Midterm #1 – Practice Exam
Question #1: Let 𝑋
1
and 𝑋
2
be independent and identically distributed random variables with probability density function 𝑓
𝑋
(𝑥) = { 𝑒 −𝑥 𝑖𝑓 𝑥 > 0
0 𝑖𝑓 𝑥 ≤ 0
such that they come from the distribution 𝑋~𝐸𝑋𝑃(1) . Compute the joint density function of 𝑈 = 𝑋
1
and 𝑉 = 𝑋
1
𝑋
2
.
 Since 𝑋
1
and 𝑋
2
are independent, their joint density function is given by 𝑓
𝑋
1
𝑋
2
(𝑥
1
, 𝑥
2
) = 𝑓
𝑋
1
(𝑥
1
)𝑓
𝑋
2
(𝑥
2
) = 𝑒 −𝑥
1 𝑒 −𝑥
2 = 𝑒 −(𝑥
1
+𝑥
2
) whenever 𝑥
1
> 0 , 𝑥
2
> 0 . We have the transformation 𝑢 = 𝑥
1
and 𝑣 = 𝑥
1 𝑥
2
so 𝑥
1
= 𝑢 and 𝑥
2
= 𝑣 𝑥
1
= 𝑣 𝑢
, so we can
𝜕𝑥
1 compute the Jacobian as 𝐽 = 𝑑𝑒𝑡 [
𝜕𝑢
𝜕𝑥
2
𝜕𝑢
𝜕𝑥
1
𝜕𝑣
𝜕𝑥
2
] = 𝑑𝑒𝑡 [
−
1 0 𝑣 1 ] = 𝑢 2 𝑢
𝜕𝑣
1 𝑢
. This implies that the joint density 𝑓
𝑈𝑉
(𝑢, 𝑣) = 𝑓
𝑋
1
𝑋
2
(𝑢, 𝑣 𝑢
) |𝐽| =
1 𝑢 𝑒
−(𝑢+ 𝑣 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 𝑓
𝑋
(𝑥) , its moment generating function is given by 𝑀
𝑋
(𝑡) = 𝐸(𝑒 𝑡𝑋 ) = ∫
∞
−∞ 𝑒 𝑡𝑥 𝑓
𝑋
(𝑥) 𝑑𝑥 . Since 𝑋~𝑁(0,1) , we know that its density function is given by 𝑓
𝑋
(𝑥) =
1
√2𝜋 𝑒
− 𝑥2
2
. We can therefore compute
𝑀
|𝑋|
(𝑡) = 𝐸(𝑒 𝑡|𝑋| ) = ∫
∞
−∞ 𝑒 𝑡|𝑥| 1
√2𝜋 𝑒
− 𝑥2
2 𝑑𝑥 =
1
√2𝜋
∫
∞
−∞ 𝑒 𝑡|𝑥|− 𝑥2
2 𝑑𝑥 =
2
√2𝜋
0
∞
∫ 𝑒 𝑡𝑥− 𝑥2
2 𝑑𝑥 =
2
√2𝜋 0
∞
∫ 𝑒
2𝑡𝑥−𝑥2
2 𝑑𝑥 =
2
√2𝜋 0
∞
∫ 𝑒 𝑡2−(𝑥−𝑡)2
2 𝑑𝑥 =
2
√2𝜋 0
∞
∫ 𝑒 𝑡2
2 𝑒
−
(𝑥−𝑡)2
2 𝑑𝑥 . Separating out the 𝑡 variable and making the substitution 𝑢 = 𝑥 − 𝑡
and 𝑑𝑢 = 𝑑𝑥
then yields
2𝑒 𝑡2
2
∫
∞
−𝑡
1
√2𝜋 𝑒
− 𝑢2
2 𝑑𝑢 = 2𝑒 𝑡2
2
[Φ(∞) − Φ(−𝑡)] = 2𝑒 𝑡2
2
[1 − Φ(−𝑡)] = 2𝑒 𝑡2
2
Φ(𝑡) . Thus, the
Moment Generating Function of
𝑌 = |𝑋|
is given by
𝑀
|𝑋|
(𝑡) = 2Φ(𝑡)𝑒 𝑡2
2
.

