MATH 105 921 Solutions to Probability Exercisesathena/P1S1.pdf · 2012-07-16 · MATH 105 921 Solutions to Probability Exercises (c) Let X 3 be the random variable that counts the

MATH 105 921 Solutions to Probability Exercises


Problem 1. Suppose we flip a fair coin once and observe either T for ”tails” or H for”heads”. Let X1 denote the random variable that equals 0 when we observe tails and equals1 when we observe heads. (This is called a Bernoulli random variable.)

(a) Make a table of the PDF of X1 and calculate E(X1) and Var(X1).

Solution: We have the following PDF table of X1:

x Pr(X1 = x)0 1

2

1 12

The expected value is:

E(X1) =

(1

2

)0 +

(1

2

)1 =

1

2.

The variance is:

Var(X1) =

(1

2

)(0− 1

2

)2

+

(1

2

)(1− 1

2

)2

=1

4.

(b) Now suppose that we flip a fair coin twice. We will observe one of four events: TT, TH,HT, or HH. Let X2 be the random variable that counts the number of heads we observe.(So, X2 can take the values 0, 1 or 2.) Redo part (a) for this new random variable.


x Pr(X2 = x)0 1

4

1 12

2 14


E(X2) =

(1

4

)0 +

(1

2

)1 +

(1

4

)2 = 1.

The variance is:

Var(X2) =

(1

4

)(0− 1)2 +

(1

2

)(1− 1)2 +

(1

4

)(2− 1)2 =

1

2.

Page 1 of 10


(c) Let X3 be the random variable that counts the number of heads we observe after threesuccessive flips of a fair coin. Redo part (a) for this new random variable.


x Pr(X3 = x)0 1

8

1 38

2 38

3 18


E(X3) =

(1

8

)0 +

(3

8

)1 +

(3

8

)2 +

(1

8

)3 =

3

2.

The variance is:

Var(X3) =

(1

8

)(0− 3

2

)2

+

(3

8

)(1− 3

2

)2

+

+

(3

8

)(2− 3

2

)2

+

(1

8

)(3− 3

2

)2

=3

4.

(d) Do you notice any patterns in your calculations? Make a guess for the pdf table, theexpectation and the variance of X4, the random variable that counts the number ofheads we observe after four successive flips of a fair coin, and then verify your guess bydirect computation.


x Pr(X4 = x)0 1

16

1 14

2 38

3 14

4 116


E(X4) =

(1

16

)0 +

(1

4

)1 +

(3

8

)2 +

(1

4

)3 +

(1

16

)4 = 2.

Page 2 of 10


The variance is:

Var(X4) =

(1

16

)(0− 2)2 +

(1

4

)(1− 2)2 +

+

(3

8

)(2− 2)2 +

(1

4

)(3− 2)2 +

(1

16

)(4− 2)2

= 1.

Problem 5. Recall the example of rolling a six-sided die. This is an example of a discreteuniform random variable, so named because the probability of observing each distinct out-come is the same, or uniform, for all outcomes. Let Y be the discrete uniform randomvariable that equals the face-value after a roll of an eight-sided die. (The die has eight faces,each with number 1 through 8.) Calculate E(Y ), Var(Y ), and StdDev(Y ).

Solution: We have the following PDF table of Y :

x Pr(Y = x)1 1

8

2 18

3 18

4 18

5 18

6 18

7 18

8 18


E(Y ) =8∑

x=1

xPr(Y = x) =1

8

(8∑

x=1

x

)=

1

8

(8(9)

2

)=

9

2.

The variance is:

Var(Y ) =8∑

x=1

(x− E(Y ))2 Pr(Y = x) =1

8

(8∑

x=1

(x− 9

2

)2)

=1

8(42) =

21

4.

The standard deviation is:

StdDev(Y ) =√

Var(Y ) =

√21

4=

√21

2.

Page 3 of 10


Problem 9. Are the following functions PDF’s?

(a) f(x) =

{12x2(x− 1) if 0 < x < 1

0 if otherwise.

Solution: f(x) is not a PDF because it fails the property that f(x) ≥ 0 for allx ∈ R. In fact, for any 0 < x < 1, we have that 12x2 > 0 (being an even power), butx− 1 < 0. So, for any 0 < x < 1,

f(x) = 12x2(x− 1) < 0.

For example, f

(1

2

)= −3

2< 0.

(b) f(x) =

{1− |x| if |x| ≤ 1

0 if otherwise..

Solution: f(x) is a PDF because it satisfies the following two properties:

• If |x| ≤ 1, then f(x) = 1 − |x| ≥ 0, and if |x| > 1, then f(x) = 0 ≥ 0, trivially.Thus, f(x) ≥ 0 for all x ∈ R.

• Observe that we may rewrite f(x) in more explicit form as:

f(x) =

1 + x if −1 < x < 0

1− x if 0 < x < 1

0 if otherwise.

