微乙

(1)

992微乙^01-05班期末考解答和評分標準

1. (10%) 解微分方程 ^{y = y(t)} 滿足 ^y⁰ ^{= y(1}− y) , y(0) = 1 2 。 Sol:

By assumption, dy

dt = y(1− y) ⇒ dy

y(1− y) = dt ⇒ Z

dt =

Z dy

y(1− y) (1%)

= Z 1

y + 1

1− ydy (2%)

⇒ t + c = ln |y| − ln |1 − y| for some constant c (2%) By assumption y(0) = 1

2, ⇒ 0 + c = ln |1

2| − ln |1 − 1

2| = 0 ⇒ c = 0 (2%) and y(0) = 1

2 > 0, 1− y(0) = 1

2 > 0,⇒ |y| = y, |1 − y| = 1 − y for t close to 0 s.t. 0 < y(t) < 1.

⇒ e^t = e^{ln y}^−ln(1−y) = y

1− y (1%)

⇒ 0 = (1 − y)e^t− y = y(−e^t− 1) + e^t

⇒ y(t) = e^t

1 + e^t (2%) But then 0 < e^t

1 + e^t < 1 for all t, i.e. y(t) = e^t

1 + e^t is a solution deﬁned for all t∈ R.

2. (15%) 解微分方程 ^x^dy

dx = y + x²sin x , y(π 2) = π

2 。 Sol:

Solve the equation

xy⁰ = y + x²sin x =⇒ y⁰− 1

xy = x sin x

Conside the integral factor

e^R⁻^x¹^dx = 1

x (5%) Then

d dx(y1

x) = sin x (3%)

=⇒ y1

x =− cos x + c

=⇒ y = x(− cos x + c) (4%)

(2)

By

y(π 2) = π

2 =⇒ y(π 2) = π

2(0 + c) = π 2 so

c = 1 (3%) Hence

y = x(1− cos x)

3. (15%) 令g(x, y) = P(X = x, Y = y) , X 取值 ¹ 或^{2 , Y} 取值 ¹或 ²或 ³。若已知

g(1, 1) = 2

11 , g(1, 2) = 3

11 , g(1, 3) = 1 11 , g(2, 1) = 1

11 , g(2, 2) = 3

11 , g(2, 3) = 1 11 , 求 (a) E(X) , (b) Var(X)。

Sol:

(a)

E(X) = 1· P(X = 1) + 2 · P(X = 2)

= 1· ( 2 11 + 3

11 + 1 11)

| {z }

(3%)

+ 2· ( 1 11 + 3

11+ 1 11)

| {z }

(3%)

= 16

11 (1%) (b)

Var(X) = E(X²)− E²(X)

= 1²· ( 2 11+ 3

11+ 1 11)

| {z }

(3%)

+ 2²· ( 1 11+ 3

11+ 1 11)

| {z }

(3%)

−256

|{z}121

(2%)

= 30 121

4. (15%) 若 ^X1 , X2 為白努利單次試驗 ^{, X}1 、 ^X2 取值 ¹的機率為 ²

5 , X1 、 ^X2 取值 ⁰的機率為 ³ 5 且 ^X1 , X₂ 獨立。令 ^{Y = X}1+ X₂ , Z = X₁− X2 , 求

(a) P(Y = 1) , P(Z = 1) 及P(Y = 1, Z = 1) 之值。

(3)

(b) 試說明 ^Y 與^Z 是否獨立。 Sol:

(a)

P(Y = 1) = P({X1 = 1, X2 = 0} ∪ {X1 = 0, X2 = 1})

= P(X₁ = 1, X₂ = 0) + P(X₁ = 0, X₂ = 1)

= P(X₁ = 1)P(X₂ = 0) + P(X₁ = 0)P(X₂ = 1) (because X₁, X₂ are indep.)

= (2 5)(3

5) + (3 5)(2

5) = 12 25

(3%)

P(Z = 1) = P(X₁ = 1, X₂ = 0)

= P(X₁ = 1)P(X₂ = 0) (because X₁, X₂ are indep.)

= (2 5)(3

5) = 6 25

(3%)

P(Y = 1, Z = 1) = P(X1 = 1, X2 = 0)

= P(X₁ = 1)P(X₂ = 0) (because X₁, X₂ are indep.)

= (2 5)(3

5) = 6 25

(4%) (b) Y and Z are not independent, because

P(Y = 1, Z = 1) = 6

25 6= (12 25)( 6

25) = P(Y = 1)P(Z = 1)

(5%) 5. (15%) 設^fX(x) = 2x

k² , 0≤ x ≤ k , 是^X 的機率密度函數 ^{; X} 取值在^{[0, k]} 。

(a) 求^{E(X) (}以 ^k 表示⁾。 ^(b) 求 ^{Var(X) (}以 ^k 表示⁾ 。 ^(c) 若 Var(X) = 2 , 求^k 之值。 Sol:

(a)

E(X) = Z k

0

x2x

k² dx = 1 k²

Z k 0

2x²dx = 2

3k. (5%)

(4)

(b)

Var(X) = E(X²)− (E(X))² (4%) E(X²) =

Z _k

0

x²2x

k² dx = 2 k²

Z _k

0

x³dx = k² 2 Var(X) = k²

2 −4

9k² = 1

18k². (4%) (c)

Var(X) = 1

18k² = 2 ⇒ k = 6. (2%)

If the answer of (b) is wrong but the answer of (c) is correct corresponding to it, then you may get 1pt in (c).

6. (15%) 設^{X , Y} 為獨立之隨機變數且其機率密度函數為

f_X(x) =







2⁻¹x²e^−x , x≥ 0, 0 , x < 0.

; f_Y(y) =







e^−y , y ≥ 0, 0 , y < 0.

若 Z = X + Y , 求^fZ(z) 。 Sol:

f_Z(z) = Z _∞

−∞

f_X(x)f_Y(z− x) dx (4%)

= Z _z

0

f_X(x)f_Y(z− x) dx

= Z _z

0

1

2x²e^−xe^−(z−x)dx (5%)

= Z z

0

1

2x²e^−zdx

= e^−z Z z

0

1

2x²dx (4%)

= e^−zz³

6 (z > 0) (2%)

7. (15%) 一本 ³⁰⁰ 頁的書有 ¹⁵ 個錯誤且其分佈近似於一個 ^Poisson過程。求 (a) 100頁中恰有 ⁵ 個錯誤之機率。

(b) 10頁中至少有 ¹ 個錯誤之機率。

(5)

Sol:

(a)

Poisson(k, λ, T ) = (λT )^k

k! e^−λT (2%) (1)

λ = 15

300, m = 100× λ = 5. (2%) (2)

Poisson(5, 15

300, 100) = (3%) (3)

= 5⁵

5!e⁻⁵. (1%) (4)

You may only get points if you have get points in previous equation. (i.e. If you have a mistake in line 2, you may not get points in line 3).

(b)

1− Poisson(0, 0.05, 10) = 1 − 0.5⁰e^−0.5

0! (5)

= 1− e^−0.5 (6)

Total 7 points. If you have already write down Eq (5) but you have a wrong answer. Then you may get 6 points. Other mistake will depend on your work to give your corresponded grade.