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IIT−JAM Mathematical Statistics (MS) 2005
1. Let
𝑷 = [
𝟏 𝟐 𝟑 𝟒 𝟓 𝟏𝟐 𝟑 𝟒 𝟖 𝟔 𝟑𝟐𝟒
𝟒𝟕
𝟔 𝟕 𝟏𝟎 𝟑𝟏𝟎 𝟏𝟒 𝟏𝟔 𝟕
]
Then the rank of the matrix P is
(a) 1
(b) 2
(c) 3
(d) 4
Solution: (c)
[
1 2 3 4 5 10 −1 −2 0 −4 100
0−1
0−2
−1−2
0−4
13
]
𝑅2′ = 𝑅2 − 2𝑅1
𝑅3′ = 𝑅3 − 2𝑅1
𝑅4′ = 𝑅4 − 4𝑅1
~ [
1 2 3 4 5 10 −1 −2 0 −4 100
00
00
−12
00
12
] 𝑅4′ = 𝑅4 − 𝑅2
~ [
1 2 3 4 5 10 −1 −2 0 −4 100
00
00
−10
0 10 0
] 𝑅4′ = 𝑅4 − 2𝑅3
𝑅𝑎𝑛𝑘 = 3
2. Consider the following system of linear equations:
𝒙 + 𝒚 + 𝒛 = 𝟑,
𝒙 + 𝒂𝒛 = 𝒃,
𝒚 + 𝟐𝒛 = 𝟑
This system has infinite number of solutions if
(a) 𝒂 = −𝟏, 𝒃 = 𝟎
(b) 𝒂 = 𝟏, 𝒃 = 𝟐
(c) 𝒂 = 𝟎, 𝒃 = 𝟏
(d) 𝒂 = −𝟏, 𝒃 = 𝟏
Solution: (a)
[1 1 11 0 𝑎0 1 2
|3𝑏3] ~ [
1 1 10 −1 𝑎 − 10 1 2
|3
𝑏 − 33
] 𝑅2′ = 𝑅2 − 𝑅1
~ [1 1 10 −1 𝑎 − 10 1 𝑎 + 1
|3
𝑏 − 3𝑏
] 𝑅3′ = 𝑅3 + 𝑅2
⇒ 𝑎 + 1 = 0; 𝑏 = 0
⇒ 𝑎 = −1, 𝑏 = 0
3. Six identical fair dice are thrown independently. Let S denote the number of dice showing
even numbers on their upper faces. Then the variance of the random variable S is
(a) ½
(b) 1
(c) 3/2
(d) 3
Solution: (c) 𝑃(𝑆 = 𝑥) = (6𝑥) (
1
2)𝑥(1
2)6−𝑥
; 𝑥 = 0, 1, 2, 3, 4, 5, 6
⇒ 𝑆 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑏𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑑𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑛 = 6, 𝑝 = 1/2
⇒ 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝑛𝑝(1 − 𝑝) = 6 ×1
2×1
2=3
2
4. Let 𝑿𝟏, 𝑿𝟐, … , 𝑿𝟐𝟏 be a random sample from a distribution having the variance 5. Let
�̅� =𝟏
𝟐𝟏∑𝑿𝒊 𝒂𝒏𝒅 𝑺 =∑(𝑿𝒊 − �̅�)
𝟐
𝟐𝟏
𝒊=𝟏
𝟐𝟏
𝒊=𝟏
Then the value of ∑(𝑺) is
(a) 5
(b) 100
(c) 0.25
(d) 105
Solution: (b)
𝐸 [1
𝑛 − 1∑(𝑥𝑖 − �̅�)
2
𝑛
𝑖=1
] = 𝜎2 ⇒ 𝐸 [1
20∑(𝑥𝑖 − �̅�)
2
21
𝑖=1
] = 5 ⇒ 𝐸(𝑆) = 20 × 5 = 100
5. Let X and Y be independent standard normal random variables. Then the distribution of
𝑼 = (𝑿 − 𝒀
𝑿 + 𝒀)𝟐
𝒊𝒔
(a) chi−square with 2 degrees of freedom
(b) chi−square with 1 degree of freedom
(c) F with (2, 2) degrees of freedom
