ST112: Probability B Quiz 2 solutions

1. Let X and Y be jointly continuous random variables with joint density


 1 · (x + 3y) · e−(x+y) if x > 0 and y > 0;
4
fX,Y (x, y) =
0 otherwise.

Determine P(X > Y ).

Solution. Note that {X > Y } = {(X, Y ) ∈ A}, where A := {(x, y) ∈ R2 : x > y}. Hence,
ZZ Z ∞Z x
1
P(X > Y ) = fX,Y (x, y) dy dx = (x + 3y)e−(x+y) dy dx
A 0 0 4
Z ∞  Z x Z x 
1 −x −y −y
= e x e dy + 3 ye dy dx
4 0 0 0
  −y x x
1 ∞ −x
Z x
e−y
Z  
e −y
= e x +3 y· +3 e dy dx
4 0 −1 0 −1 0 0

1 ∞ −x
Z
−(4x + 3)e−x + (x + 3) dx


= e
4 0
 Z ∞ Z ∞ 
1 −2x −x
= − (4x + 3)e dx + (x + 3)e dx
4 0 0
∞ Z ∞ ∞ Z ∞
e−2x e−x
   
1 −2x −x
= − (4x + 3) · −2 e dx + (x + 3) + e dx
4 −2 0 0 −1 0 0
  −2x ∞  −x ∞ 
1 3 e e
= − −2 +3+
4 2 −2 0 −1 0
 
1 3 3
= − − 1 + 3 + 1 = = 0.375.
4 2 8

1
2. Let X and Y be independent random variables. Asume that X has doubly exponential
distribution with parameter λ > 0, that is,
λ −λ|x|
fX (x) = ·e , x ∈ R,
2
and Y has Gamma distribution with parameters 2 and λ > 0, that is,

λ2 · y · e−λy if y > 0;
fY (y) =
0 otherwise.

Mark all the correct answers.

(a) X + Y has Gamma distribution with parameters 3 and λ.

λ λ3 2 −λz
(b) For z > 0, fX+Y (z) = 8
+ 4
z e .
λ3 2 −λz
(c) For z > 0, fX+Y (z) = 2
z e .
λ λz
(d) For z < 0, fX+Y (z) = 8
e .
(e) For z < 0, fX+Y (z) = 0.

Solution. Since X and Y are independent, to compute the density of X + Y we can use
the convolution formula
Z ∞
fX+Y (z) = fX (x) · fY (z − x) dx
−∞
Z z
λ −λ|x| 2
= ·e · λ · (z − x) · e−λ(z−x) dx.
−∞ 2

For z 6 0 we have
z
λ3
Z
fX+Y (z) = · eλx · (z − x) · e−λz+λx dx
2 −∞
z
λ3 −λz
Z
= ·e · (z − x) · e2λx dx
2 −∞
z Z z 2λx
λ3 −λz e2λx
 
e
= ·e · (z − x) · + dx
2 2λ −∞ −∞ 2λ
 2λx z
λ3 −λz e
= ·e ·
2 (2λ)2 −∞

λ3 eλz λ
= · 2 = · eλz .
2 4λ 8
For z > 0, we split the integral as

λ3 0 λx λ3 z
Z Z
−λ(z−x)
fX+Y (z) = e · (z − x) · e dx + (z − x) · e−λz dx.
2 −∞ 2 0

2
Now, the first integral gives

λ3 0 λx
Z
e · (z − x) · e−λ(z−x) dx
2 −∞
 Z 0 Z 0
λ3 −λz

2λx 2λx
= ·e · z e dx − x · e dx
2 −∞ −∞
 2λx 0 0 Z 0 2λx !
λ3 −λz e2λx

e e
= ·e · z − x· + dx
2 2λ −∞ 2λ −∞ −∞ 2λ

 2λx 0 !
λ3 −λz z e
= ·e · +
2 2λ (2λ)2 −∞

λ3 −λz λ3 −λz
   
z 1 z 1
= ·e · + = ·e · + .
2 2λ 4λ2 4 λ 2λ2

The second integral is

Z z z
λ3 −λz λ3 −λz (z − x)2 λ3 2 −λz

·e · (z − x) dx = ·e · − = ·z ·e .
2 0 2 2 0 4

Hence, for z > 0,

λ3 −λz
 
z 1 2
fX+Y (z) = ·e · + +z .
4 λ 2λ2
Correct alternatives: d
Grading: +10 points for alternative d, -2.5 points for each incorrect alternative marked
(grade cannot be lower than zero)

3
3. Let X and Y be independent discrete random variables, each with Geometric distribution
of parameter p ∈ (0, 1), that is, the probability mass functions pX and pY of X and Y
are 
(1 − p)k−1 · p if k ∈ N;
pX (k) = pY (k) =
0 otherwise.
Mark all correct answers.