Question #3: Let 𝑋
1
and 𝑋
2
be independent random variables with respective probability density functions 𝑓
𝑋
1
(𝑥) = { 𝑒 −𝑥 𝑖𝑓 𝑥 > 0
0 𝑖𝑓 𝑥 ≤ 0
and 𝑓
𝑋
2
(𝑥) = {
1
2 𝑒
−
(𝑥−1)
2
𝑖𝑓 𝑥 > 1
0 𝑖𝑓 𝑥 ≤ 1
. Calculate the probability density function of the transformed random variable 𝑌 = 𝑋
1
+ 𝑋
2
.
 Since 𝑋
1
and 𝑋
2
are independent random variables, their joint density is 𝑓
𝑋
1
𝑋
2
(𝑥
1
, 𝑥
2
) = 𝑓
𝑋
1
(𝑥
1
)𝑓
𝑋
2
(𝑥
2
) =
1
2 𝑒 −𝑥
1 𝑒
−
(𝑥2−1)
2
=
1
2 𝑒
−
(2𝑥1+𝑥2−1)
2
if 𝑥
1
> 0 and 𝑥
2
> 1 .
We have 𝑦 = 𝑥
1
+ 𝑥
2
and generate 𝑧 = 𝑥
1
so 𝑥
1
= 𝑧 and 𝑥
2
= 𝑦 − 𝑥
1
= 𝑦 − 𝑧 , so we
𝜕𝑥
1 can compute the Jacobian as
𝐽 = 𝑑𝑒𝑡 [
𝜕𝑦
𝜕𝑥
2
𝜕𝑦
𝜕𝑥
1
𝜕𝑧
𝜕𝑥
2
] = 𝑑𝑒𝑡 [
0 1
1 −1
] = −1
. This implies
𝜕𝑧 that the joint density 𝑓
𝑌𝑍
(𝑦, 𝑧) = 𝑓
𝑋
1
𝑋
2
(𝑧, 𝑦 − 𝑧)|𝐽| =
1
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 find the PDF 𝑓
𝑌
(𝑦) = ∫
∞
−∞ 𝑓
𝑌𝑍
(𝑦, 𝑧) 𝑑𝑧 =
1
2 0 𝑦−1
∫ 𝑒
−
(𝑧+𝑦−1)
2 𝑑𝑧 =
1
2 𝑒
−
(𝑦−1)
2
0 𝑦−1
∫ 𝑒
− 𝑧
2 𝑑𝑧 =
1
2 𝑒
−
(𝑦−1)
2
[−2𝑒
− 𝑧
2
]
0 𝑦−1
= −𝑒
−
(𝑦−1)
2
[𝑒
− 𝑧
2 𝑦−1
]
0
= −𝑒
−
(𝑦−1)
2
(𝑒
−
(𝑦−1)
2
− 1) = 𝑒
1−𝑦
2
(1 − 𝑒
1−𝑦
2
) .
Thus, we find that 𝑓
𝑌
(𝑦) = 𝑒
1−𝑦
2
(1 − 𝑒
1−𝑦
2
) whenever 𝑦 > 0 and zero otherwise.
Question #4: Let 𝑋
1
, … , 𝑋 𝑛
be independent and identically distributed random variables from a cumulative distribution function 𝐹
𝑋
(𝑥) = {
1 −
1 𝑥
𝑖𝑓 𝑥 > 1
0 𝑖𝑓 𝑥 ≤ 1
. Find the limiting distribution of the transformed first order statistic given by 𝑋 𝑛
1:𝑛
.
 We can calculate that 𝐹
𝑋 𝑛
1:𝑛
(𝑥) = 𝑃(𝑋 𝑛
1:𝑛
≤ 𝑥) = 𝑃 (𝑋
1:𝑛
≤ 𝑥 𝑛
1
) = 1 − 𝑃 (𝑋
1:𝑛
> 𝑥
1 𝑛
) =
1 − 𝑃 (𝑋 𝑖
> 𝑥 𝑛
1
) 𝑛
= 1 − [1 − 𝐹
𝑋
(𝑥 𝑛
1
)] 𝑛
= 1 − [1 − (1 −
1
1 𝑥 𝑛
)] 𝑛
= 1 −
1 𝑥
when 𝑥 > 1 .