(d) F with (1, 1) degrees of freedom
Solution: (d)
𝑥(1)2
𝑥(1)2 = 𝐹(1,1)
6. In three independent throws of a fair dice, let X denote the number of upper faces showing
six. Then the value of 𝑬(𝟑 − 𝑿)𝟐 is
(a) 20/3
(b) 2/3
(c) 5/2
(d) 5/12
Solution: (a)
𝑊ℎ𝑒𝑛 𝑋 = 1, 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 (3
1) (1
6) (5
6)2
= 25/72
𝑊ℎ𝑒𝑛 𝑋 = 2, 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 (3
2) (1
6)2
(5
6) =
5
72
𝑊ℎ𝑒𝑛 𝑋 = 3, 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 (3
3) (1
6)3
=1
216
𝑊ℎ𝑒𝑛 𝑋 = 0, 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 (3
0) (5
6)3
=125
216
𝐸(3 − 𝑋)2 =∑(3 − 𝑥)2𝑃(𝑋 = 𝑥)
=100
72+5
72+ 0 +
125
24=160
24=20
3
7. Let
𝑷 = [
𝟏 𝟎 𝟏 + 𝒙 𝟏 + 𝒙𝟎 𝟏 𝟏 𝟏𝟏𝟏
𝟏 + 𝒙𝟏 + 𝒙
𝟎𝟏 + 𝒙
𝟏 + 𝒙𝟎
]
Then the determinant of the matrix P is
(a) 𝟑(𝒙 + 𝟏)𝟑
(b) 𝟑(𝒙 + 𝟏)𝟐
(c) 𝟑(𝒙 + 𝟏)
(d) (𝒙 + 𝟏)(𝟐𝒙 + 𝟏)
Solution: (a)
|
1 0 1 + 𝑥 1 + 𝑥0 1 1 100
1 + 𝑥1 + 𝑥
−(1 + 𝑥)0
0(1 + 𝑥)
| Expanding this w.r.t. 1st column, we will get a 3×3 matrix, then
perform the given operations:
𝑅3′ = 𝑅3 − 𝑅1 & 𝑅4 ′ = 𝑅4 − 𝑅1
det (𝑃) = |1 1 1
1 + 𝑥 −(1 + 𝑥) 01 + 𝑥 0 −(1 + 𝑥)
| = 3(1 + 𝑥)2
8. The area of the region
{(𝒙, 𝒚) ∶ 𝟎 ≤ 𝒙, 𝒚 ≤ 𝟏,𝟑
𝟒≤ 𝒙 + 𝒚 ≤
𝟑
𝟐} 𝒊𝒔
(a) 9/16
(b) 7/16
(c) 13/32
(d) 19/32
Solution: (d) 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑟𝑒𝑎 = 1 × 1 −1
2×3
4×3
4−1
2×1
2×1
2 = 1 −
9
32=
1
8=
19
32
9. Let E, F and G be three events such that the events E and F are mutually exclusive,
𝑷(𝑬 ∪ 𝑭) = 𝟏,𝑷(𝑬 ∩ 𝑮) =𝟏
𝟒 𝒂𝒏𝒅 𝑷(𝑮) =
𝟕
𝟏𝟐
Then 𝑷(𝑬 ∪ 𝑭)equals
(a) 1/12
(b) ¼
(c) 5/12
(d) 1/3
Solution: (d)
𝑃(𝐹 ∩ 𝐺) = 𝑃(𝐺) − 𝑃(𝐸 ∩ 𝐺) =7
12−1
4=
1
3
10. Let X and Y have the joint probability mass function
𝑷(𝑿 = 𝒙, 𝒀 = 𝒚) =𝟏
𝟑𝒙, 𝒚 = 𝟏, 𝟐,… , 𝒙; 𝒙 = 𝟏, 𝟐, 𝟑.
Then the value of the conditional expectation 𝑬(𝒀|𝑿 = 𝟑) is
(a) 1
(b) 2
(c) 1.5
(d) 2.5
Solution: (b)
𝐸(𝑌|𝑋 = 3) =∑𝑦 . 𝑃(𝑌 = 𝑦;𝑋 = 3)
𝑃(𝑋 = 3)
𝑃(𝑋 = 𝑥) =1
3𝑥(1 + 1 +⋯+ 𝑥 𝑡𝑖𝑚𝑒𝑠) =
1
3
= ∑𝑦1
3=1 + 2 + 3
3 = 2
3
𝑦=1
.