(a) X + Y has Geometric distribution with parameter 1 − (1 − p)2 .

(b) min{X, Y } has Geometric distribution with parameter 1 − (1 − p)2 .
(c) P(X + Y = k) = (k − 1)p2 (1 − p)k−2 for k = 2, 3, . . .
(d) P(max{X, Y } 6 2) = p2 (2 − p)2 .
(e) None of the above.

Solution. Since X and Y are independent, we can apply the discrete convolution formula,
which gives
X
P(X + Y = k) = pX (j) · pY (k − j)
j

k−1
X
= (1 − p)j−1 · p · (1 − p)k−j−1 · p
j=1

k−1
X
2
=p · (1 − p)k−2
j=1

= p2 · (1 − p)k−2 · (k − 1).

Here we can already see that X + Y is not Geometric. We now compute

P(min{X, Y } > k) = P(X > k, Y > k)

= P(X > k) · P(Y > k) = P(X > k)2
∞
!2
X
= (1 − p)j−1 · p
j=k

∞
!2
X
= p2 · (1 − p)j−k+k−1
j=k

∞
!2
X
= p2 · (1 − p)2(k−1) · (1 − p)`
`=0
 2
2 2(k−1) 1
= p · (1 − p) · = (1 − p)2(k−1) .
1 − (1 − p)

4
Hence,

P(min{X, Y } = k) = P(min{X, Y } > k) − P(min{X, Y } > k + 1)

= (1 − p)2(k−1) − (1 − p)2k

= (1 − p)2(k−1) · 1 − (1 − p)2
 

k−1 
= (1 − p)2 · 1 − (1 − p)2 .


Hence, min{X, Y } is Geometric with parameter 1 − (1 − p)2 .

Correct alternatives: b, c, d
Grading: 0 points if “None of the above” was marked. Marking (b), (c), (d) is worth
0.33 each, marking (a) gives a penalty of 0.33 (but grade cannot be negative)

5
4. Let X be a continuous random variable with log-normal distribution, that is,

 √1 · e− 12 (log x)2 if x > 0;
x 2π
fX (x) =
0 otherwise.
√
Determine the value of A = 2π · E[| log X|].
Solution. We have
√ √ Z ∞
| log x| − 1 (log x)2
A= 2π · E[|X|] = 2π · √ ·e 2 dx
0 x 2π
Z ∞
1 2
= |y| · e− 2 y dy
−∞
Z ∞ Z 0
− 12 y 2 1 2
= y·e dy − y · e− 2 y dy
0 −∞
Z ∞ Z ∞
1 2 1 2
= z · e− 2 z dz + z · e− 2 z dz
0 0
Z ∞
1 2
=2 z · e− 2 z dz
0
∞ ∞
e−w/2
Z 
= e−w/2 dw = = 2.
0 −1/2 0

6
5. Let X be a discrete random variable with Binomial distribution with parameters 4 and p ∈
(0, 1). Consider the random variable Y defined as

Y = cos( 12 πX).

Mark all the correct answers.

(a) Y has Bernoulli distribution with parameter p.
(b) Y is discrete uniform over the interval [−1, 1].
(c) E[Y ] = p2 (1 − p)2 (p2 + (1 − p)2 − 6).
(d) E[Y ] = 0.
(e) E[Y 2 ] = p2 (1 − p)2 (p2 + (1 − p)2 + 6).
(f) Var(Y ) = 12p2 (1 − p)2 .
Solution. Note that Im(X) = {0, 1, 2, 3, 4}, as X is Binomial with parameters 4 and p.
Hence,
• Y = 1 if and only if X = 0, 4;
• Y = 0 if and only if X = 1, 3;
• Y = −1 if and only if X = 2.
Therefore, Im(Y ) = {−1, 0, 1}, so Y is not Bernoulli. We compute

pY (1) = P(X = 0) + P(X = 4) = (1 − p)4 + p4 ;

pY (0) = P(X = 1) + P(X = 3) = 4(1 − p)3 p + 4p3 (1 − p);
pY (−1) = P(X = 2) = 6p2 (1 − p)2 .