Since there is no dependence on 𝑛 , we may simply compute the desired limiting distribution as lim 𝑛→∞
𝐹
𝑋 𝑛
1:𝑛
(𝑥) = { lim 𝑛→∞ lim 𝑛→∞
(1 −
1 𝑥
) 𝑖𝑓 𝑥 > 1
(0) 𝑖𝑓 𝑥 ≤ 1
= {
1 −
1 𝑥
𝑖𝑓 𝑥 > 1
0 𝑖𝑓 𝑥 ≤ 1
= 𝐹
𝑋
(𝑥) .

Question #5: Let 𝑋
1
, … , 𝑋 𝑛
be independent and identically distributed random variables with density function 𝑓
𝑋
(𝑥) = {
1 𝑖𝑓 𝑥 ∈ (0,1)
0 𝑖𝑓 𝑥 ∉ (0,1)
. That is, they are sampled from a random variable distributed as
𝑋~𝑈𝑁𝐼𝐹(0,1)
. Approximate
𝑃(∑
90 𝑖=1
𝑋 𝑖
≤ 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
12
. Then if we let 𝑌 = ∑ 90 𝑖=1
𝑋 𝑖
, we have 𝐸(𝑌) = 𝐸(∑ 90 𝑖=1
𝑋 𝑖
) = ∑ 90 𝑖=1
𝐸(𝑋 𝑖
) = ∑ 90 𝑖=1
1
2
=
90
= 45 and
2
𝑉𝑎𝑟(𝑌) = 𝑉𝑎𝑟(∑ 90 𝑖=1
𝑋 𝑖
) = ∑ 90 𝑖=1
𝑉𝑎𝑟(𝑋 𝑖
) = ∑ 90 𝑖=1
1
12
90
=
12
= 7.5
so that 𝑠𝑑(𝑌) = √7.5
.
Thus
𝑃(∑
90 𝑖=1
𝑋 𝑖
≤ 77) = 𝑃(𝑌 ≤ 77) = 𝑃 (
𝑌−𝐸(𝑌) 𝑠𝑑(𝑌)
≤
77−𝐸(𝑌) 𝑠𝑑(𝑌)
) = 𝑃 (
𝑌−45
√7.5
≤
77−45
) ≈
√7.5
𝑃 (𝑍 ≤
32
√7.5
) = 𝑃(𝑍 ≤ 11.68) = Φ(11.68) ≈ 0.99
where 𝑍~𝑁(0,1) .
Question #6: If 𝑋~𝐸𝑋𝑃(1) such that 𝑓
𝑋
(𝑥) = { 𝑒 −𝑥 𝑖𝑓 𝑥 > 0
0 𝑖𝑓 𝑥 ≤ 0
, what transformation of 𝑋 will 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 𝐹
𝑌
(𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(1 − 𝑒 −𝑋 ≤ 𝑦) =
𝑃(𝑋 ≤ −𝑙𝑛 (1 − 𝑦)) = 𝐹
𝑋
(−𝑙 𝑛(1 − 𝑦)) and then by differentiating we obtain 𝑓
𝑌
(𝑦) = 𝑑 𝑑𝑦
[𝐹
𝑋
(−𝑙 𝑛(1 − 𝑦))] = 𝑓
𝑋
(−𝑙 𝑛(1 − 𝑦)) (−
−1
1−𝑦
) =
1−𝑦
1−𝑦
= 1 so that 𝑌~𝑈𝑁𝐼𝐹(0,1) .
 Since we need to end up with 1−𝑦
2(1−𝑦)
=
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 therefore compute 𝐹
𝑍
(𝑧) = 𝑃(𝑍 ≤ 𝑧) = 𝑃(1 − 2𝑒 −𝑋 ≤ 𝑧) = 𝑃 (𝑋 ≤ −𝑙 𝑛 (
1−𝑧
2
)) =
𝐹
𝑋
(−𝑙 𝑛 (
1−𝑧
2
)) so after differentiating we have 𝑓
𝑍
(𝑧) = 𝑑 𝑑𝑦
[𝐹
𝑋
(−𝑙 𝑛 (
1−𝑧
2
))] = 𝑓
𝑋
(−𝑙 𝑛 (
1−𝑧
2
)) (−
2
1−𝑦
(−
1
2
) ) =
1−𝑧
2(1−𝑧)
=
1
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 𝑌 𝑛
= −𝑋 where 𝑋~𝑁(0,1) . Then 𝑌 𝑛
~𝑁(0,1) so 𝑌 𝑛
→ 𝑑 𝑋 . However, we have 𝑃(|𝑌 𝑛
− 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 𝑌 𝑛
↛ 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 𝑋
1
, … , 𝑋 𝑛
are independent and identically distributed with finite mean 𝜇 and variance 𝜎 2 , then the sample mean 𝑋̅ → 𝑝 𝜇 .