11. Let 𝑿𝟏 𝒂𝒏𝒅 𝑿𝟐 be independent random variables with respective moment generating function
𝑴𝟏(𝒕) = (𝟑
𝟒+𝟏
𝟒𝒆𝒕)
𝟑
𝒂𝒏𝒅 𝑴𝟐(𝒕) = 𝒆𝟐(𝒆𝒕−𝟏) , − ∞ < 𝑡 < ∞
Then the value of 𝑷(𝑿𝟏 + 𝑿𝟐 = 𝟏) is
(a) 𝟖𝟏
𝟔𝟒𝒆−𝟐
(b) 𝟐𝟕
𝟔𝟒𝒆−𝟐
(c) 𝟏𝟏
𝟔𝟒𝒆−𝟐
(d) 𝟐𝟕
𝟑𝟐𝒆−𝟐
Solution: (a)
𝐸[𝑒𝑡𝑥] = (3
4+1
4𝑒𝑡)
3
⇒ ∫ 𝑒𝑡𝑥𝑓(𝑥)𝑑𝑥 = (3
4+1
4𝑒𝑡)
3∞
−∞
𝑃(𝑋1 = 0; 𝑋2 = 1) + 𝑃(𝑋1 = 1; 𝑋2 = 0)
= (3
4)3 21𝑒−2
1!+ 3𝐶1 (
3
4)2
(1
4)𝑒−220
0!= (
54
64+27
64) 𝑒−2 =
81
64𝑒−2
12.
𝐥𝐢𝐦𝒏→∞
[𝟏
𝟐𝒏𝟐𝚪(
𝒏𝟐) ∫ 𝒆−
𝒕𝟐𝒕𝒏𝟐−𝟏
∞
𝒏−√𝟐𝒏
𝒅𝒕] 𝒆𝒒𝒖𝒂𝒍𝒔
(a) 0.5
(b) 0
(c) 0.0228
(d) 0.1587
Solution: (a)
lim𝑛→∞
⇒ 𝑐ℎ𝑖 𝑠𝑞𝑢𝑎𝑟𝑒 ⇒ 𝑛𝑜𝑟𝑚𝑎𝑙
𝑚𝑒𝑎𝑛 𝑡𝑜 ∞ = 0.5
𝑛 − √2𝑛 = 𝑛 − 1. Standard deviation.
13. Let 𝑿𝟏 𝒂𝒏𝒅 𝑿𝟐 be two independent random variables having the same mean 𝜽. Suppose
that 𝑬(𝑿𝟏 − 𝜽)𝟐 = 𝟏 𝒂𝒏𝒅 𝑬(𝑿𝟐 − 𝜽)
𝟐 = 𝟐. For estimating 𝜽, consider the
estimators 𝑻𝜶(𝑿𝟏, 𝑿𝟐) = 𝜶𝑿𝟏 − (𝟏 − 𝜶)𝑿𝟐, 𝒂 ∈ [𝟎, 𝟏]. The value of 𝜶 ∈ [𝟎, 𝟏],for which the
variance of 𝑻𝜶(𝑿𝟏, 𝑿𝟐) is minimum, equals
(a) 2/3
(b) ½
(c) ¼
(d) ¾
Solution: (a)
𝑉(𝛼𝑋1 − (1 − 𝛼)𝑋2) = 𝛼2𝑉(𝑋1) + (1 − 𝛼)
2𝑉(𝑋2) = 𝛼2 + 2(1 − 𝛼)2 = 3𝛼2 − 4𝛼 + 2
= 3 [𝛼2 −4
3𝛼 +
2
3] = 3 [(𝛼 −
2
3)2
+2
9]
⇒ 𝛼 =2
3 𝑓𝑜𝑟 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒.
