University of Toronto Scarborough

Department of Computer & Mathematical Sciences

STAB52H3 Introduction to Probability

Term Test 2

November 16, 2020

Duration: 60 minutes

Examination aids allowed: Open notes-books

Instructions:

• Read the questions carefully and answer only what is being asked.

• Answer all questions directly on the examination paper; use the last pages if you need

more space, and provide clear pointers to your work.

• Show your intermediate work, and write clearly and legibly.

Question: 1 2 3 4 5 6 7 8 9 10 Total

Points: 20 20 20 20 20 20 20 20 20 20 200

Score:

1. (20 points) Consider two random variables X and Y that are binary valued, i.e., the set

of possible values is {0, 1}. Let the marginal distribution of X be

P(X = 0) = p, P(X = 1) = 1− p.

It is given that the conditional distribution of Y given X is

P(Y = 0|X = 0) = 3

4

, P(Y = 1|X = 0) = 1

4

,

P(Y = 1|X = 1) = 1

2

, P(Y = 0|X = 1) = 1

2

.

(a) (10 points) Find the value of p ∈ [0, 1] such that the marginal distributions of X

and Y are the same.

(b) (10 points) For the value of p given in (a), find the conditional distribution of X

given Y .

Solution: (a) By the law of total probability, we have

P(Y = 0) = P(Y = 0|X = 0)P(X = 0) + P(Y = 0|X = 0)P(X = 0)

=

3

4

p+

1

2

(1− p).

We require that X and Y have the same marginal distribution. So

P(Y = 0) =

3

4

p+

1

2

(1− p) = p = P(X = 0).

Solving gives p = 2

3

. Since X and Y are binary valued, if p = 2

3

then they have the

same marginal distributions.

(b) We have

P(X = 0|Y = 0) = P(Y = 0|X = 0)P(X = 0)

P(Y = 0)

= P(Y = 0|X = 0) = 3

4

,

P(X = 1|Y = 0) = 1− P(X = 0|Y = 0) = 1

4

,

P(X = 1|Y = 1) = P(Y = 1|X = 1)P(X = 1)

P(Y = 1)

= P(Y = 1|X = 1) = 1

2

,

P(X = 0|Y = 1) = 1− P(X = 1|Y = 1) = 1

2

.

(Note that the transition probabilities from X to Y are the same as those from Y to

X.)

Page 2 of 10 (contd...)

2. (20 points) Consider two random variables X and Y that are binary valued, i.e., the set

of possible values is {0, 1}. Let the marginal distribution of X be

P(X = 0) = p, P(X = 1) = 1− p.

It is given that the conditional distribution of Y given X is

P(Y = 0|X = 0) = 2

3

, P(Y = 1|X = 0) = 1

3

,

P(Y = 1|X = 1) = 1

2

, P(Y = 0|X = 1) = 1

2

.

(a) (10 points) Find the value of p ∈ [0, 1] such that the marginal distributions of X

and Y are the same.

(b) (10 points) For the value of p given in (a), find the conditional distribution of X

given Y .

Solution: (a) By the law of total probability, we have

P(Y = 0) = P(Y = 0|X = 0)P(X = 0) + P(Y = 0|X = 0)P(X = 0)

=

2

3

p+

1

2

(1− p).

We require that X and Y have the same marginal distribution. So

P(Y = 0) =

2

3

p+

1

2

(1− p) = p = P(X = 0).

Solving gives p = 3

5

. Since X and Y are binary valued, if p = 3

5

then they have the

same marginal distributions.

(b) We have

P(X = 0|Y = 0) = P(Y = 0|X = 0)P(X = 0)

P(X = 0)

= P(Y = 0|X = 0) = 2

3

,

P(X = 1|Y = 0) = 1− P(X = 0|Y = 0) = 1

3

,

P(X = 1|Y = 1) = P(Y = 1|X = 1)P(X = 1)

P(Y = 1)

= P(Y = 1|X = 1) = 1

2

,

P(X = 0|Y = 1) = 1− P(X = 1|Y = 1) = 1

2

.