So, ∫ ∞−∞

f(x) dx =

∫ 0

−1(1 + x) dx+

∫ 1

0

(1− x) dx

=

(x+

x2

2

)|0−1 +

(x− x2

2

)|10

= 0 + 1− 1

2+ 1− 1

2− 0 = 1.

Thus,

∫ ∞−∞

f(x) dx = 1.

(c) f(x) =

{π2

cos(πx) if |x| < 12

0 if otherwise.

Page 4 of 10


Solution: f(x) is a PDF because it satisfies the following two properties:

• If |x| ≤ 1

2, then −π

2≤ πx ≤ π

2, which means that cos(πx) ≥ 0. Therefore,

f(x) =π

2cos(πx) ≥ 0 for |x| ≤ 1

2. If |x| > 1

2, then f(x) = 0 ≥ 0 trivially. Thus,

f(x) ≥ 0 for all x ∈ R.

• We have that:∫ ∞−∞

f(x) dx =

∫ 12

− 12

π

2cos(πx) dx

=1

2sin(πx) |

12

− 12

=1

2sin(π

2

)− 1

2sin(−π

2

)=

1

2+

1

2= 1.

Thus,

∫ ∞−∞

f(x) dx = 1.

(d) f(x) =

{12

if |x| ≤ 2

0 if otherwise.

Solution: f(x) is not a PDF because it fails to satisfy the property that

∫ ∞−∞

f(x) dx =

1. In fact, we have that: ∫ ∞−∞

f(x) dx =

∫ 2

−2

1

2dx

=1

2x |2−2

=1

2(2− (−2)) = 2.

Thus,

∫ ∞−∞

f(x) dx = 2 6= 1.

Problem 14. What is the CDF of the density function1

π(1 + x2)?

Page 5 of 10


Solution: Recall that given a PDF f(x), we can find the CDF as follows:

F (x) =

∫ x

−∞f(t) dt

= lima→−∞

∫ x

a

1

π(1 + t2)dt

= lima→−∞

1

πarctan t |xa

= lima→−∞

1

πarctanx− 1

πarctan a

=1

πarctanx+

1

2.

Thus, the CDF of the given density function is F (x) =1

πarctanx+

1

2for any x ∈ R.

Problem 15. Show that p(x) =e−x

(1 + e−x)2on x ∈ R is a PDF.

Solution: To show that p(x) is a PDF, we need to verify the following two properties:

• We observe that e−x > 0 (being an exponential function), and (1+e−x)2 > 0 (beingan even power) for all x ∈ R. Thus,

p(x) =e−x

(1 + e−x)2> 0 for all x ∈ R.

• We have that: ∫ ∞−∞

p(x) dx =

∫ 0

−∞p(x) dx+

∫ ∞0

p(x) dx,

if both improper integrals on the right hand side converge. Using direct substitutionwith u = 1 + e−x, and du = −e−x dx, we get:∫

e−x

(1 + e−x)2dx =

∫−u−2 du = u−1 + C = (1 + e−x)−1 + C

Thus,∫ 0

−∞p(x) dx = lim

a→−∞

∫ 0

a

p(x) dx = lima→−∞

(1 + e−x)−1 |0a= lima→−∞

1

2− 1

1 + e−a=

1

2,∫ ∞

0

p(x) dx = limb→∞

∫ b

0

p(x) dx = limb→∞

(1 + e−x)−1 |b0= limb→∞

1

1 + e−b− 1

2=

1

2,∫ ∞

−∞p(x) dx =

1

2+

1

2= 1.

Page 6 of 10


So,

∫ ∞−∞

p(x) dx = 1.

Therefore, p(x) is a PDF.

Problem 19. Let Z be a standard normal random variable (the classic ”bell curve”,given by the density:

φ(x) =1√2πe−

x2

2 .

Verify that E(Z) = 0.

Solution: We have that:

E(Z) =

∫ ∞−∞

xφ(x) dx =

∫ 0

−∞xφ(x) dx+

∫ ∞0

xφ(x) dx,

if both improper integrals on the right hand side converge. Using direct substitution

with u = −x2

2and du = −x dx to find an antiderivative of xφ(x), we get:∫

xφ(x) dx =

∫1√2πxe−

x2

2 dx =

∫− 1√

2πeu du = − 1√

2πeu + C = − 1√

2πe−

x2

2 + C

Thus, ∫ 0

−∞xφ(x) dx = lim

a→−∞

∫ 0

a

xφ(x) dx = lima→−∞

− 1√2πe−

x2

2 |0a= −1√2π,∫ ∞

0

xφ(x) dx = limb→∞

∫ b

0

xφ(x) dx = limb→∞− 1√

2πe−

x2

2 |b0=1√2π,

E(Z) = − 1√2π

+1√2π

= 0.

Therefore, E(Z) = 0, as desired.

Problem 20. Expectations (and variances) need not be finite. Let X be a Cauchy

random variable given by the PDF: p(x) =1

π(1 + x2), for x ∈ R.

