Download Midterm #1 Practice Problems

Midterm #1 – Practice Exam
Question #1: Let 𝑋1 and 𝑋2 be independent and identically distributed random variables
with probability density function 𝑓𝑋 (𝑥) = {
𝑒 −𝑥 𝑖𝑓 𝑥 > 0
such that they come from the
0
𝑖𝑓 𝑥 ≤ 0
distribution 𝑋~𝐸𝑋𝑃(1). Compute the joint density function of 𝑈 = 𝑋1 and 𝑉 = 𝑋1 𝑋2.

Since 𝑋1 and 𝑋2 are independent, their joint density function is given by
𝑓𝑋1 𝑋2 (𝑥1 , 𝑥2 ) = 𝑓𝑋1 (𝑥1 )𝑓𝑋2 (𝑥2 ) = 𝑒 −𝑥1 𝑒 −𝑥2 = 𝑒 −(𝑥1 +𝑥2 ) whenever 𝑥1 > 0, 𝑥2 > 0. We
𝑣
𝑣
have the transformation 𝑢 = 𝑥1 and 𝑣 = 𝑥1 𝑥2 so 𝑥1 = 𝑢 and 𝑥2 = 𝑥 = 𝑢, so we can
1
𝜕𝑥1
compute the Jacobian as 𝐽 =
𝜕𝑢
𝑑𝑒𝑡 [𝜕𝑥
2
𝜕𝑢
𝜕𝑥1
𝜕𝑣
]
𝜕𝑥2
1
= 𝑑𝑒𝑡 [− 𝑣
𝑢2
𝜕𝑣
𝑣
0
1
1 ] = . This implies that the
𝑢
𝑢
𝑣
1
joint density 𝑓𝑈𝑉 (𝑢, 𝑣) = 𝑓𝑋1 𝑋2 (𝑢, 𝑢) |𝐽| = 𝑢 𝑒 −(𝑢+𝑢) whenever 𝑢 > 0 and 𝑣 > 0.
Question #2: Let 𝑋 be a standard normal random variable. Compute the Moment Generating
Function (MGF) of 𝑌 = |𝑋| using the standard normal distribution function Φ(z).

For some continuous random variable 𝑋 with density 𝑓𝑋 (𝑥), its moment generating
∞
function is given by 𝑀𝑋 (𝑡) = 𝐸(𝑒 𝑡𝑋 ) = ∫−∞ 𝑒 𝑡𝑥 𝑓𝑋 (𝑥) 𝑑𝑥. Since 𝑋~𝑁(0,1), we know
that its density function is given by 𝑓𝑋 (𝑥) =
𝑀|𝑋| (𝑡) = 𝐸(𝑒
∞
2
∫ 𝑒
√2𝜋 0
2𝑡𝑥−𝑥2
2
𝑡|𝑋|
)=
𝑑𝑥 =
𝑥2
∞
1
−
∫−∞ 𝑒 𝑡|𝑥| √2𝜋 𝑒 2
2
∞
∫ 𝑒
√2𝜋 0
𝑡2 −(𝑥−𝑡)2
2
𝑑𝑥 =
𝑑𝑥 =
2
1
√2𝜋
𝑥2
𝑒 − 2 . We can therefore compute
2
∞ 𝑡|𝑥|−𝑥
2
𝑒
∫
√2𝜋 −∞
1
∞
𝑡2
𝑒 2 𝑒−
∫
0
√2𝜋
(𝑥−𝑡)2
2
𝑑𝑥 =
2
∞ 𝑡𝑥−𝑥
2
𝑒
∫
√2𝜋 0
2
𝑑𝑥 =
𝑑𝑥. Separating out the 𝑡
variable and making the substitution 𝑢 = 𝑥 − 𝑡 and 𝑑𝑢 = 𝑑𝑥 then yields
𝑡2
∞ 1
𝑢2
𝑡2
𝑡2
𝑡2
2𝑒 2 ∫−𝑡
𝑒 − 2 𝑑𝑢 = 2𝑒 2 [Φ(∞) − Φ(−𝑡)] = 2𝑒 2 [1 − Φ(−𝑡)] = 2𝑒 2 Φ(𝑡). Thus, the
√2𝜋
𝑡2
2
Moment Generating Function of 𝑌 = |𝑋| is given by 𝑀|𝑋| (𝑡) = 2Φ(𝑡)𝑒 .
Question #3: Let 𝑋1 and 𝑋2 be independent random variables with respective probability
1 −
𝑒 −𝑥 𝑖𝑓 𝑥 > 0
𝑒
(𝑥)
(𝑥)
density functions 𝑓𝑋1
={
and 𝑓𝑋2
= {2
0
𝑖𝑓 𝑥 ≤ 0
0
(𝑥−1)
2
𝑖𝑓 𝑥 > 1
. Calculate the
𝑖𝑓 𝑥 ≤ 1
probability density function of the transformed random variable 𝑌 = 𝑋1 + 𝑋2.