(Note that the transition probabilities from X to Y are the same as those from Y to

X.)

Page 3 of 10 (contd...)

3. (20 points) Let (X, Y ) be a random vector whose distribution is given by

P((X, Y ) = (0, 0)) = P((X, Y ) = (0, 1)) = P((X, Y ) = (1, 1)) =

1

3

.

(a) (4 points) Find the marginal distributions of X and Y , and give their name and

parameters.

(b) (16 points) Find the correlation coefficient Cor(X, Y ) = ρXY between X and Y .

Solution: (a) We have X ∼ Bernoulli(1

3

) and Y ∼ Bernoulli(2

3

).

(b) The covariance is given by

Cov(X, Y ) = E[XY ]− E[X]E[Y ].

We have

E[XY ] =

1

3

0 · 0 + 1

3

0 · 1 + 1

3

1 · 1 = 1

3

,

Recall that if Z ∼ Bernoulli(p) then E[Z] = p and Var(Z) = p(1− p). So

E[X] =

1

3

, E[Y ] =

2

3

,

and we have

Cov(X, Y ) =

1

3

− 1

3

2

3

=

1

9

.

Also

Var(X) =

1

2

2

3

=

2

9

, Var(Y ) =

2

3

1

3

=

2

9

.

Thus

Cor(X, Y ) =

Cov(X, Y )√

Var(X)

√

Var(Y )

=

1

2

.

4. (20 points) Let (X, Y ) be a random vector whose distribution is given by

P((X, Y ) = (0, 1)) = P((X, Y ) = (1, 0)) = P((X, Y ) = (1, 1)) =

1

3

.

(a) (4 points) Find the marginal distributions of X and Y , and give their name and

parameters.

(b) (16 points) Find Var(X + Y ).

Page 4 of 10 (contd...)

Solution: (a) We have X ∼ Bernoulli(2

3

) and Y ∼ Bernoulli(2

3

).

(b) We have

Var(X + Y ) = Var(X) + 2Cov(X, Y ) + Var(Y ).

The covariance is given by

Cov(X, Y ) = E[XY ]− E[X]E[Y ].

We have

E[XY ] =

1

3

0 · 1 + 1

3

1 · 0 + 1

3

1 · 1 = 1

3

,

Recall that if Z ∼ Bernoulli(p) then E[Z] = p and Var(Z) = p(1− p). So

E[X] =

2

3

, E[Y ] =

2

3

,

and we have

Cov(X, Y ) =

1

3

− 2

3

2

3

=

−1

9

.

Also

Var(X) = Var(Y ) =

2

3

1

3

=

2

9

.

Thus

Var(X + Y ) =

2

9

− 21

9

+

2

9

=

2

9

.

5. (20 points) Consider two independent Exponential RVs: X ∼ Exponential(1) and Y ∼

Exponential(2). Find the probability P(X > Y ).

Solution:

The joint PDF is fX,Y (x, y) = fX(x)fY (y) = e

−x2e−2y, ∀x, y > 0. The probability is

given by the integral of the joint PDF over the region R = {(x, y) ∈ R : x > y} =

Page 5 of 10 (contd...)

{(x, y) ∈ R : 0 < x, 0 < y < x}

P(X > Y ) =

∫∫

R

fX,Y (x, y)dxdy

=

∫ ∞

x=0

[∫ x

y=0

2e−xe−2ydy

]

dx

=

∫ ∞

x=0

e−x

[∫ x

y=0

2e−2ydy

]

dx

=

∫ ∞

x=0

e−x

[−e−2y]x

0

dx

=

∫ ∞

x=0

e−x

[

1− e−2x] dx

=

∫ ∞

x=0

e−xdx−

∫ ∞

x=0

e−3xdx

= 1− [−1

3

e−3x

]∞

x=0

= 1− [0 + 1

3

] = 2

3

.