14. Let 𝒙𝟏 = 𝟑, 𝒙𝟐 = 𝟒, 𝒙𝟑 = 𝟑. 𝟓, 𝒙𝟒 = 𝟐. 𝟓 be the observed values of a random sample from the
probability density function
𝒇(𝒙|𝜽) =𝟏
𝟑[𝟏
𝜽𝒆𝒙𝜽 +
𝟏
𝜽𝟐𝒆−𝒙𝜽𝟐 + 𝒆𝒙] , 𝒙 > 0, 𝜃 > 0
Then the method of moments estimate (MME) of 𝜽 is
(a) 3.5
(b) 4
(c) 2.5
(d) 1
Solution: (b)
𝐿(𝑥, 𝜃) =1
3[1
𝜃𝑒𝑥𝜃 +
1
𝜃2𝑒−𝑥𝜃2 + 𝑒𝑥]
𝜕𝐿
𝜕𝜃= 0
⇒−1
𝜃2𝑒𝑥/𝜃 −
𝑥
𝜃3𝑒𝑥/𝜃 −
2
𝜃3𝑒−𝑥/𝜃
2
=−𝑒
𝑥𝜃
𝜃3(𝑥 + 𝜃) +
2
𝜃5𝑒−𝑥𝜃2(𝑥 − 𝜃2)
Which is increasing, so 𝜃 = 𝑥𝑚𝑎𝑥 = 4
15. Let 𝒙𝟏 = −𝟐, 𝒙𝟐 = 𝟏, 𝒙𝟑 = 𝟑, 𝒙𝟒 = −𝟒 be the observed values of a random sample from the
distribution having the probability density function.
𝒇(𝒙|𝜽) =𝒆−𝒛
𝒆𝜽 − 𝒆−𝜽, −𝜽 ≤ 𝒙 ≤ 𝜽, 𝜽 > 0.
Then the maximum likelihood estimate of 𝜽 is
(a) 3
(b) 0.5
(c) 4
(d) Any value between 1 and 2
Solution: (c)
𝐿(𝑥, 𝜃) =∑𝑒−𝑥𝑖
𝑒𝜃 − 𝑒−𝜃 ; 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑏𝑢𝑡 𝑓𝑜𝑟 − 𝜃 ≤ 𝑥 ≤ 𝜃, 𝑖. 𝑒. , 𝜃 = 4
16. Let X and Y be independent and identically distributed uniform random variables over the
interval (0,1) and let 𝑺 = 𝑿 + 𝒀. Find the probability that the quadratic equation 𝟗𝒙𝟐 + 𝟗𝑺𝒙 +
𝟏 = 𝟎 has no real root.
Solution:
81𝑆2 − 36 < 0 𝑔𝑖𝑣𝑒𝑠 𝑆2 < (6
9)2
⇒ −2
3< 𝑆 <
2
3
𝑃𝑟𝑜𝑏.=
12 ×
23 ×
23
1 × 1=2
9
17. Find the number of real roots of the polynomial 𝒇(𝒙) = 𝒙𝟓 + 𝒙𝟑 − 𝟐𝒙 + 𝟏.
Solution: 𝑓(𝑥) = 𝑥5 + 𝑥3 − 2𝑥 + 1.
Using Descartes rule:
f(x) = + + − + At most 2 positive.
𝑓(−𝑥) = −− + + At most 1 negative
𝑓′(𝑥) = 5𝑥4 + 3𝑥2 − 2 = 5 [𝑥4 +3
5𝑥2 −
2
5] = 5 [(𝑥2 +
3
10)2
− (7
10)2
]
𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑓′(𝑥) = 0
⇒ 𝑥2 =4
10=2
5𝑥 = ±√
2
5
We know complex roots occur in pairs. So, the given equation has 3 real roots.
18. Consider the 𝒏 × 𝒏 matrix
𝑷 =
[ 𝟐
𝒏 + 𝟏
𝟏
𝒏 + 𝟏
𝟏
𝒏 + 𝟏⋯
𝟏
𝒏 + 𝟏𝟏
𝒏 + 𝟏
𝟐
𝒏 + 𝟏
𝟏
𝒏 + 𝟏⋯
𝟏
𝒏 + 𝟏𝟏
𝒏 + 𝟏⋮𝟏
𝒏 + 𝟏
𝟏
𝒏 + 𝟏⋮𝟏
𝒏 + 𝟏
𝟐
𝒏 + 𝟏⋮𝟏
𝒏 + 𝟏
⋯⋮⋯
𝟏
𝒏 + 𝟏⋮𝟐
𝒏 + 𝟏]
Let I be the 𝒏 × 𝒏 identity matrix and let J be the 𝒏 × 𝒏 matrix whose all entries are 1. Express the
matrix P as 𝒂𝑰 + 𝒃𝑱, for suitable constants a and b. It is known that the inverse of P is of the form
𝒄𝑰 − 𝒅𝑱, for some constants c and d. Find the constants c and d in terms of n.