Since 𝑋1 and 𝑋2 are independent random variables, their joint density is
1
𝑓𝑋1 𝑋2 (𝑥1 , 𝑥2 ) = 𝑓𝑋1 (𝑥1 )𝑓𝑋2 (𝑥2 ) = 2 𝑒 −𝑥1 𝑒 −
(𝑥2 −1)
2
1
= 2 𝑒−
(2𝑥1 +𝑥2 −1)
2
if 𝑥1 > 0 and 𝑥2 > 1.
We have 𝑦 = 𝑥1 + 𝑥2 and generate 𝑧 = 𝑥1 so 𝑥1 = 𝑧 and 𝑥2 = 𝑦 − 𝑥1 = 𝑦 − 𝑧, so we
𝜕𝑥1
can compute the Jacobian as 𝐽 =
𝜕𝑥1
𝜕𝑦
𝑑𝑒𝑡 [𝜕𝑥
2
𝜕𝑧
]
𝜕𝑥2
𝜕𝑦
0
= 𝑑𝑒𝑡 [
1
1
] = −1. This implies
−1
𝜕𝑧
1
that the joint density 𝑓𝑌𝑍 (𝑦, 𝑧) = 𝑓𝑋1 𝑋2 (𝑧, 𝑦 − 𝑧)|𝐽| = 2 𝑒 −
(2𝑧+𝑦−𝑧−1)
2
1
= 2 𝑒−
(𝑧+𝑦−1)
2
whenever 𝑧 > 0 and 𝑦 − 𝑧 > 1 → 𝑦 > 1 + 𝑧 or 𝑧 < 𝑦 − 1. We then integrate out 𝑧 to
∞
1
𝑦−1 −(𝑧+𝑦−1)
find the PDF 𝑓𝑌 (𝑦) = ∫−∞ 𝑓𝑌𝑍 (𝑦, 𝑧) 𝑑𝑧 = 2 ∫0
1
𝑒−
2
(𝑦−1)
2
𝑧
[−2𝑒 −2 ]
𝑦−1
0
= −𝑒 −
(𝑦−1)
2
Thus, we find that 𝑓𝑌 (𝑦) = 𝑒
𝑧
[𝑒 −2 ]
𝑦−1
0
1−𝑦
2
(1 − 𝑒
𝑒
= −𝑒 −
1−𝑦
2
(𝑦−1)
2
2
(𝑒 −
1
𝑑𝑧 = 2 𝑒 −
(𝑦−1)
2
(𝑦−1)
2
− 1) = 𝑒
𝑦−1 −𝑧
∫0
1−𝑦
2
𝑒
2
(1 − 𝑒
𝑑𝑧 =
1−𝑦
2
).
) whenever 𝑦 > 0 and zero otherwise.
Question #4: Let 𝑋1 , … , 𝑋𝑛 be independent and identically distributed random variables
1
from a cumulative distribution function 𝐹𝑋 (𝑥) = {
1 − 𝑥 𝑖𝑓 𝑥 > 1
0
𝑖𝑓 𝑥 ≤ 1
. Find the limiting
𝑛
distribution of the transformed first order statistic given by 𝑋1:𝑛
.