Note that pY (0) = pY (−1) if and only if

2(1 − p)3 p + 2p3 (1 − p) + 3p2 (1 − p)2 = 0

⇔ p(1 − p)(2(1 − p)2 + 2p2 + 3p(1 − p)) = 0
⇔ p(1 − p)(2 − 4p + 2p2 + 2p2 + 3p − 3p2 ) = 0
⇔ p(1 − p)(2 − p2 − p) = 0
⇔ p = 0 or p = 1,

but both p = 0 and p = 1 are excluded. Hence, there is no value of p ∈ (0, 1) such that Y
is discrete uniform on [−1, 1].
Next, we compute

E[Y ] = −1 · pY (−1) + 0 · pY (0) + 1 · pY (1)

= −6p2 (1 − p)2 + p4 + (1 − p)4 ;
E[Y 2 ] = (−1)2 · pY (−1) + 0 · pY (0) + 12 · pY (1)
= 6p2 (1 − p)2 + p4 + (1 − p)4 ;
Var(Y ) = 6p2 (1 − p)2 + p4 + (1 − p)4 − (p4 + (1 − p)4 − 6p2 (1 − p)2 )2
= 4p(1 − p)[4p6 − 12p5 + 4p4 + 12p3 − 8p2 + 1].

7
Alternative (6) is not true for all values of p. To see this, note that

Var(Y ) − 12p2 (1 − p)2

= 4p(1 − p)[4p6 − 12p5 + 4p4 + 12p3 − 8p2 + 1] − 4p(1 − p)[3p(1 − p)]
= 4p(1 − p)[4p6 − 12p5 + 4p4 + 12p3 − 5p2 − 3p + 1],

and this is not zero for all p ∈ (0, 1).

Correct alternatives: None.
Grading: everyone who submitted the quiz was given full marks for this question, as
there was an issue with submitting the correct answer.

8
6. Let X be a continuous random variable with probability density function given by

 3 (1 − x2 ) if x ∈ (−1, 1);
4
fX (x) =
0 otherwise.

Compute the variance of X.

Solution. Note that for any k ∈ N we have
Z 1  k+1 1  k+3 1 !
k 3 k 2 3 x x
E[X ] = · x (1 − x ) dx = −
4 −1 4 k + 1 −1 k + 3 −1

3 2 − 2


4 k+1 k+3
if k is even;
=
0 otherwise.

Hence,
E[X] = 0
and    
2 32 2 2 3 10 − 6 1
Var(X) = E[X ] − E[X ] = − = = = 0.2.
4 3 5 4 15 5

9
7. Let X and Y be discrete random variables with joint mass function

 4 x2 (y + 2) if x = 1 , 1 and y = −1, 2;
25 2
pX,Y (x, y) =
0 otherwise.

Mark all the correct answers.

(a) X and Y are independent.

(b) X and Y are uncorrelated.
4 2
(c) pX (x) = 25
x for x = 12 , 1.
9
(d) E[X] = 10 .
(e) E[Y ] = 25 .
(f) E[XY ] = 63 50
.

Solution.

(a) Yes, they are independent since

pX,Y (x, y) = g(x) · h(y),

4 2
with for example g(x) = 25
x and h(y) = y + 2.
(b) Since they are independent, they are uncorrelated.
(c) We have
4 2 4 4 4
pX (x) = x · 1 + x2 · 4 = x2 · 5 = x2 .
25 25 25 5
(d) The expectation is
1 4 1 4 1 4 9
E[X] = · · +1· ·1= + = .
2 5 4 5 10 5 10

(e) We have
pX,Y (x, y) 1
pY (y) = = (y + 2).
pX (x) 5
Then,
1 1 1 8 7
E[Y ] = − (−1 + 2) + 2 · (2 + 2) = − + = .
5 5 5 5 5
(f) Since X and Y are independent,
63
E[XY ] = E[X] · E[Y ] = .
50
Correct alternatives: a, b, d, f
Grading: +2.5 for marking each correct alternative, -2.5 for marking each incorrect al-
ternative (but grade cannot be negative).