(a) Prove that E(X) does not exist (i.e. the expectation is infinite).

Page 7 of 10



E(X) =

∫ ∞−∞

xf(x) dx =

∫ 0

−∞xf(x) dx+

∫ ∞0

xf(x) dx,

if both improper integrals on the right hand side converge. Using direct substitutionwith u = 1 + x2 and du = 2x dx to find an antiderivative of xf(x), we get:∫

xf(x) dx =

∫x

π(1 + x2)dx =

∫1

2πu−1 du =

1

2πln |u|+ C =

1

2πln(1 + x2) + C

Thus,∫ 0

−∞xf(x) dx = lim

a→−∞

∫ 0

a

xf(x) dx = lima→−∞

1

2πln(1 + x2) |0a= lim

a→−∞− 1

2πln(1 + a2) = −∞

Since

∫ 0

−∞xf(x) dx diverges, the improper integral

∫ ∞−∞

xf(x) dx also diverges. There-

fore, the expectations E(X) does not exist.

Problem 21. Let p(x) =1

βe−

xβ for 0 ≤ x < ∞, β > 0, an exponential random variable

with parameter β.

(a) Show that this density integrates to 1.

Solution: We have that:∫ ∞−∞

p(x) dx =

∫ ∞0

1

βe−

xβ dx

= limb→∞

∫ b

0

1

βe−

xβ dx

= limb→∞−e−

xβ |b0

= limb→∞−e−

bβ + 1 = 1.

Thus,

∫ ∞−∞

p(x) dx = 1, as desired.

(b) Calculate Pr(X ≥ 1), E(X) and StdDev(X), for the exponential random variable X.

Page 8 of 10



Pr(X ≥ 1) =

∫ ∞1

p(x) dx

=

∫ ∞1

1

βe−

xβ dx

= limb→∞

∫ b

1

1

βe−

xβ dx

= limb→∞−e−

xβ |b1

= limb→∞−e−

bβ + e−

1β

⇒ Pr(X ≥ 1) = e−1β .

Using integration by parts with u = x, du = dx, and dv = e−xβ dx, v = −βe−

xβ , we

find that the expected value is:

E(X) =

∫ ∞−∞

xp(x) dx

=

∫ ∞0

1

βxe−

xβ dx

= limb→∞

∫ b

0

1

βxe−

xβ dx

= limb→∞

(−xe−

xβ − βe−

xβ

)|b0

= limb→∞−be−

bβ − βe−

bβ + 0 + β (using L’Hopital’s Rule)

⇒ E(X) = β.

To find StdDev(X), we first compute Var(X) as follows:

Var(X) =

∫ ∞−∞

(x− E(X))2p(x) dx =

∫ ∞0

1

β(x− β)2e−

xβ dx

=

∫ ∞0

(1

βx2e−

xβ − 2xe−

xβ + βe−

xβ

)dx

= limb→∞

∫ b

0

(1

βx2e−

xβ − 2xe−

xβ + βe−

xβ

)dx

Using integration by parts twice on the first summand and integration by parts once

Page 9 of 10


on the second summand, we get the following antiderivative of the integrand:

Var(X) = limb→∞

∫ b

0

(1

βx2e−

xβ − 2xe−

xβ + βe−

xβ

)dx

= limb→∞

(−β2e−

xβ − x2e−

xβ

)|b0

= lim→∞

(−β2e−

bβ − b2e−

bβ + β2 + 0

)(using L’Hopital’s Rule)

⇒ Var(X) = β2.

Therefore, StdDev(X) =√

Var(X) =√β2 = β, since β > 0.

Problem 24. Let X be a Laplace random variable given by the PDF: p(x) =1

2e−|x| for

x ∈ R.

(a) Verify that p(x) is in fact a valid PDF.

Solution: p(x) is a valid PDF because it satisfies the following two properties:

• For any real number x, we have that e−|x| ≥ 0 (being an exponential function).

Thus, p(x) =1

2e−|x| ≥ 0 for all x ∈ R.

• Observe that we may rewrite p(x) in more explicit form as:

p(x) =

{12ex if x < 0

12e−x if 0 ≤ x

So, ∫ ∞−∞

p(x) dx =

∫ 0

−∞

1

2ex dx+

∫ ∞0

1

2e−x dx (if both integrals converge)

= lima→−∞

∫ 0

a

1

2ex dx+ lim

b→∞

∫ b

0

1

2e−x dx

= lima→−∞

1

2ex |0a + lim

b→∞−1

2e−x |b0

= lima→−∞

1

2(1− ea) + lim

b→∞−1

2

(e−b − 1

)=

1

2+

1

2= 1.

Thus,

∫ ∞−∞

p(x) dx = 1.

Page 10 of 10

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MATH 105 921 Solutions to Probability Exercisesathena/P1S1.pdf · 2012-07-16 · MATH 105 921 Solutions to Probability Exercises (c) Let X 3 be the random variable that counts the