1
1
𝑛
𝑛 (𝑥) = 𝑃(𝑋1:𝑛 ≤ 𝑥) = 𝑃 (𝑋1:𝑛 ≤ 𝑥 𝑛 ) = 1 − 𝑃 (𝑋1:𝑛 > 𝑥 𝑛 ) =
We can calculate that 𝐹𝑋1:𝑛
1
𝑛
1
𝑛
1 − 𝑃 (𝑋𝑖 > 𝑥 𝑛 ) = 1 − [1 − 𝐹𝑋 (𝑥 𝑛 )] = 1 − [1 − (1 −
1
1
𝑥𝑛
𝑛
1
)] = 1 − 𝑥 when 𝑥 > 1.
Since there is no dependence on 𝑛, we may simply compute the desired limiting
1
lim (1 − 𝑥) 𝑖𝑓 𝑥 > 1
𝑛 (𝑥) = {𝑛→∞
distribution as lim 𝐹𝑋1:𝑛
𝑛→∞
lim (0)
𝑛→∞
𝑖𝑓 𝑥 ≤ 1
1
={
1 − 𝑥 𝑖𝑓 𝑥 > 1
0
𝑖𝑓 𝑥 ≤ 1
= 𝐹𝑋 (𝑥).
Question #5: Let 𝑋1 , … , 𝑋𝑛 be independent and identically distributed random variables
with density function 𝑓𝑋 (𝑥) = {
1 𝑖𝑓 𝑥 ∈ (0,1)
. That is, they are sampled from a random
0 𝑖𝑓 𝑥 ∉ (0,1)
variable distributed as 𝑋~𝑈𝑁𝐼𝐹(0,1). Approximate 𝑃(∑90
𝑖=1 𝑋𝑖 ≤ 77) using 𝑍~𝑁(0,1).

Since 𝑋~𝑈𝑁𝐼𝐹(0,1), we know that 𝐸(𝑋) =
0+1
2
1
= 2 and 𝑉𝑎𝑟(𝑋) =
(1−0)2
12
1
90
90
90
we let 𝑌 = ∑90
𝑖=1 𝑋𝑖 , we have 𝐸(𝑌) = 𝐸(∑𝑖=1 𝑋𝑖 ) = ∑𝑖=1 𝐸(𝑋𝑖 ) = ∑𝑖=1 2 =
1
1
= 12. Then if
90
2
= 45 and
90
90
90
𝑉𝑎𝑟(𝑌) = 𝑉𝑎𝑟(∑90
𝑖=1 𝑋𝑖 ) = ∑𝑖=1 𝑉𝑎𝑟(𝑋𝑖 ) = ∑𝑖=1 12 = 12 = 7.5 so that 𝑠𝑑(𝑌) = √7.5.
𝑌−𝐸(𝑌)
𝑃(∑90
𝑖=1 𝑋𝑖 ≤ 77) = 𝑃(𝑌 ≤ 77) = 𝑃 (
Thus
𝑃 (𝑍 ≤
32
𝑠𝑑(𝑌)
≤
77−𝐸(𝑌)
𝑠𝑑(𝑌)
) = 𝑃(
𝑌−45
√7.5
77−45
≤
√7.5
)≈
) = 𝑃(𝑍 ≤ 11.68) = Φ(11.68) ≈ 0.99 where 𝑍~𝑁(0,1).
√7.5
Question #6: If 𝑋~𝐸𝑋𝑃(1) such that 𝑓𝑋 (𝑥) = {
𝑒 −𝑥 𝑖𝑓 𝑥 > 0
, what transformation of 𝑋 will
0
𝑖𝑓 𝑥 ≤ 0
yield a random variable 𝑌~𝑈𝑁𝐼𝐹(0,1)? What about a random variable 𝑍~𝑈𝑁𝐼𝐹(−1,1)?