10
8. Let X and Y be jointly continuous random variables with joint density

 1 (x + y) if x ∈ (0, 1) and y ∈ (0, 2);
3
fX,Y (x, y) =
0 otherwise.

Compute E[sin(πXY )].

Solution. We have
Z ∞ Z ∞
E[sin(πXY )] = sin(πxy) · fX,Y (x, y) dx dy
−∞ −∞
Z 2Z 1
1
= · sin(πxy) · (x + y) dxdy
3 0 0
Z 1 Z 2 Z 2 Z 1 
1
= x sin(πxy) dy dx + y sin(πxy) dx dy
3 0 0 0 0
Z 1  2 Z 2  1 !
1 cos(πxy) cos(πxy)
= x − dy dx + y − dy
3 0 πx 0 0 πy 0
Z 1 Z 2 
1 1 1
= (1 − cos(2πx)) dx + (1 − cos(πy)) dy
3 0 π 0 π
 1  2 !
1 1 sin(2πx) 2 sin(πy) 1
= − + − = = 0.31 . . .
3 π 2π 0 π π 0 π

11
9. Let X1 , . . . , X10 be independent discrete random variables such that for all n,

1


 16
if k = 0;


 1 if k = 1;

4
pXn (k) =
1


 2
if k = 2;


 3 if k = 3.

16

Consider the random variable

10
X
Y := Xn .
n=1

Compute E[Y ].
Solution. We have, for each n,
1 1 1 3 1 9 29
E[Xn ] = 0 · +1· +2· +3· = +1+ = .
16 4 2 16 4 16 16
Then, " #
10 10
X X 29 290
E[Y ] = E Xn = E[Xn ] = 10 · = = 18.125
n=1 n=1
16 16
.

12
10. Let X be a Gaussian random variable with mean 0 and variance 4, that is,
1 2
fX (x) = √ · e−x /8 , x ∈ R.
8π
Let Y = X 2 .
Recall that if Z is a Gaussian random variable with mean µ and variance σ 2 then
Z ∞ Z ∞
1 (z−µ)2
1= fZ (z) dz = √ · e− 2σ2 dz.
−∞ 2πσ 2 −∞
Mark all the correct answers.

(a) The moment-generating function MY (t) of Y is well defined and finite for all t ∈ R.
(b) The moment-generating function MY (t) of Y is well defined and finite for t < 81 .
(c) E[Y ] = 4.
(d) Var(X) = 16.
(e) None of the above.

Solution. Notice that

Z ∞ Z ∞
1 2 1 x2 1
tY
MY (t) = E[e ] = E[e tX 2
]= √ · tx2
e ·e− x8
dx = √ · e− 2 ( 4 −2t) dx.
8π −∞ 8π −∞

The previous integral converges if and only if

1 1
− 2t > 0 ⇔ t< .
4 8
1
Hence, for t < 8
we have
Z ∞
1 x2 1−8t
MY (t) = √ · e− x ( 4 ) dx
8π −∞
Z ∞ ( )
1 1 x2
=√ ·√ · exp − 4

dx
4 2π −∞ 2 1−8t
r Z ∞ ( )
1 4 1 x2
= · · q · exp − 4

dx.
2 1 − 8t −∞ 2π 4
2 1−8t
1−8t

Now note that the expression inside the integral is the density function of a Gaussian
4
with mean 0 and variance σ 2 = 1−8t , so the integral equals 1. Therefore,

1
MY (t) = √ .
1 − 8t
To determine the moments, we compute the derivatives of MY :
1 8 4
MY0 (t) = · 3/2
= ,
2 (1 − 8t) (1 − 8t)3/2
3 8 58
MY00 (t) = 4 · · 5/2
= .
2 (1 − 8t) (1 − 8t)5/2

13
Then,
E[Y ] = MY0 (0) = 4, E[Y 2 ] = MY00 (0) = 58,
so
Var(Y ) = E[Y 2 ] − E[Y ]2 = 58 − 16 = 42.
Correct alternatives: b, c
Grading: 0 points if “None of the above” was marked. Marking (b) and (c) is worth 5
points each, marking (a) and (d) is worth -5 points each (but grade cannot be negative).

14