1. Suppose the probability function describing the life (in hours) of an ordinary 60 Watt bulb is
f (x) =
1 −x/1000
e
,
1000
x > 0.
Let A be the event that the bulb lasts 500 hours or more, and let B be the event that it lasts between 400 and
800 hours.
(a) What is the probability of A? (2%)
Ans. P (A) = e−0.5 = 0.6056.
Sol) In fact, X ∼ Exp(0.001). Hence, F (x) = 1 − e−0.001x , x > 0.
P (A) = P (X > 500) = e−0.5 = 0.6056.
(b) What is the probability of B? (3%)
Ans. P (B) = e−0.4 − e−0.8 = 0.2210.
Sol)
P (B) = P (400 < X ≤ 800) = e−0.4 − e−0.8 = 0.2210.
(c) What is P (A|B)? (5%)
Ans. P (A|B) =
e−0.5 −e−0.8
e−0.4 −e−0.8
= 0.7713.
Sol)
P (A|B) = P (X > 500|400 < X ≤ 800) =
=
P (500 < X ≤ 800)
P (400 < X ≤ 800)
e−0.5 − e−0.8
= 0.7713.
e−0.4 − e−0.8
2. Let X be a continuous random variable with pdf
½
kx−7
0
fX (x) =
x≥1
otherwise
(a) Find the value of k. (5%)
Ans. k = 6.
Sol)
Z
1 =
∞
k
x−7 dx =
1
⇒k
¯∞
x−6 ¯¯
k
=
−6 ¯1
6
= 6
(b) Compute the expectation E[X]. (5%)
Ans. E[X] = 65 .
Sol)
Z
E[X] = 6
∞
−6
x
1
1
¯∞
6
6 −5 ¯¯
dx = − x ¯ =
5
5
1
(c) Compute the variance V ar[X]. (5%)
Ans. V ar[X] =
6
100 .
Sol)
¯∞
6 −4 ¯¯
6
6
x dx = − x ¯ =
4
4
1
1
µ ¶2
6
6
6
E[X 2 ] − (E[X])2 = −
=
.
4
5
100
Z
2
E[X ] =
V ar[X] =
∞
−5
(d) Let Y = X 2 . Compute the pdf fY (y). (5%)
½
3y −4 y ≥ 1
Ans. fY (y) =
0
otherwise
Sol) It is clear that I(Y ) = [1, ∞). Consider y ≥ 1 below.
FY (y)
2
= P (Y ≤ y) = P (X ≤ y) = P (1 ≤ X ≤
½
3y −4 y ≥ 1
⇒ fY (y) =
0
otherwise
√
Z
√
y) = 6
1
y
√
y
x−7 dx = −x−6 |1
= 1 − y −3 .
3. Let X and Y be continuous random variables with joint pdf
f (x, y) = xe−(x+y) ,
x > 0 and y > 0.
(a) Find the marginal pdf’s of X and Y . (5%)
Ans. fX (x) = xe−x , x > 0 and fY (y) = e−y , y > 0.
Sol)
Z
∞
xe−(x+y) dy
Z ∞
−x
= xe
e−y dy
fX (x) =
0
0
= xe−x , x > 0;
Z ∞
fY (y) =
xe−(x+y) dx
0
Z ∞
−y
= e
xe−x dx
0
= e−y ,
y > 0.
(b) Are X and Y independent random variables? Why or why not? (5%)
Ans. Yes, X and Y are independent, since f (x, y) = fX (x)fY (y) for all x and y.
Sol)
(c) Let Z = X + Y . Find the pdf Z. (5%)
Sol) In fact, X ∼ Γ(2, 1) and Y ∼ Exp(1) ≡ Γ(1, 1), and X and Y are independent, so X + Y ∼ Γ(3, 1).
Therefore,
1
fZ (z) = z 2 e−z , z > 0.
2
4. The random variables X and Y have a joint density function given by
½
ke−2x /x 0 ≤ x < ∞, 0 ≤ y ≤ x
f (x, y) =
0
otherwise
2
(a) Find k. (5%)
Ans. k = 2.
Sol)
Z
1 =
∞
0
⇒k
Z
x
k
Z
x−1 e−2x dydx = k
0
∞
e−2x dx =
0
k
2
= 2
(b) Compute Cov(X, Y ). (Hint: To obtain E[X] and E[Y ], you needn’t to compute fX (x) and fY (y) explicitly.) (5%)
Ans. Cov(X, Y ) = 0.
Sol)
Z
E[X] =
E[Y ] =
Z
∞
x
Z
∞
e−2x dx = 1
Z0 ∞ Z 0∞
Z0 ∞ Z x
−1 −2x
2
y
x e
yx−1 e−2x dydx
dxdy = 2
2
x
0
Z
x−1 e−2x dydx = 2
y
Z
∞
Z
x
0
∞
0
1
x−1 e−2x
ydydx =
xe−2x dx =
4
0
Z0 ∞ Z x
Z ∞0
Z x
Z
2
ye−2x dydx = 2
e−2x
ydydx =
= 2
E[XY ] =
0
Cov(X, Y ) =
0
0
0
∞
x2 e−2x dx =
0
1
4
E[XY ] − E[X]E[Y ] = 0
5. A lottery is designed so that each ticket has a 10% chance of paying $2, a 4% chance of paying $5, a 1% chance
of paying $10, and an 85% chance of paying nothing. You buy a ticket, and call its value X.