Solution:
𝑃 =1
𝑛 + 1𝐼 +
1
𝑛 + 1𝐽
𝑃𝑃−1 = (𝑎𝐼 + 𝑏𝐽)(𝑐𝐼 − 𝑑𝐽)
𝐼 = 𝑎𝑐𝐼 + (𝑏𝑐 − 𝑎𝑑)𝐽 − 𝑏𝑑𝐽2 = 𝑎𝑐𝐼 + (𝑏𝑐 − 𝑎𝑑)𝐽 − (𝑛 + 1)𝑏𝑑𝐽
⇒ 𝑎𝑐 = 1 & 𝑏𝑐 − 𝑎𝑑 − (𝑛 + 1)𝑏𝑑 = 0
⇒ 𝐶 = (𝑛 + 1)
1
𝑛 + 1(𝑛 + 1) −
1
(𝑛 + 1)𝑑 − (𝑛 + 1)
1
𝑛 + 1𝑑 = 0
⇒ 1 =𝑛 + 2
𝑛 + 1𝑑 ⇒ 𝑑 =
𝑛 + 1
𝑛 + 2
⇒ 𝑃−1 = (𝑛 + 1)𝐼 −(𝑛 + 1)
(𝑛 + 2)𝐽
19. Let 𝒇(𝒙) = (𝒙 + 𝟏)|𝒙𝟐 − 𝟏|, −∞ < 𝑥 < ∞. Verify the differentiability of the function f(.) at the
points 𝒙 = −𝟏 𝒂𝒏𝒅 𝒙 = 𝟏.
Solution:
𝑓(𝑥) = (𝑥 + 1)(𝑥2 − 1) ; 𝑥 ≥ 1 𝑜𝑟 𝑥 ≤ −1
= (𝑥 + 1)(1 − 𝑥2) ; −1 ≤ 𝑥 ≤ 1
𝑓′(𝑥) = 3𝑥2 + 2𝑥 − 1 ; 𝑥 ≥ 1 𝑜𝑟 𝑥 ≤ −1
= −3𝑥2 − 2𝑥 + 1 ; −1 ≤ 𝑥 ≤ 1
⇒ 𝑓(𝑥)𝑖𝑠 𝑛𝑜𝑡 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = −1 𝑎𝑛𝑑 𝑥 = 1
20. There are four urns labeled 𝑼𝟏, 𝑼𝟐, 𝑼𝟑 𝒂𝒏𝒅 𝑼𝟒, each containing 3 blue and 5 red balls. The fifth
urn, labeled 𝑼𝟓, contains 4 blue and 4 red balls. An urn is selected at random from these five
urns and a ball is drawn at random from it. Given that the selected ball is red, find the probability
that it came from the urn 𝑼𝟓.
Solution:
15×48
15×48 +
45×58
=4
24=1
6
21. Let
𝑭(𝒙) =
{
𝟎, 𝒊𝒇 𝒙 > 0
𝒙𝟐
𝟏𝟎𝒊𝒇 𝟎 ≤ 𝒙 < 1
𝒙 + 𝟐
𝟖𝒄(𝟔𝒙 − 𝒙𝟐 − 𝟏)
𝟐𝟏,
𝒊𝒇 𝟏 ≤ 𝒙 < 2𝒊𝒇 𝟐 ≤ 𝒙 ≤ 𝟑𝒊𝒇 𝒙 > 3
Find the value of c for which F(.) is a cumulative distribution function of a random variable X.
Also evaluate 𝑷(𝟏 ≤ 𝑿 ≤ 𝟐).