Let 𝑌 = 1 − 𝑒 −𝑋 , so we can compute that 𝐹𝑌 (𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(1 − 𝑒 −𝑋 ≤ 𝑦) =
𝑃(𝑋 ≤ −𝑙𝑛(1 − 𝑦)) = 𝐹𝑋 (−𝑙 𝑛(1 − 𝑦)) and then by differentiating we obtain 𝑓𝑌 (𝑦) =
𝑑
𝑑𝑦

[𝐹𝑋 (−𝑙 𝑛(1 − 𝑦))] = 𝑓𝑋 (−𝑙 𝑛(1 − 𝑦)) (−
Since we need to end up with
1−𝑦
2(1−𝑦)
−1
1−𝑦
1−𝑦
) = 1−𝑦 = 1 so that 𝑌~𝑈𝑁𝐼𝐹(0,1).
1
= 2 to conclude that 𝑍~𝑈𝑁𝐼𝐹(−1,1), it is clear
that we must make the transformation 𝑍 = 1 − 2𝑒 −𝑋 to obtain this result. We can
1−𝑧
therefore compute 𝐹𝑍 (𝑧) = 𝑃(𝑍 ≤ 𝑧) = 𝑃(1 − 2𝑒 −𝑋 ≤ 𝑧) = 𝑃 (𝑋 ≤ −𝑙 𝑛 (
1−𝑧
𝐹𝑋 (−𝑙 𝑛 (
2
1−𝑧
𝑓𝑋 (−𝑙 𝑛 (
2
𝑑
1−𝑧
)) so after differentiating we have 𝑓𝑍 (𝑧) = 𝑑𝑦 [𝐹𝑋 (−𝑙 𝑛 (
2
1
1−𝑧
1
2
2
)) (− 1−𝑦 (− 2) ) = 2(1−𝑧) = 2 so that we conclude 𝑍~𝑈𝑁𝐼𝐹(−1,1).
)) =
))] =
Question #7: Does convergence in probability imply convergence in distribution? Does
convergence in distribution imply convergence in probability? Give a counterexample if not.

We have that 𝐶𝑃 ⇒ 𝐶𝐷, but that 𝐶𝐷 ⇏ 𝐶𝑃. As a counterexample, consider 𝑌𝑛 = −𝑋
where 𝑋~𝑁(0,1). Then 𝑌𝑛 ~𝑁(0,1) so 𝑌𝑛 →𝑑 𝑋. However, we have 𝑃(|𝑌𝑛 − 𝑋| > 𝜖) =
𝑃(|−𝑋 − 𝑋| > ϵ) = 𝑃(2|𝑋| > ϵ) = 𝑔(𝜖) > 0. This expression won’t converge to zero
as 𝑛 → ∞ since there is no dependence on 𝑛 in the probability expression. This
implies that there is no convergence in probability, that is 𝑌𝑛 ↛𝑝 𝑋.
Question #8: State the law of large numbers (LLN), including its assumptions. Then state
the central limit theorem (CLT), including its assumptions.

Law of Large Numbers: If 𝑋1 , … , 𝑋𝑛 are independent and identically distributed with
finite mean 𝜇 and variance 𝜎 2 , then the sample mean 𝑋̅ →𝑝 𝜇.