(a) What is the expectation E[X]? (5%)
Ans. E[X] = 0.5.
Sol)
E[X] =
2 × 0.1 + 5 × 0.04 + 10 × 0.01 + 0 × 0.85 = 0.5.
(b) Compute the variance V ar[X] and the standard deviation σX . (5%)
√
Ans. V ar[X] = 2.15, σX = 2.15 = 1.4663.
Sol)
E[X 2 ] =
V ar[X] =
σX
22 × 0.1 + 52 × 0.04 + 102 × 0.01 + 0 × 0.85 = 2.4
E[X 2 ] − (E[X])2 = 2.4 − 0.52 = 2.15
√
=
2.15 = 1.4663
(c) Suppose you buy 100 lottery tickets, where each ticket is independent of the others. Let Y be the total
value of all 100 tickets put together. Compute E[Y ] and σy . (5%)
√
Ans. E[Y ] = 50, σY = 215 = 14.663.
P100
Sol) Let Xi denote the ith ticket bought. Then, E[Xi ] = 0.5, V ar[Xi ] = 2.15 and Y = i=1 Xi . Because
Xi ’s are independent, we have
" 100 # 100
X
X
E[Y ] = E
Xi =
E[Xi ] = 50
i=1
V ar[Y ]
⇒ σY
= V ar
=
√
" 100
X
#
i=1
Xi =
i=1
215 = 14.663
3
100
X
i=1
V ar[Xi ] = 215
2
6. Suppose that a random variable X has variance σX
= 2 and moment generating function MX (t) = a+b(e−2t +e2t ),
−∞ < t < ∞.
(a) Find the values of a and b. (5%)
Ans. a = 0, b = 12 .
Sol)
E[X]
E[X 2 ]
t+
t + ···
1!
2!
= a + b(e−2t + e2t )
µ
¶
µ
¶
2t 4t2
2t 4t2
= a+b 1−
+
± ··· + b 1 +
+
+ ···
1!
2!
1!
2!
8bt2
+ ···
= (a + 2b) +
2!
⇒ E[X] = 0
⇒ E[X 2 ] = 8b
MX (t) =
1+
(1)
(2)
2
Since σX
= E[X 2 ] − (E[X])2 = 8b = 2, we have b = 1/4. Comparing Eq. (1) and (2), we immediately have
1
a = 2.
(b) What is the distribution of X ? Give your reasoning and justification. (5%)
 1
 2 x=0
1
x ∈ {−2, 2}
Ans. X is a discrete random variable with pmf pX (x) =
 4
0 otherwise
+ 14 (e−2t + e2t ). Consider the following pmf for random variable
 1
 2 x=0
1
x ∈ {−2, 2}
pX (x) =
(3)
 4
0 otherwise
Sol) From Part (a), we have MX (t) =
X
1
2
Then,
MX (t) = E[eXt ] = pX (0)e−0t + pX (−2)e−2t + pX (2)e2t =
1 1 −2t
+ (e
+ e2t )
2 4
This implies that X is a discrete random variable with pmf of Eq. (3).
7. A certain standardized placement exam is made up of 200 multiple choice questions, where for each question the
test-taker must choose the correct answer from 4 possible answers.
(a) If a student selects his answers randomly, and X is the number of correct answers that he selects, then
what is the probability distribution of the random variable X? (2%)
Ans. X ∼ B(200, 0.25).
Sol) Let Xi , i = 1, . . . , 200 represent the correctness of the randomly chosen answer for the ith problem,
i.e., Xi = 1 represents that the answer is correct, and Xi = 0 represents that the answer is incorrect. It
is easily seen that Xi ∼ B(1, 0.25), and Xi ’s are independent. Then, the total number of correct answers
P200
X = i=0 Xi . So, X ∼ B(200, 0.25).
(b) What are E[X] and V ar[X]? (5%)
Ans. E[X] = 50 and V ar[X] = 37.5.
Sol)
(c) Write a precise mathematical formula for the probability that X is between 40 (included) and 60 (included).
Do not need to evaluate this sum. (3%)
Ans. P (40 ≤ X ≤ 60) =
¶
60 µ
X
200
0.25x × 0.75200−x .
x
x=40
Sol)
4
(d) Estimate the probability in part b. using Chebychevs theorem. (5%)
Ans. P (40 ≤ X ≤ 60) ≥ 0.625.
Sol) Since µX = 50, P (40 ≤ X ≤ 60) = P (|X − µX | ≤ 10). By Chebychevs theorem,
P (|X − µX | ≤ 10) ≥ 1 −
37.5
= 0.625.
102
(e) Estimate the probability in part b. using the binomial approximation to the binomial. (5%)
Ans. P (40 ≤ X ≤ 60) = 0.9137.
Sol) By central limit theorem, X ∼ B(200, 0.25) ≈ N (50, 37.5). Considering compensation,
µ
¶
µ
¶
60.5 − 50
39.5 − 50
√
√
P (40 ≤ X ≤ 60) = Φ
−Φ
= Φ(1.715) − Φ(−1.715) = 0.9137.
37.5
37.5
5