6. (20 points) Consider two independent and identically distributed Exponential RVsX, Y ∼

Exponential(1). Find the probability P(X > 2Y ).

Solution:

The joint PDF is fX,Y (x, y) = fX(x)fY (y) = e

−xe−y, ∀x, y > 0. The probability is

given by the integral of the joint PDF over the region R = {(x, y) ∈ R : x > 2y} =

Page 6 of 10 (contd...)

{(x, y) ∈ R : 0 < x, 0 < y < x/2}

P(X > 2Y ) =

∫∫

R

fX,Y (x, y)dxdy

=

∫ ∞

x=0

[∫ x

y=0

e−xe−ydy

]

dx

=

∫ ∞

x=0

e−x

[∫ x/2

y=0

e−ydy

]

dx

=

∫ ∞

x=0

e−x

[−e−y]x/2

0

dx

=

∫ ∞

x=0

e−x

[

1− e−x/2] dx

=

∫ ∞

x=0

e−xdx−

∫ ∞

x=0

e−3x/2dx

= 1− [−2

3

e−3x/2

]∞

x=0

= 1− [0 + 2

3

] = 1

3

.

7. Let the RV X follow Uniform(0, 1) distribution, and define the new RV Y = 1

1+X

.

(a) (3 points) What is the range of possible values of Y ?

(b) (10 points) Find the CDF of Y .

(c) (7 points) Find the PDF of Y .

Solution:

(a) Since X ranges in [0, 1], then Y = 1

1+X

will take values between 1

1+0

= 1 and

1

1+1

= 1

2

, i.e. Y ∈ [1

2

, 1]

(b) The CDF of the Uniform(0,1) is the identity function, FX(x) = x, ∀x ∈ (0, 1).

We have

FY (y) = P(Y ≤ y) = P

(

1

1 +X

≤ y

)

= P

(

1 +X ≥ 1

y

)

= P

(

X ≥ 1

y

− 1

)

= 1− P

(

1 +X <

1

y

)

= 1− FX

(

1

y

− 1

)

= 1−

(

1

y

− 1

)

= 2− 1

y

, ∀y ∈ [1

2

, 1]

Page 7 of 10 (contd...)

(c) There are two ways to find the PDF of Y :

a) Differentiate the CDF of Y form the previous part:

fY (y) =

d

dy

FY (y) =

d

dy

(2− y−1) = y−2 = 1

y2

, ∀y ∈ [1

2

, 1]

b) Use the PDF method:

h(x) = (1 + x)−1 ⇒

{

h−1(y) = y−1 − 1

h′(x) = −(1 + x)−2

fY (y) =

fX(h

−1(y))

|h′(h−1(y))| =

1

| − (1 + h−1(y))−2|

=

1

(1 + y−1 − 1)−2 =

1

y2

, ∀y ∈ [1

2

, 1]

8. Let the RV X follow Uniform(0, 1) distribution, and define the new RV Y = 1−X2.

(a) (3 points) What is the range of possible values of Y ?

(b) (10 points) Find the CDF of Y .

(c) (7 points) Find the PDF of Y .

Solution:

(a) Since X ranges in [0, 1], then Y = 1 −X2 will take values between 1 − 02 = 1

and 1− 12 = 0, i.e. Y ∈ [0, 1]

(b) The CDF of the Uniform(0,1) is the identity function, FX(x) = x, ∀x ∈ (0, 1).