Solution:
𝑓(𝑥) =
{
0 ; 𝑥 < 0𝑥
5; 0 ≤ 𝑥 < 1
1
8𝑐(3 − 𝑥)
0
; 1 ≤ 𝑥 < 2; 2 ≤ 𝑥 ≤ 3; 𝑥 > 3
⇒ ∫𝑥
5𝑑𝑥 +∫
1
8𝑑𝑥 + ∫ 𝐶(3 − 𝑥)𝑑𝑥
3
2
2
1
1
0
= 1
⇒1
10+1
8+ 𝐶 (3 −
5
2) = 1
⇒𝐶
2= 1 −
1
8−1
10=31
40
⇒ 𝐶 =31
20
𝑃(1 ≤ 𝑥 < 2) = ∫1
8𝑑𝑥 =
1
8
2
1
22. Let A, B and C be pair wise independent events such that 𝑷(𝑨 ∩ 𝑩) = 𝟎. 𝟏 𝒂𝒏𝒅 𝑷(𝑩 ∩ 𝑪) =
𝟎. 𝟐. Show that 𝑷(𝑨𝑪 ∪ 𝑪) ≥ 𝟕/𝟖.
Solution:
𝑃(𝐴)𝑃(𝐵) = 0.1
𝑃(𝐵)𝑃(𝐶) = 0.2
𝑃(𝐴𝐶 ∪ 𝐶) = 𝑃(𝐴𝐶) + 𝑃(𝐶) − 𝑃(𝐴𝐶)𝑃(𝐶)
= 1 − 𝑃(𝐴) + 𝑃(𝐶) − 𝑃(𝐶)
= 1 − 𝑃(𝐴) + 𝑃(𝐴)𝑃(𝐶)
= 1 − 𝑃(𝐴)[1 − 𝑃(𝐶)]
= 1 − 𝑃(𝐴)[1 − 2𝑃(𝐴)]
= 1 − 𝑃(𝐴) + 2[𝑃(𝐴)]2
= 2 [(𝑃(𝐴))2−1
2𝑃(𝐴) +
1
2]
= 2 [{𝑃(𝐴) −1
4}2
+7
16]
= 2 [𝑃(𝐴) −1
4]2
+7
8 ≥
7
8
23. Let the random variables X and Y have the joint probability density function
𝒇(𝒙, 𝒚) = {𝒄𝒆−(𝒙+𝒚) , 𝒚 > 𝑥 > 0
𝟎, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
Evaluate c and 𝑬(𝒀|𝑿 = 𝟐)
Solution:
∫ ∫ 𝑐𝑒−(𝑥)𝑒−𝑦𝑑𝑥 𝑑𝑦𝑦
0
∞
0
= ∫ 𝑐(1 − 𝑒−𝑦)𝑒−𝑦𝑑𝑦 ∞
0
= 𝑐 [∫ 𝑒−𝑦𝑑𝑦 − ∫ 𝑒−2𝑦𝑑𝑦∞
0
∞
0
] = 𝑐 [1 −1
2]
𝑆𝑜,𝑐
2= 1
⇒ 𝑐 = 2
𝑓(𝑥) = ∫ 2𝑒−𝑥𝑒−𝑦𝑑𝑦 ∞
𝑥
= 2𝑒−2𝑥 𝑥 > 0
𝐸(𝑌|𝑋 = 2) = ∫2𝑦𝑒−(2+𝑦)
2𝑒−4𝑑𝑦
∞
2
= ∫ 𝑦𝑒−(𝑦−2)𝑑𝑦∞
0
= −𝑦𝑒−(𝑦−2) − 𝑒−(𝑦−2) |∞
0= 2 − 1 = 1
24. Find the area of the region enclosed between the curves
𝒚 = (𝒙 − 𝟐)𝟐 𝒂𝒏𝒅 𝒚 = 𝟒 − 𝒙𝟐
Solution:
(𝑥 − 2)2 = 4 − 𝑥2 ⇒ 2𝑥2 − 4𝑥 = 0 ⇒ 2𝑥(𝑥 − 2) = 0 ⇒ 𝑥 = 0, 2
𝐴 = ∫ (𝑥 − 2)2 − 4 + 𝑥2 𝑑𝑥2
0
=(𝑥 − 2)2
3− 4𝑥 +
𝑥3
3 |2
0 = 𝐴𝑛𝑠𝑤𝑒𝑟
25. Let �̅� be the sample mean of a random sample of size 10 from a distribution having the
probability density function
𝒇(𝒙|𝜽) =𝟏
𝜽𝒆𝒙𝜽 , 𝒙 > 0, 𝜃 > 0.