Central Limit Theorem: If 𝑋1 , … , 𝑋𝑛 are independent and identically distributed with
finite mean 𝜇 and variance 𝜎 2 , then
[∑𝑛
𝑖=1 𝑋𝑖 ]−𝑛𝜇
𝜎 √𝑛
𝑋̅ −𝜇
→𝑑 𝑁(0,1) and 𝜎/
√𝑛
→𝑑 𝑁(0,1).
Question #10: Let 𝑋1 and 𝑋2 be independent random variables with Cumulative Distribution
Function (CDF) given by 𝐹𝑋 (𝑥) = 1 − 𝑒 −𝑥
2 /2
for 𝑥 > 0. Find the joint probability density
function 𝑓𝑌1 𝑌2 (𝑦1 , 𝑦2 ) of the random variables 𝑌1 = √𝑋1 + 𝑋2 and 𝑌2 = 𝑋
𝑋2
1 +𝑋2

𝑥2
𝑑
.
𝑥2
We first find the density 𝑓𝑋 (𝑥) = 𝑑𝑥 𝐹𝑋 (𝑥) = −𝑒 − 2 (−𝑥) = 𝑥𝑒 − 2 for 𝑥 > 0. Then since
𝑋1 and 𝑋2 are independent random variables, their joint density function is
2
𝑓𝑋1 𝑋2 (𝑥1 , 𝑥2 ) = 𝑓𝑋1 (𝑥1 )𝑓𝑋2 (𝑥2 ) = [𝑥1 𝑒
𝑥
− 1
2
2
] [𝑥2 𝑒
We then have that 𝑦1 = √𝑥1 + 𝑥2 and 𝑦2 = 𝑥
𝑥
− 2
2
𝑥2
1 +𝑥2
2
] = 𝑥1 𝑥2 𝑒
2
𝑥 +𝑥
− 1 2
2
for 𝑥1 > 0, 𝑥2 > 0.
so that 𝑥2 = 𝑦2 (𝑥1 + 𝑥2 ) = 𝑦2 𝑦12 and
𝑥1 = 𝑦12 − 𝑥2 = 𝑦12 − 𝑦2 𝑦12 . These computations allow us to compute the Jacobian as
𝜕𝑥1
𝜕𝑥1
𝜕𝑦
𝜕𝑦2
]
𝜕𝑥2
𝐽 = 𝑑𝑒𝑡 [𝜕𝑥1
2
𝜕𝑦1
= 𝑑𝑒𝑡 [
𝜕𝑦2
2𝑦1 − 2𝑦1 𝑦2
2𝑦1 𝑦2
−𝑦12
3
3
3
3
2 ] = 2𝑦1 − 2𝑦1 𝑦2 + 2𝑦1 𝑦2 = 2𝑦1 .
𝑦1
This
implies that the joint density function is 𝑓𝑌1 𝑌2 (𝑦1 , 𝑦2 ) = 𝑓𝑋1 𝑋2 (𝑦12 − 𝑦2 𝑦12 , 𝑦2 𝑦12 )|𝐽| =
(𝑦12 − 𝑦2 𝑦12 )(𝑦2 𝑦12 )𝑒 −
2 2
2 2
(𝑦2
1 −𝑦2 𝑦1 ) +(𝑦2 𝑦1 )
2
2𝑦13 whenever 𝑦12 − 𝑦2 𝑦12 > 0, 𝑦2 𝑦12 > 0. This
work also reveals that 𝑌1 and 𝑌2 are not independent since their joint probability
density function 𝑓𝑌1 𝑌2 (𝑦1 , 𝑦2 ) cannot be factored into two functions of 𝑦1 and 𝑦2 .
Question #11: Let 𝑋1 , 𝑋2 , 𝑋3 be independent random variables and suppose that we know
𝑋1 ~𝑁(2,4), 𝑋2 ~𝑁(3,9), and 𝑋1 + 𝑋2 + 𝑋3 ~𝑁(4,16). What is the distribution of 𝑋3?