We have

FY (y) = P(Y ≤ y) = P

(

1−X2 ≤ y)

= P

(

X2 ≥ 1− y) = P(X ≤√1− y)

= 1− P

(

X <

√

1− y

)

= 1− FX

(√

1− y

)

= 1−

√

1− y, ∀y ∈ [0, 1]

(c) There are two ways to find the PDF of Y :

a) Differentiate the CDF of Y form the previous part:

fY (y) =

d

dy

FY (y) =

d

dy

(

1−

√

1− y

)

=

d

dy

(

1− (1− y)12

)

=

1

2

(1− y)−12 = 1

2

√

1− y , ∀y ∈ [0, 1]

Page 8 of 10 (contd...)

b) Use the PDF method:

h(x) = 1− x2 ⇒

{

h−1(y) =

√

1− y

h′(x) = −2x

fY (y) =

fX(h

−1(y))

|h′(h−1(y))| =

1

| − 2(√1− y)|

=

1

2(

√

1− y) , ∀y ∈ [0, 1]

9. Consider two RVs X, Y where X has marginal PDF fX(x) =

{

x/2, x ∈ (0, 2)

0, otherwise

and

Y has conditional PDF fY |X(y|x) =

{

1/x, y ∈ (0, x)

0, otherwise

(i.e. Y given X is uniformly

distributed in (0, X).)

(a) (10 points) Find the marginal PDF of Y ; are X and Y independent? (justify your

answer).

(b) (10 points) Find the conditional PDF of X given Y , and identify its distribution

(i.e. give its name and parameters).

Solution:

(a) The joint PDF of X, Y is given by

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

{

x

2x

= 1

2

, 0 < y < x < 2

0, otherwise

The marginal PDF of Y is found by integrating out X from the joint PDF

fY (y) =

∫ ∞

x=−∞

fX,Y (x, y)dx =

{∫ 2

x=y

1

2

dx = 2−y

2

, y ∈ (0, 2)

0, otherwise

The RVs are NOT independent, since the conditional and marginal PDFs of Y

are different (moreover, the conditional range of Y depends on X).

(b) The conditional PDF of X given Y is

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

fY (y)

=

{

1/2

(2−y)/2 =

1

2−y , x ∈ (y, 2)

0, otherwise

Note that the conditional PDF of X does not depend on X (is flat), and is

actually Uniform(y, 2).

Page 9 of 10 (contd...)

10. Consider two RVs X, Y where X has marginal PDF fX(x) =

{

2x, x ∈ (0, 1)

0, otherwise

and Y has

conditional PDF fY |X(y|x) =

{

1/x, y ∈ (0, x)

0, otherwise

(i.e. Y given X is uniformly distributed

in (0, X).)

(a) (10 points) Find the marginal PDF of Y ; are X and Y independent? (justify your

answer).

(b) (10 points) Find the value of the conditional probability P(X > 3

4

|Y = 1

2

).

Solution:

(a) The joint PDF of X, Y is given by

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

{

2x/x = 2, 0 < y < x < 1

0, otherwise

The marginal PDF of Y is found by integrating out X from the joint PDF

fY (y) =

∫ ∞

x=−∞

fX,Y (x, y)dx =

{∫ 1

x=y

2dx = 2(1− y), y ∈ (0, 1)

0, otherwise

The RVs are NOT independent, since the conditional and marginal PDFs of Y

are different (moreover, the conditional range of Y depends on X).

(b) Note that the joint distribution is Uniform over the triangle 0 < y < x < 1. If

you condition on Y = 1/2, then X will range uniformly over (1/2, 1); hence, the

conditional probability of X > 3/4 is P (X > 3

4

|Y = 1

2

) = 1/2. You can verify

this more formally using the conditional PDF of X given Y

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

fY (y)

=

{

2

2(1−y) =

1

1−y , x ∈ (y, 1)

0, otherwise

Note that the conditional PDF of X does not depend on X (is flat), and is

actually Uniform(y, 1). For Y = 1/2, we have fX|Y (x|1/2) = 11−1/2 = 2,∀x ∈

(1/2, 1).

P(X > 3

4

|Y = 1

2

) =

∫ 1

3/4

fX|Y (x|1/2)dx

=

∫ 1

3/4

2dx = 2(1− 3/4) = 1/2

Page 10 of 10 Total marks: 200

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