Using Chebyshev’s inequality, show that
𝑷(𝟎 < �̅� ≤ 𝟐𝜽) ≥ 𝟎. 𝟗
Solution:
𝑓(𝑥|𝜃) =1
𝜃𝑒−𝑥/𝜃 ; 𝜃 > 0; 𝑥 > 0
⇒ 𝑖𝑡 𝑖𝑠 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟1
𝜃,
ℎ𝑒𝑛𝑐𝑒 𝑚𝑒𝑎𝑛 𝑤𝑖𝑙𝑙 𝑏𝑒 1
1/𝜃 𝑎𝑛𝑑, 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
1
1/𝜃2 𝑖. 𝑒. , 𝜃 𝑎𝑛𝑑 𝜃2
So, mean = 𝜃 𝑎𝑛𝑑 𝑠. 𝑑. = 𝜃
So, by Chebyshev’s inequality
𝑃(0 < �̅� < 2𝜃) > 1 −1
10= 1 − 0.1
⇒ 𝑃(0 < �̅� ≤ 2𝜃) ≥ 0.9
26. Let X be a continuous random variable having the probability density function
𝒇(𝒙) = {𝟐
𝟐𝟓(𝒙 + 𝟐) ,−𝟐 ≤ 𝒙 ≤ 𝟑
𝟎, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
Find the cumulative distribution function 𝒀 = 𝑿𝟐. Hence derive the expression for probability
density function of Y.
Solution:
Let 𝐹(𝑥) = 𝑃(𝑌 = 𝑋2 ≤ 𝑥) = 𝑃(𝑋2 ≤ 𝑥) = 𝑃(−√𝑥 ≤ 𝑋 ≤ √𝑥)
𝐹(𝑥) =
{
0 𝑥 < 0
∫2
25(𝑥 + 2)𝑑𝑥 +∫
2
25(𝑥 + 2)𝑑𝑥
√𝑥
0
0
−√𝑥
; 0 ≤ 𝑥 < 4
∫2
25(𝑥 + 2)𝑑𝑥
√𝑥
−2
1
4 ≤ 𝑥 < 9𝑥 ≥ 9
⟹ 𝐹(𝑥) =
{
0; 𝑥 < 0
8√𝑥
250 ≤ 𝑥 < 4
𝑥 − 4
25+4(√𝑥 + 2)
251;
; 4 ≤ 𝑥 < 9𝑥 ≥ 9
𝑝. 𝑑. 𝑓. 𝑜𝑓 𝑌 = 𝑋2
𝑓(𝑥) =
{
0; 𝑥 ≥ 04
25√𝑥0 ≤ 𝑥 ≤ 4
1
25+
2
25√𝑥0;
; 4 ≤ 𝑥 ≤ 9𝑥 ≥ 9
27. Let X be a single observation from the probability density function
𝒇(𝒙|𝜽) = (𝜽 + 𝟏)𝒙𝜽 , 𝟎 < 𝑥 < 1,
Where 𝜽 ∈ {𝟏, 𝟐} is unknown. Find the most powerful test of level
𝜶 =𝟏𝟑
𝟒𝟗 𝒇𝒐𝒓 𝒕𝒆𝒔𝒕𝒊𝒏𝒈 𝑯𝟎 ∶ 𝜽 = 𝟏 𝒂𝒈𝒂𝒊𝒏𝒔𝒕 𝑯𝟏 ∶ 𝜽 = 𝟐
What is the power of the test?
