Mathematical Statistics
Hints/Solutions to Assignment 7
1. We have 𝑝𝑋 (−1) =
5
12
, 𝑝𝑋 (0) =
1
6
5
, 𝑝𝑋 (1) =
5
12
.
5
So 𝐸 (𝑋) = 0, 𝐸 (𝑋 2 ) = 6 , 𝑉𝑎𝑟(𝑋) = 6 .
5
5
1
Also 𝑝𝑌 (−2) = 12 , 𝑝𝑋 (−1) = 12 , 𝑝𝑋 (2) = 6 .
11
11
37
So 𝐸 (𝑌) = − 12 , 𝐸 (𝑌 2 ) = 4 , 𝑉𝑎𝑟(𝑌) = 18 .
𝐸 (𝑋𝑌) = 0, 𝐶𝑜𝑣(𝑋, 𝑌) = 0, 𝜌𝑋,𝑌 = 0.
Further, the joint pmf of (𝑈, 𝑉 ) is obtained as
U
1
V
4
1
5
0
1
1
12
1
12
1
3
1
2
5
5
5
39
35
We have 𝑝𝑈 (0) = 6 , 𝑝𝑈 (1) = 6 . So 𝐸 (𝑈) = 6 , 𝐸 (𝑈 2 ) = 6 , 𝑉𝑎𝑟(𝑈) = 36 .
5
7
11
Also 𝑝𝑉 (1) = 12 , 𝑝𝑉 (4) = 12 . So 𝐸 (𝑉 ) = 4 , 𝐸 (𝑉 2 ) = 4 , 𝑉𝑎𝑟(𝑉 ) = 16 .
𝐸 (𝑈𝑉 ) =
7
, 𝐶𝑜𝑣(𝑈, 𝑉 ) =
3
1
24
, 𝜌𝑈,𝑉 =
1
5√7
.
2. 𝑊 = 𝑍 2 = (𝑋1 − 𝑋2 )2 + (𝑌1 − 𝑌2 )2 .
1
𝑍2
1
Now 2 (𝑋1 − 𝑋2 ) ∼ 𝑁 (0, 1), 2 (𝑌1 − 𝑌2 ) ∼ 𝑁(0, 1). So 𝑇 = 2 ∼ 𝜒22 .
√
√
1
Then 𝑓𝑇 (𝑡) = 2 𝑒
𝑡
−
2
1
𝑤
, 𝑡 > 0. So 𝑓𝑊 (𝑤) = 4 𝑒 − 4 , 𝑤 > 0.
3. The joint pdf of (𝑋1 , 𝑋2 ) is
𝑋
𝑓𝑋1 ,𝑋2 (𝑥1 , 𝑥2 ) = 𝜆2 𝑒 −(𝑥1+𝑥2 ) , 𝑥1 > 0, 𝑥2 > 0. 𝑌1 = 𝑋1 , 𝑌2 = 𝑋1 + 𝑋2 .
2
𝑦1 𝑦2
𝑦2
The inverse transformation is 𝑥1 = 1+𝑦 , 𝑥2 = 1+𝑦 .
1
1
The Jacobian of transformation is
𝑦2
𝑦1
(1+𝑦1 )2
𝐽=|
𝑦2
− (1+𝑦 )2
1
1+𝑦1
𝑦2
1 | = (1+𝑦 )2 . The joint pdf of (𝑌1 , 𝑌2 ) is
1
1+𝑦1
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) =
𝜆2 𝑦2 𝑒 −𝜆𝑦2
(1+𝑦1 )2
, 𝑦1 > 0, 𝑦2 > 0.
The marginal pdfs of 𝑌1 and 𝑌2 are
1
𝑓𝑌1 (𝑦1 ) = (1+𝑦 )2 . 𝑦1 > 0 and 𝑓𝑌2 (𝑦2 ) = 𝜆2 𝑦2 𝑒 −𝜆𝑦2 , 𝑦2 > 0 .
1
Clearly 𝑌1 and 𝑌2 are independent.
4. The joint pdf of (𝑋1 , 𝑋2 ) is
1
1
2
2
𝑓𝑋1 ,𝑋2 (𝑥1 , 𝑥2 ) = 2𝜋 𝑒 −2(𝑥1 +𝑥2 ) , 𝑥1 ∈ ℝ, 𝑥2 ∈ ℝ.
𝑋
𝑌1 = 𝑋12 + 𝑋22 , 𝑌2 = 𝑋1 .
2
The inverse transformation is 𝑥1 = ±𝑦2 √
𝑦1
, 𝑥2 = ±√
1+𝑦22
𝑦1
1+𝑦22
.
The Jacobian of transformation is
±
𝐽=
|
|
±
𝑦2
±√
2√𝑦1 (1+𝑦22)
1
∓𝑦2 √
2√𝑦1 (1+𝑦22)
𝑦1
(1+𝑦22 )
3
𝑦1
(1+𝑦22)
1
|
= ∓ 2(1+𝑦 2 ) .
|
2
3
The joint pdf of (𝑌1 , 𝑌2 ) is
𝑦1
1
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = 2𝜋(1+𝑦 2 ) 𝑒 − 2 , 𝑦1 > 0, 𝑦2 ∈ ℝ .