 Central Limit Theorem: If 𝑋
1
, … , 𝑋 𝑛
are independent and identically distributed with finite mean 𝜇 and variance 𝜎 2 , then
[∑ 𝑛 𝑖=1
𝑋 𝑖
]−𝑛𝜇 𝜎√𝑛
→ 𝑑 𝑁(0,1) and 𝑋̅−𝜇 𝜎/√𝑛
→ 𝑑 𝑁(0,1) .
Question #10: Let 𝑋
1
and 𝑋
2
be independent random variables with Cumulative Distribution
Function (CDF) given by 𝐹
𝑋
(𝑥) = 1 − 𝑒 −𝑥
2
/2 for 𝑥 > 0 . Find the joint probability density function 𝑓
𝑌
1
𝑌
2
(𝑦
1
, 𝑦
2
) of the random variables 𝑌
1
= √𝑋
1
+ 𝑋
2
and 𝑌
2
=
𝑋
1
𝑋
2
+𝑋
2
.
 We first find the density 𝑓
𝑋
(𝑥) = 𝑑 𝑑𝑥
𝐹
𝑋
(𝑥) = −𝑒
− 𝑥2
2
(−𝑥) = 𝑥𝑒
− 𝑥2
2
for 𝑥 > 0 . Then since
𝑋
1
and 𝑋
2
are independent random variables, their joint density function is 𝑓
𝑋
1
𝑋
2
(𝑥
1
, 𝑥
2
) = 𝑓
𝑋
1
(𝑥
1
)𝑓
𝑋
2
(𝑥
2
) = [𝑥
1 𝑒
− 𝑥1
2
] [𝑥
2 𝑒
− 𝑥2
2
] = 𝑥
1 𝑥
2 𝑒
−
2+𝑥 2
2
2
for 𝑥
1
> 0, 𝑥
2
> 0 .
We then have that 𝑦
1
= √𝑥
1
+ 𝑥
2
and 𝑦
2
= 𝑥
1 𝑥
2
+𝑥
2
so that 𝑥
2
= 𝑦
2
(𝑥
1
+ 𝑥
2
) = 𝑦
2 𝑦 2
1
and 𝑥
1
= 𝑦 2
1
− 𝑥
2
= 𝑦 2
1
− 𝑦
2 𝑦 2
1
. These computations allow us to compute the Jacobian as

𝜕𝑥
1
𝐽 = 𝑑𝑒𝑡 [
𝜕𝑦
1
𝜕𝑥
2
𝜕𝑦
1
𝜕𝑥
1
𝜕𝑦
2
𝜕𝑥
2
] = 𝑑𝑒𝑡 [
2𝑦
1
2𝑦
1 𝑦
2
𝜕𝑦
2
− 2𝑦
1 𝑦
2
−𝑦 2
1 𝑦 2
1
] = 2𝑦 3
1
− 2𝑦 3
1 𝑦
2
+ 2𝑦 3
1 𝑦
2
= 2𝑦 3
1
. This implies that the joint density function is 𝑓
𝑌
1
𝑌
2
(𝑦
1
, 𝑦
2
) = 𝑓
𝑋
1
𝑋
2
(𝑦 2
1
− 𝑦
2 𝑦 2
1
, 𝑦
2 𝑦 2
1
)|𝐽| =
(𝑦 2
1
− 𝑦
2 𝑦 2
1
)(𝑦
2 𝑦 2
1
)𝑒
−
2−𝑦2𝑦12)
2
2
2)
2
2𝑦 3
1
whenever 𝑦 2
1
− 𝑦
2 𝑦 2
1
> 0, 𝑦
2 𝑦 2
1
> 0 . This work also reveals that 𝑌
1
and 𝑌
2
are not independent since their joint probability density function 𝑓
𝑌
1
𝑌
2
(𝑦
1
, 𝑦
2
) cannot be factored into two functions of 𝑦
1
and 𝑦
2
.
Question #11: Let 𝑋
1
, 𝑋
2
, 𝑋
3
be independent random variables and suppose that we know
𝑋
1
~𝑁(2,4) , 𝑋
2
~𝑁(3,9) , and 𝑋
1
+ 𝑋
2
+ 𝑋
3
~𝑁(4,16) . What is the distribution of 𝑋
3
?
