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