2
We have added twice as two inverse images are there. The marginal pdfs of 𝑌1 and 𝑌2
are given by
1
𝑦1
1
𝑓𝑌1 (𝑦1 ) = 2 𝑒 − 2 , 𝑦1 > 0 and 𝑓𝑌2 (𝑦2 ) = 𝜋(1+𝑦 2 ) , 𝑦2 ∈ ℝ .
2
Clearly 𝑌1 and 𝑌2 are independent.
5. The joint pdf of (𝑋1 , 𝑋2 ) is
𝜆𝑛1 +𝑛2
𝑛 −1 𝑛 −1
𝑓𝑋1 ,𝑋2 (𝑥1 , 𝑥2 ) = Γ(𝑛 )Γ(𝑛 ) 𝑒 −𝜆(𝑥1+𝑥2 ) 𝑥1 1 𝑥2 2
1
2
, 𝑥1 > 0, 𝑥2 > 0
𝑋
1
𝑌1 = 𝑋 +𝑋
and 𝑌2 = 𝑋1 + 𝑋2 .
1
2
The inverse transformation is 𝑥1 = 𝑦1 𝑦2 , 𝑥2 = 𝑦2 (1 − 𝑦1 ).
The Jacobian of transformation is
𝑦2
𝑦1
𝐽 = |−𝑦 1 − 𝑦 | = 𝑦2 .
2
1
The joint pdf of (𝑌1 , 𝑌2 ) is
𝜆𝑛1 +𝑛2
𝑛 −1
𝑛 +𝑛 −1
𝑓𝑌1 ,𝑌2 (𝑦1 , 𝑦2 ) = Γ(𝑛 )Γ(𝑛 ) 𝑦1 1 (1 − 𝑦1 )𝑛2 −1 𝑦2 1 2 𝑒 −𝜆𝑦2 , 0 < 𝑦1 < 1, 𝑦2 > 0 .
1
2
The marginal pdfs of 𝑌1 and 𝑌2 are given by
Γ(𝑛 +𝑛 )
𝑛 −1
𝑓𝑌1 (𝑦1 ) = Γ(𝑛 1)Γ(𝑛2 ) 𝑦1 1 (1 − 𝑦1 )𝑛2 −1 , 0 < 𝑦1 < 1
1
2
and
𝜆𝑛1 +𝑛2
𝑛 +𝑛2−1 −𝜆𝑦2
𝑓𝑌2 (𝑦2 ) = Γ(𝑛 +𝑛 ) 𝑦2 1
1
2
𝑒
, 𝑦2 > 0 .
Clearly 𝑌1 and 𝑌2 are independent.
6. Similar to Q. 5. Also 𝑌1 , 𝑌2 , … , 𝑌𝑛 are independent.
7. ln 𝑊 = 2 + 2 ln 𝑋1 + 4 ln 𝑋2 + 4 ln 𝑋3 ∼ 𝑁(30, 28).
𝑃(𝑊 ≤ 𝑀 ) = 0.5 ⇒ 𝑃(ln 𝑊 ≤ ln 𝑀 ) = 0.5
⇒ ln 𝑀 = 30 ⇒ 𝑀 = 𝑒 30 .
𝑃(𝐿 ≤ 𝑊 ≤ 𝑅) = 0.90 ⇒ 𝑃(ln 𝐿 ≤ ln 𝑊 ≤ ln 𝑅) = 0.90.
So 𝐿 = 𝑒 30−1.645×√28 and 𝑅 = 𝑒 30+1.645×√28 .
1
8. 𝜌 = 2 , 𝜇1 = 0, 𝜇2 = 0, 𝜎1 = 1, 𝜎2 = 1. 𝑉 (𝑋 − 𝑌) = 1 ,
𝑋 + 𝑌 ∼ 𝑁(0,3) .
So 𝑃 (−1 < 𝑋 + 𝑌 < 2) = Φ(1.15) − Φ(−0.58) = 0.8749 − 0.2810 = 0.5939.
9. Let 𝑋 → length of Section A, 𝑌 →length of Section B . Then
𝑋 ∼ 𝑁(20, 0.03), 𝑌 ∼ 𝑁 (14, 0.01). 𝑋 + 𝑌 ∼ 𝑁(34, 0.04).
𝑃(33.6 < 𝑋 + 𝑌 < 34.4) = 𝑃 (−2 < 𝑍 < 2) = 2Φ(2) − 1 = 0.9544.
𝑢
10. The inverse transformation is 𝑥 = √𝑢𝑣 , 𝑦 = √ 𝑣 .
The Jacobian of transformation is
𝑣
1
𝐽 = ||
2
𝑢
√𝑢
√𝑣
1
−√𝑣 3
√𝑢𝑣
𝑢
|| = − 1 .
2𝑣
The joint pdf of (𝑈, 𝑉) is
1
𝑓𝑈,𝑉 (𝑢, 𝑣) = 2𝑢2𝑣 , 𝑢 > max(𝑣, 𝑣 −1 ) , 𝑣 > 0.
The marginal pdfs of 𝑈 and 𝑉 are given by
ln 𝑢
𝑓𝑈 (𝑢) = 𝑢2 , 𝑢 > 1 and
1
,
0<𝑣<1
,
2𝑣 2
𝑣>1
𝑓𝑉 (𝑣) = { 21
.