Math 180C MIDTERM 2 DUE 05/22/20, 11:00pm

Write your name and PID on the top of EVERY PAGE.

Write the solutions to each problem on separate pages.

CLEARLY INDICATE on the top of each page the number of

the corresponding problem.

Remember this exam is graded by a human being. Write your

solutions NEATLY AND COHERENTLY, or they risk not re-

ceiving full credit.

You may assume that all transition probability functions are

STATIONARY.

You are allowed to use the textbook, lecture notes and your

personal notes. You are not allowed to use the electronic devices

(except for accessing the online version of the textbook) or outside

assistance. Outside assistance includes but is not limited to other

people, the internet and unauthorized notes.

Math 180C MIDTERM 2, Page 2 of 5 DUE 05/22/20, 11:00pm

1. (25 points) Let Y > 0 be a random variable having Gamma distribution with parameters

2 and λ, i.e., the p.d.f. of Y is given by

fY (y) = λ

2ye−λy, y > 0, (1)

and let X ∼ Unif[0, Y ] be a random variable uniformly distributed on [0, Y ].

It is given that E(X) = 1.

(a) (10 points) Determine the unknown parameter λ. [Hint. Compute E(X) with

unknown parameter λ.]

(b) (5 points) Determine

P (X ≤ t |Y = y) =

, t < 0,

, 0 ≤ t < y,

, t ≥ y.

(2)

for y > 0.

(c) (10 points) Using the results from (a) and (b), compute P (X ≤ t) and determine

the marginal distribution of X.

Solution.

(a) Compute E(X) by conditioning on the value of Y

E(X) =

∫ ∞

0

E(X |Y = y)fY (y)dy. (3)

For any y > 0, E(X |Y = y) = y

2

, therefore, using the integration by parts,

E(X) =

∫ ∞

0

y

2

λ2ye−λydy = −λ

2

∫ ∞

0

y2de−λy =

∫ ∞

0

yλe−λydy =

1

λ

. (4)

We conclude that λ = 1.

(b) Given Y = y, X has uniform distribution on [0, y], so

P (X ≤ t |Y = y) =

0, t < 0,

t

y

, 0 ≤ t < y,

1, t ≥ y.

(5)

(c) Using the results from parts (a) and (b), we compute P (X ≤ t) for t > 0 by

Math 180C MIDTERM 2, Page 3 of 5 DUE 05/22/20, 11:00pm

conditioning on the value of Y

P (X ≤ t) =

∫ ∞

0

P (X ≤ t |Y = y)fY (y)dy (6)

=

∫ t

0

fY (y)dy +

∫ ∞

t

t

y

fY (y)dy (7)

=

∫ t

0

ye−ydy +

∫ ∞

t

t

y

ye−ydy (8)

= (−te−t + 1− e−t) + te−t (9)

= 1− e−t. (10)

We see that X has exponential distribution with rate 1.

2. (25 points) Certain device consists of two components, A and B. Whenever one of the

components fails, the whole device is immideately replaced by a new one. Components

A and B are the only components that can fail.

Suppose that the lifetimes of components A and B (in days) are independent random

variables both having exponential distributions with rate λ. Let N(t) be the renewal

process counting the number of the replacements of the device on the time interval [0, t].

(a) (10 points) Express the interrenewal times in terms of the lifetimes of components

A and B (hint: this is not a sum) and compute the distribution of the interrenewal

times.

(b) (10 points) Determine an asymptotic expression for the mean age of the device at

time t in the long run.

(c) (5 points) What is the long run probability that the device will fail within next 24

hours?

Solution. Denote by Xi and Yi, i ≥ 1, the random variables describing the lifetimes

of the components A and B correspondingly. Then all Xi and Yi, i ≥ 1, are i.i.d. with

exponential distribution with rate λ.

(a) Denote by Zi the lifetime of the i-th device. Then Zi = min{Xi, Yi} is the in-

terrenewal time of the process that counts the number of the replacement of the

device. Since Xi and Yi are independent exponentially distributed random vari-

ables, Zi ∼ Exp(2λ).

(b) Let δt denote the age of the device at time t. Then, using the limit theorem for the

average age, we have

lim

t→∞

E(δt) =

Var(Z1) + (E(Z1))

2

2E(Z1)

=

(2λ)−2 + (2λ)−2

2λ−1

=

1

2λ

. (11)

Math 180C MIDTERM 2, Page 4 of 5 DUE 05/22/20, 11:00pm

(c) We have to compute the probability that δt ≤ 1 in the limit as t→∞. The limiting

density of δt is given by

f(x) = 2λe−2λx, (12)

therefore

lim

t→∞

P (δt ≤ 1) =

∫ 1

0

2λe−2λxdx = 1− e−2λ. (13)

3. (25 points) The climate of a certain tropical country is characterized by the alternating

periods of rain and (sunny) periods without precipitations. Let (Xi)i≥0 and (Yi)i≥1 be

the random variables describing the lengths of the consecutive rainy and sunny periods of

time correspondingly, and assume that (Xi)i≥0 and (Yi)i≥0 are two independent families

of i.i.d. continuous random variables.

We start the observation at the beginning of one of the rainy periods and count the

number of times the weather changes from sunny to rainy. Suppose that

E(X1) = α, Var(X1) = 2α

2, E(Y1) = β, Var(Y1) = 2β

2 (14)

for some α > 0, β > 0.

(a) (15 points) Determine an asymptotic expression (linear and constant terms) of the

expected number of times the weather changes from sunny to rainy on the interval

[0, t] for t 1.

(b) (10 points) What is the long run average fraction of time that the weather in this

country is sunny?

Solution.

(a) Denote by N(t) the renewal process that counts the number of times the weather

changes from sunny to rainy. Then the interrenewal times are given by Xi + Yi.

Denote

µ = E(Xi + Yi) = α + β, σ

2 = Var(Xi + Yi) = 2(α

2 + β2). (15)

Then the asymptotic behavior of E(N(t)) (see (7.17) in the textbook) yields that

for t 1

E(N(t)) ≈ t

α + β

+

σ2 − µ2

2µ2

=

t

α + β

+

(α− β)2

2(α + β)2

. (16)

(b) The long run fraction of sunny days is

β

α + β

. (17)

Math 180C MIDTERM 2, Page 5 of 5 DUE 05/22/20, 11:00pm

4. (25 points) Let X1, X2, . . . be i.i.d. random variables having exponential distribution

with rate 1, i.e., X1 ∼ Exp(1).

(a) (10 points) Let Y be an exponential random variable with rate λ. Compute

MY (t) = E(e

tY ) =

{

, t < λ,

, t ≥ λ (18)

for t ∈ (−∞,∞). (Recall that MY (t) is called the moment generating function of

Y ).

(b) (15 points) Using the result from (a) show that for any t < 1, the process (Zn)n≥1

defined by

Zn = (1− t)net

∑n

i=1Xi , Z0 = 1 (19)

is a nonegative martingale.

Solution.

(a) The moment generating function of Y is given by

E(etY ) =

∫ ∞

0

etyλe−λydy =

∫ ∞

0

λe−(λ−t)y =

{

λ

λ−t , t < λ,

+∞, t ≥ λ. (20)

(b) Using the result of part (a) with λ = 1, for t < 1

E((1− t)e−Xi) = 1. (21)

Therefore,

Zn =

n∏

i=1

(1− t)e−Xi (22)

is a product of i.i.d. nonnegative random variables with mean 1, which defines a

(nonnegative) multiplicative martingale.