Solution: 𝛼 = 𝑝(𝑟𝑒𝑗𝑒𝑐𝑡 |𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒)
𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒 ⇒ 𝜃 = 1 ⇒ 𝑓(𝑥|1) = 2𝑥, 0 < 𝑥 < 1
Let critical region be (k, 1), then
∫ 𝑓(𝑥)𝑑𝑥 = 𝛼 = ∫ 2𝑥 𝑑𝑥1
𝑘
= 𝛼1
𝑘
⇒ 1− 𝑘2 =13
49
⇒ 𝑘2 =36
49
⇒ 𝑘 =6
7
So, acceptance region is (0, 6/7)
𝛽 = 𝑝 (𝑎𝑐𝑐𝑒𝑝𝑡𝑠 |𝐻1 𝑖𝑠 𝑡𝑟𝑢𝑒)
𝐻1 𝑖𝑠 𝑡𝑟𝑢𝑒 𝑔𝑖𝑣𝑒𝑠 𝜃 = 2
⇒ 𝑓(𝑥|2) = 3𝑥2; 0 < 𝑥 < 1
⇒ 𝛽 = ∫ 3𝑥2𝑑𝑥6/7
0
=216
343
⇒ 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡𝑒𝑠𝑡, 1 − 𝛽 = 1 — 216
343 =
133
343 .
28. Let 𝑿𝟏, 𝑿𝟐, … , 𝑿𝒏 be a random sample from a population having the probability density
function
𝒇(𝒙|𝜽, 𝝈) = {𝟏
𝝈𝒆−(𝒙−𝜽)
𝝈𝟐 , 𝒙 > 0
𝟎, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
−∞ < 𝜃 < ∞, 𝜎 > 0
Find the method of moments and maximum likelihood estimators of 𝜽 𝒂𝒏𝒅 𝝈.
Solution: Likelihood function,
𝐿(𝑥, 𝜃) =∏1
𝜎𝑒−(𝑥𝑖−𝜃)𝜎2
𝑛
𝑖=1
=1
𝜎𝑛 𝑒−(∑𝑥𝑖−𝑛𝜃)
𝜎2
⇒ log 𝐿 = −𝑛 log𝜎 −∑𝑥𝑖 + 𝑛𝜃
𝜎2
⇒1
𝐿.𝜕𝐿
𝜕𝜃=𝑛
𝜎2> 0 (𝑎𝑙𝑤𝑎𝑦𝑠)
⇒ 𝑔𝑖𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔.
Hence 𝜃 = 𝑥(𝑛) .
29. Let 𝒇(𝒙) = 𝟑(𝒙 − 𝟐)𝟑 − (𝒙 − 𝟐), 𝟎 ≤ 𝒙 ≤ 𝟐𝟎.
Let 𝒙𝟎 𝒂𝒏𝒅 𝒚𝟎 be the points of the global minima and global maxima, respectively, of f(.) in the
interval [0, 20]. Evaluate 𝒇(𝒙𝟎) + 𝒇(𝒚𝟎).
Solution:
𝑓(𝑥) = 3(𝑥 − 2)3 − (𝑥 − 2)
⇒ 𝑓′(𝑥) = 9(𝑥 − 2)2 − 1 = 9𝑥2 − 36𝑥 + 8
𝑁𝑜𝑤 𝑓′(𝑥) = 0
⇒ 9𝑥2 − 36𝑥 + 8 = 0
⇒ 𝑥 =36 ± √1296 − 9 × 8 × 4
18
=36 ± 12√7
18= 2 ±
2√7
3
𝑓′′(𝑥) = 18𝑥 − 36 = 18(𝑥 − 2)
𝑓′′ (2 +2√7
3) > 0
⇒ 𝑥 = 2 +2√7
3 𝑖𝑠 𝑙𝑜𝑐𝑎𝑙 𝑚𝑖𝑛𝑖𝑚𝑎 𝑝𝑜𝑖𝑛𝑡
𝑓′′ (2 −2√7
3) < 0
⇒ 𝑥 = 2 −2√7
3 𝑖𝑠 𝑙𝑜𝑐𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑎 𝑝𝑜𝑖𝑛𝑡.
Clearly 𝑥 = 0 & 𝑥 = 20 are global minima & global maxima point (given in the question).
⇒ 𝑥0 = 0 & 𝑦0 = 20
𝑓(𝑥0) = 𝑓(0) = 3(−2)3 − (−2) = −22
𝑓(𝑦0) = 𝑓(20) = 3(18)3 − (18) = 18[3(18)2 − 1] = 17478
𝑆𝑜, 𝑓(𝑥0) + 𝑓(𝑦0) = 𝑓(0) + 𝑓(20) = 22 + 17478 = 17456