Stat560
Solution to Final Exam
1. True/False. Reason the answers.
(a) If two continuous random variables jointly have a rectangular support then they must
be independent.
Answer: False. Counterexample: for X and Y having joint p.m.f. as follows, the
support is rectangular. However, they are dependent.
y
x
0
1
0
1
fX (x)
1
4
1
6
1
4
1
3
1
2
1
2
fY (y)
5
12
7
12
See also Problem 5 for another counterexample.
(b) The probability that a continuous random variable assumes a particular value is zero.
Answer: True.
2. (Multiple Choise). Reason the answer. Assume random variable X has mean 3 and variance
4. If Y = −X + 5, then
a) µY = 2, σY2 = 1
b) µY = 8, σY2 = 4
c) µY = 2, σY2 = 4
d) µY = 8, σY2 = 1
2
Answer: (c). µY = −µX + 5 = 2. σY2 = σX
= 4.
3. Suppose that the random variable X be exponentially distributed with parameter λ = 2. Use
any method to determine the distribution of Y = exp(−λX).
1
distribution function method: For 0 < y < 1,
FY (y) = P (Y ≤ y) = P (exp(−2X) ≤ y) = P (−2X ≤ lny)
lny
lny
= P (X ≥ − ) = exp −2 · (− ) = y.
2
2
Consequently, fY (y) = FY0 (y) = 1 for 0 < y < 1. That is, Y ∼ U (0, 1).
transformation method: For 0 < y < 1,
y = exp(−2x) ⇒ x = −
lny
2
1
dx
=−
and hence
dy
2y
lny 1 fY (y) = fX (− )− 2
2y
1
= 2y ·
= 1.
2y
4. According to a genetics theory, a certain cross of guinea pigs will result in red, black, and
white offspring in the ratio 8:4:4.
(a) Find the probability that among eight offspring five will be red, two black, and one
white. (Hint: multinomial distribution)
8
The probabilities ofan offspring being red, black, and white are pr = 8+4+4
= 12 ,
4
= 14 , respectively. Hence the probability that among eight offspring
pb = pw = 8+4+4
five will be red, two black, and one white is
5 2 8
1
1
1
21
= 0.0820.
·
·
=
5, 2, 1
2
4
4
256
(b) Of 50 offspring what is the probability that at least 30 will be either black or white?
Use normal approximation. (Hint: binomial distribution)
2
Of 50 offspring, N , the number of black or white ones follows a binomial distribution
b(n = 50, p = 12 ). The mean and the standard deviation of this number are, respecq
1
tively, µ = 50 · 2 = 25, and σ = 50 · 12 · 12 = 3.5355. Hence the desired probability
can be approximated by
P (N ≥ 30) = P (N > 29.5)
continuity correction
29.5 − 25
≈ P Z>
= P (Z > 1.27) = 1 − Φ(1.27)
3.5355
= 1 − 0.8980 = 0.1020.
5. If the joint distribution of X and Y is given by
y
1
2
3
x
0
a
1
2
1
8
1
32
1
32
a
1
4
3
16
3
a
a
1
32
a
1
32
(a) Find the value of a in the table.
a = 15 (1 − 81 − 14 −
3
36
−4·
1
)
32
=
1
.
16
(b) Compute the marginal p.m.f. of X.
The row margins in the table below show the p.m.f. of X:
y
1
x
2
3
fY (y)
0
1
2
3
fX (x)
1
16
1
32
1
32
1
8
1
8
1
16
1
32
7
32
1
4
3
16
1
16
1
2
1
16
1
16
1
32
5
32
1
2
11
32
5
32
(c) Are X and Y independent?
Although the support of (X, Y ) is rectangular, X and Y are dependent since, for
instance,
1
1
7
f (1, 1) = 6=
·
= fX (1)fY (1).
8
2
32
3