 We use the Moment Generating Function (MGF) technique to note that since the three random variables are independent, we have
𝑀
𝑋
1
+𝑋
2
+𝑋
3
(𝑡) = 𝑀
𝑋
1
(𝑡)𝑀
𝑋
2
(𝑡)𝑀
𝑋
3
(𝑡) →
𝑀
𝑋
3
(𝑡) =
𝑀
𝑋1+𝑋2+𝑋3
(𝑡)
𝑀
𝑋1
(𝑡)𝑀
𝑋2
(𝑡)
= 𝑒
4𝑡+16𝑡2/2
(𝑒 2𝑡+4𝑡2/2 )(𝑒 3𝑡+9𝑡2/2 )
= 𝑒
4𝑡+16𝑡2/2 𝑒 5𝑡+13𝑡2/2
= 𝑒 −𝑡+3𝑡
2
/2 . This reveals that the random variable 𝑋
3
~𝑁(−1,3) since 𝑀
𝐴
(𝑡) = 𝑒 𝜇𝑡+𝜎
2 𝑡
2
/2 when 𝐴~𝑁(𝜇, 𝜎 2 ) .
Question #12: Let 𝑋
1
, … , 𝑋
5
be independent 𝑁(𝜇, 𝜎 2 ) random variables. Give examples of random variables 𝑌
1
, … , 𝑌
4
defined in terms of the random variables 𝑋
1
, … , 𝑋
5
such that we have 𝑌
1
~𝑁(0,1) , 𝑌
2
~𝜒 2 (3) , 𝑌
3
~𝐹(1,3) , and 𝑌
4
~𝑡(3) .
 We have that 𝑌
1
=
𝑋
1
−𝜇 𝜎
~𝑁(0,1) , 𝑌
2
= ∑ 3 𝑖=1
𝑌 𝑖
2 = ∑ 3 𝑖=1
(
𝑋 𝑖
−𝜇
)
2 𝜎
~𝜒 2 (3) , 𝑌
3
=
𝑌
2
1
/1
𝑌
2
/3
=
(
𝑋1−𝜇
) 𝜎
2
/1
[∑
3 𝑖=1
(
𝑋𝑖−𝜇 𝜎
)
2
]/3
~𝐹(1,3) , and 𝑌
4
=
𝑌
1
√𝑌
2
/3
=
(
𝑋1−𝜇 𝜎
)
√[∑
3 𝑖=1
(
𝑋𝑖−𝜇
) 𝜎
2
]/3
~𝑡(3) . These random variables are constructed since if some 𝑋~𝑁(𝜇, 𝜎 2 ) , then 𝑍 =
𝑋−𝜇
~𝑁(0,1) . This implies that 𝜎
𝑌 = 𝑍 2 ~𝜒 2 (1)
, so that
𝑈 = ∑ 𝑛 𝑖=1
𝑌 𝑖
~𝜒 2 (𝑛)
. Finally, we note that
𝑉 =
𝑈/𝑛
~𝐹(𝑛, 𝑘)
𝑆/𝑘 where 𝑆~𝜒 2 (𝑘) and that 𝑇 =
𝑍
√𝑈/𝑛
~𝑡(𝑛) .

Question #13: If 𝑋
1
and 𝑋
2
are independent 𝐸𝑋𝑃(1) random variables such that their densities are 𝑓(𝑥) = 𝑒 −𝑥 1{𝑥 > 0} , then find the joint density of 𝑈 =
𝑋
1
𝑋
2
and 𝑉 = 𝑋
1
+ 𝑋
2
.
