electrical and Computer Systems Spring 2012

Problem 3:

a) Since X and Y are independent, their joint density is given by

fX Y(x, y) = fX(x)fY(y) = exp((x + y))u(x)u(y) .

b) The transformation can be written in matrix form asUV

= A

XY

, (1)

where

A =

1 11 1

.

The transformation (1) is one-to-one and its inverse is given byXY

=

h1(U, V)h2(U, V)

= A1

UV

withA1 =

1

2

1 11 1

.

The Jacobian J(u, v) of the transformation (1) is given by

J = | det A1| = 12

,

and the joint density of U and V can be expressed as

fU V(u, v) = J(u, v)fX Y(h1(u, v), h2(u, v))

=1

2exp(u)u(u + v)u(u v)

=

12 exp(u) u > |v|

0 otherwise .

c) The marginal density of U is

fU(u) =

fU V(u, v)dv

=1

2exp(u)u(u)

uu

dv = u exp(u)u(u) ,

which is an Erlang distribution of order 1 and parameter = 1, as expected since U is the

sum of two exponential random variables with parameter = 1. The marginal density ofV is

fV(v) =

fU V(u, v)du

=1

2

|v|

exp(u)du = 12

exp(|v|) ,

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EEC161 Probabilistic Analysis of Electrical and Computer Systems Spring 2012

so that V has a Laplace distribution with parameter = 1.

d) The joint densityfU,V(u, v) = fU(u)fV(v) ,

so U and V are not independent.

e) The random variables U and V have for meanmUmV

= A

mXmY

=

1 11 1

11

=

20

,

and for covarianceKU KU V

KV U KV

= E

U mUV mV

U mU V mV

= A

1 00 1

AT =

2 00 2

.

Since KU V = E[(UmU)(V mV)] = 0, we conclude that U and V are uncorrelated eventhough they are not independent.

Problem 4

a) By the central limit theorem, for large n, the random variable

Yn =Sn mXn

KXn=

Sn 3n3

n

is N(0, 1) distributed. For n = 100, we find therefore

P[Sn > 360] = P[Yn >60

30] = P[Yn > 2]

= Q(2) = 2.275.102 ,

so there is roughly only a 2% chance the parking lot capacity will be exceeded after 100days.

b) On day N, we have

P[SN > 360] = P

YN >

360 3N3

N = Q

120 N

N

= 1

2,

so that120 N

N= Q1(1/2) = 0

and thus N = 120.

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