We use the Moment Generating Function (MGF) technique to note that since the three
random variables are independent, we have 𝑀𝑋1 +𝑋2 +𝑋3 (𝑡) = 𝑀𝑋1 (𝑡)𝑀𝑋2 (𝑡)𝑀𝑋3 (𝑡) →
𝑀𝑋3 (𝑡) =
𝑀𝑋1 +𝑋2 +𝑋3 (𝑡)
𝑀𝑋1 (𝑡)𝑀𝑋2 (𝑡)
=
2
𝑒 4𝑡+16𝑡 /2
2
2
(𝑒 2𝑡+4𝑡 /2 )(𝑒 3𝑡+9𝑡 /2 )
=
2
𝑒 4𝑡+16𝑡 /2
2
𝑒 5𝑡+13𝑡 /2
= 𝑒 −𝑡+3𝑡
the random variable 𝑋3 ~𝑁(−1,3) since 𝑀𝐴 (𝑡) = 𝑒 𝜇𝑡+𝜎
2 𝑡 2 /2
2 /2
. This reveals that
when 𝐴~𝑁(𝜇, 𝜎 2 ).
Question #12: Let 𝑋1 , … , 𝑋5 be independent 𝑁(𝜇, 𝜎 2 ) random variables. Give examples of
random variables 𝑌1 , … , 𝑌4 defined in terms of the random variables 𝑋1 , … , 𝑋5 such that we
have 𝑌1 ~𝑁(0,1), 𝑌2 ~𝜒 2 (3), 𝑌3 ~𝐹(1,3), and 𝑌4 ~𝑡(3).

We have that 𝑌1 =
𝑋1 −𝜇
𝜎
𝑋𝑖 −𝜇 2
~𝑁(0,1), 𝑌2 = ∑3𝑖=1 𝑌𝑖2 = ∑3𝑖=1 (
2
𝑋 −𝜇
( 1 ) /1
𝜎
𝑋 −𝜇 2
3
[∑𝑖=1( 𝑖 ) ]/3
𝜎
~𝐹(1,3), and 𝑌4 =
𝑌1
√𝑌2 /3
𝑋 −𝜇
( 1 )
𝜎
=
2
𝜎
) ~𝜒 2 (3), 𝑌3 =
𝑌2 /3
=
~𝑡(3). These random variables
𝑋 −𝜇
√[∑3𝑖=1( 𝑖 ) ]/3
𝜎
are constructed since if some 𝑋~𝑁(𝜇, 𝜎 2 ), then 𝑍 =
𝑋−𝜇
𝜎
~𝑁(0,1). This implies that
𝑌 = 𝑍 2 ~𝜒 2 (1), so that 𝑈 = ∑𝑛𝑖=1 𝑌𝑖 ~𝜒 2 (𝑛). Finally, we note that 𝑉 =
where 𝑆~𝜒 2 (𝑘) and that 𝑇 =
𝑌12 /1
𝑍
√𝑈/𝑛
~𝑡(𝑛).
𝑈/𝑛
𝑆/𝑘
~𝐹(𝑛, 𝑘)
Question #13: If 𝑋1 and 𝑋2 are independent 𝐸𝑋𝑃(1) random variables such that their
𝑋
densities are 𝑓(𝑥) = 𝑒 −𝑥 1{𝑥 > 0}, then find the joint density of 𝑈 = 𝑋1 and 𝑉 = 𝑋1 + 𝑋2.
2