 By independence, 𝑓
𝑋
1
𝑋
2
(𝑥
1
, 𝑥
2
) = 𝑓
𝑋
1
(𝑥
1
)𝑓
𝑋
2
(𝑥
2
) = [𝑒 −𝑥
1 ][𝑒 −𝑥
2 ] = 𝑒 −(𝑥
1
+𝑥
2
) if 𝑥
1
> 0 and 𝑥
2
> 0 . We have 𝑢 = 𝑥
1 𝑥
2
and 𝑣 = 𝑥
1
+ 𝑥
2
so 𝑥
1
= 𝑢𝑥
2
= 𝑢(𝑣 − 𝑥
1
) , which implies that 𝑥
1
+ 𝑢𝑥
1
= 𝑢𝑣 so 𝑥
1
= 𝑢𝑣
1+𝑢
. Then 𝑥
2
= 𝑣 − 𝑥
1 𝑢𝑣
= 𝑣 −
1+𝑢
= 𝑣+𝑢𝑣−𝑢𝑣
1+𝑢
= 𝑣
1+𝑢
. This
𝜕𝑥
1 allows us to compute the Jacobian as 𝐽 = 𝑑𝑒𝑡 [
𝜕𝑢
𝜕𝑥
2
𝜕𝑢
𝜕𝑥
1
𝜕𝑣
𝜕𝑥
2
] = 𝑑𝑒𝑡 [
𝜕𝑣 𝑣(1+𝑢)−𝑢𝑣
(1+𝑢) 2
−𝑣
(1+𝑢) 2 𝑢
1+𝑢
1
] =
1+𝑢 𝑣(1+𝑢)−𝑢𝑣
(1+𝑢) 2
1
(
1+𝑢
) + 𝑢
1+𝑢 𝑣
(
(1+𝑢) 2
) = 𝑣
(1+𝑢) 3 𝑢𝑣
+
(1+𝑢) 3
= 𝑣(1+𝑢)
(1+𝑢) 3
= 𝑣
(1+𝑢) 2
. Thus, the joint density is 𝑓
𝑈𝑉
(𝑢, 𝑣) = 𝑓
𝑋
1
𝑋
2 𝑢𝑣
(
1+𝑢
, 𝑣
1+𝑢
) |𝐽| = 𝑣
(1+𝑢) 2 𝑒 𝑢𝑣
−(
1+𝑢
+ 𝑣
1+𝑢
)
= 𝑣
(1+𝑢) 2 𝑒 −𝑣 whenever we have that 𝑢 > 0, 𝑣 > 0 and zero otherwise.
Question #14: If 𝑋
1
, 𝑋
2
, 𝑋
3
are a random sample of size 𝑛 = 2 from the 𝑈𝑁𝐼𝐹(0,1) distribution, then a) find the joint density of the order statistics 𝑌
1
, 𝑌
2
, 𝑌
3
, b) find the marginal densities of 𝑌
1
and 𝑌
3
, and c) find the mean of the sample range 𝑅 = 𝑌
3
− 𝑌
1
. a) We have 𝑓
𝑌
1
𝑌
2
𝑌
3
(𝑦
1
, 𝑦
2
, 𝑦
3
) = 3! ∏ 3 𝑖=1 𝑓
𝑋
(𝑦 𝑖
) = 6 ∏ 3 𝑖=1
1 = 6 if 0 < 𝑦
1
< 𝑦
2
< 𝑦
3
< 1 . b) We have 𝑓
𝑌
1
(𝑦
1
) = 3𝑓
𝑋
(𝑦
1
)[1 − 𝐹
𝑋
(𝑦
1
)] 2 = 3(1 − 𝑦
1
) 2 if 0 < 𝑦
1
< 1 and that 𝑓
𝑌
3
(𝑦
3
) = 3𝑓
𝑋
(𝑦
3
)[𝐹
𝑋
(𝑦
3
)] 2 = 3𝑦 2
3
whenever 0 < 𝑦
3
< 1 . c) We have 𝐸(𝑅) = 𝐸(𝑌
3
) − 𝐸(𝑌
1
) , so we compute 𝐸(𝑌
3
) = ∫ 𝑦
3
3𝑦 2
3 𝑑𝑦
3
=
3
4
[𝑦 3
3
] 1
0
=
3
4 and 𝐸(𝑌
1
) = ∫ 𝑦
1
3(1 − 𝑦
1
) 2 𝑑𝑦
1
0
1
= 3 ∫ 𝑦
1
− 2𝑦 2
1
+ 𝑦 3
1 𝑑𝑦
1
= 3 [ 𝑦
1
2
−
2𝑦
3
1
3
+ 𝑦
4
1
4
1
]
0
=
1
4
.
Then, we have that 𝐸(𝑅) = 𝐸(𝑌
3
) − 𝐸(𝑌
1
) =
3
4
−
1
4
=
1
2
. Alternatively, we could have found 𝑓
𝑌
1
𝑌
3
(𝑦
1
, 𝑦
3
) and computed 𝐸(𝑅) as a double integral.