By independence, 𝑓𝑋1 𝑋2 (𝑥1 , 𝑥2 ) = 𝑓𝑋1 (𝑥1 )𝑓𝑋2 (𝑥2 ) = [𝑒 −𝑥1 ][𝑒 −𝑥2 ] = 𝑒 −(𝑥1 +𝑥2 ) if 𝑥1 > 0
𝑥
and 𝑥2 > 0. We have 𝑢 = 𝑥1 and 𝑣 = 𝑥1 + 𝑥2 so 𝑥1 = 𝑢𝑥2 = 𝑢(𝑣 − 𝑥1 ), which implies
2
𝑢𝑣
𝑢𝑣
that 𝑥1 + 𝑢𝑥1 = 𝑢𝑣 so 𝑥1 = 1+𝑢. Then 𝑥2 = 𝑣 − 𝑥1 = 𝑣 − 1+𝑢 =
𝜕𝑥1
allows us to compute the Jacobian as 𝐽 =
𝜕𝑥1
𝜕𝑢
𝑑𝑒𝑡 [𝜕𝑥
2
𝜕𝑣
]
𝜕𝑥2
𝜕𝑢
𝑣(1+𝑢)−𝑢𝑣
1
(1+𝑢)
(1+𝑢)2
𝑢
𝑣
𝑣
𝑢𝑣
+ 1+𝑢 ((1+𝑢)2 ) = (1+𝑢)3 + (1+𝑢)3 =
𝑢𝑣
𝑣
𝜕𝑣
𝑣(1+𝑢)
(1+𝑢)3
𝑣
𝑣
1+𝑢
𝑣
= 1+𝑢. This
𝑣(1+𝑢)−𝑢𝑣
(1+𝑢)2
𝑑𝑒𝑡 [ −𝑣
(1+𝑢)2
= (1+𝑢)2 .
𝑢𝑣
𝑣
=
𝑣+𝑢𝑣−𝑢𝑣
Thus,
𝑢
1+𝑢
]
1
=
1+𝑢
the
joint
𝑣
density is 𝑓𝑈𝑉 (𝑢, 𝑣) = 𝑓𝑋1 𝑋2 (1+𝑢 , 1+𝑢) |𝐽| = (1+𝑢)2 𝑒 −(1+𝑢+1+𝑢) = (1+𝑢)2 𝑒 −𝑣 whenever
we have that 𝑢 > 0, 𝑣 > 0 and zero otherwise.
Question #14: If 𝑋1 , 𝑋2 , 𝑋3 are a random sample of size 𝑛 = 2 from the 𝑈𝑁𝐼𝐹(0,1)
distribution, then a) find the joint density of the order statistics 𝑌1 , 𝑌2 , 𝑌3 , b) find the marginal
densities of 𝑌1 and 𝑌3 , and c) find the mean of the sample range 𝑅 = 𝑌3 − 𝑌1 .
a) We have 𝑓𝑌1 𝑌2 𝑌3 (𝑦1 , 𝑦2 , 𝑦3 ) = 3! ∏3𝑖=1 𝑓𝑋 (𝑦𝑖 ) = 6 ∏3𝑖=1 1 = 6 if 0 < 𝑦1 < 𝑦2 < 𝑦3 < 1.
b) We have 𝑓𝑌1 (𝑦1 ) = 3𝑓𝑋 (𝑦1 )[1 − 𝐹𝑋 (𝑦1 )]2 = 3(1 − 𝑦1 )2 if 0 < 𝑦1 < 1 and that
𝑓𝑌3 (𝑦3 ) = 3𝑓𝑋 (𝑦3 )[𝐹𝑋 (𝑦3 )]2 = 3𝑦32 whenever 0 < 𝑦3 < 1.
1
3
3
c) We have 𝐸(𝑅) = 𝐸(𝑌3 ) − 𝐸(𝑌1 ), so we compute 𝐸(𝑌3 ) = ∫0 𝑦3 3𝑦32 𝑑𝑦3 = 4 [𝑦33 ]10 = 4
1
1
𝑦
and 𝐸(𝑌1 ) = ∫0 𝑦1 3(1 − 𝑦1 )2 𝑑𝑦1 = 3 ∫0 𝑦1 − 2𝑦12 + 𝑦13 𝑑𝑦1 = 3 [ 21 −
3
1
1
2𝑦13
3
+
𝑦14
1
1
] = 4.
4
0
Then, we have that 𝐸(𝑅) = 𝐸(𝑌3 ) − 𝐸(𝑌1 ) = 4 − 4 = 2. Alternatively, we could have
found 𝑓𝑌1 𝑌3 (𝑦1 , 𝑦3 ) and computed 𝐸(𝑅) as a double integral.