982微積分乙01-05班期末考解答與評分標準
1. (10%) (有移民的人口模型) 令ρ 為移民率 (移民數/單位時間) 為一常數 。 設 P (t)表示在時間 t 之 人口總數 ,滿足 P (t0) = P0 及P0(t) = λP (t) + ρ , 其中 λ > 0 為常數 。 求 P (t)的一般式 。 Sol:Z t
t0
P0(t)
P (t) + ρλ dt = Z t
t0
λ dt
lnP (t) + ρλ
P0+λρ = λ(t− t0) P (t) = −ρ
λ + (P0+ ρ
λ)eλ(t−t0)
Write down the exact solution: 10 pts.
Write down the general solution: 8 pts.
Write down or derive the formula: 5 pts.
Others : give you some points accorddng to the wrong extent.
2. (10%) 解微分方程 dy
dt = 2ty + 4t , y(0) = 1 。 Sol:
eR−2tdt = e−t2 (6 pts) e−t2y0− 2tye−t2 = 4te−t2
⇒ e−t2y = Z
4te−t2dt =−2e−t2 + C
⇒ y = Cet2 − 2 (3 pts)
y(0) = 1⇒ 1 = C − 2 ⇒ C = 3 ⇒ y(t) = 3et2 − 2 (1 pt)
3. (10%) 解微分方程 dy
dt = (y− 1)(y − 2) , y(0) = 0 。
Sol:
dy
dt = (y− 1)(y − 2)
⇒ Z
( 1
(y− 1)(y − 2))dy = Z
dt (3 pts)
⇒ Z
( 1
y− 2 − 1
y− 1)dy = Z
dt (2 pts)
⇒ log(|y − 2|) − log(|y − 1|) = t + c
⇒ log(¯¯
¯¯y− 2 y− 1
¯¯¯¯) = t + c
⇒¯¯
¯¯y− 2 y− 1
¯¯¯¯ = et+c
⇒ y = 2− c1et
1− c1et (3 pts for any one of above forms)
⇒ y = 2− 2et
1− 2et (Since y(0) = 0) (2 pts)
4. (10%) 令V 為一隨機變數 。 而 V = k , k ≥ 2 , 表示在連續進行白努利試驗時 , 到第 k 次才出現 兩次 + 事件 。 計算 P(V = k) 。(設 p為一次白努利試驗出現 +之機率 , 而q = 1− p 。)
Sol:
P (V = k) = C1k−1pqk−2· p = C1k−1p2qk−2 (10 pts) Any other answers similar to the correct one receives at most 4 points.
You will get 4 points if your answer has one place different to the correct one.
You will get 2 points if your answer has two places different to the correct one.
You will get 0 point if your answer has more than two places different to the correct one.
5. (20%) 設X , Y 為隨機變數取值 1或 2 。 已知
P(X = 1, Y = 1) = 4
19, P(X = 1, Y = 2) = 2
19, P(X = 2, Y = 1) = 7
19, P(X = 2, Y = 2) = 6 19 求 P(X = 1) , P(X = 2) , E(X)及 Var(X) 。(各5%)
Sol:
(a)
P(X = 1) = P(X = 1, Y = 1) + P(X = 1, Y = 2) (1 pt)
= 4 19+ 2
19 (2 pts)
= 6
19 (2 pts) (b)
P(X = 2) = P(X = 2, Y = 1) + P(X = 2, Y = 2) (1 pt)
= 7 19+ 6
19 (2 pts)
= 13
19 (2 pts) (c)
E(X) = 1· P(X = 1) + 2 · P(X = 2) (1 pt)
= 6
19+ 2·13
19 (2 pts)
= 32
19 (2 pts) (d) Var(X) = E(X2)− E(X)2 (1 pt)
E(X2) = 12· 6
19+ 22· 13 19 = 58
19 (2 pts) Var(X) = 58
19 − (32
19)2 = 78
361 (2 pts) 6. (15%) 設隨機變數 X 之機率密度函數為
f (t) =
ct(1− t) 若0≤ t ≤ 1
0 其他
其中 c 為常數 。 求 c之值 , 再求 E(X) , Var(X) 。(各5%) Sol:
Since f (x) is a random function, 1 =
Z ∞
f (x) dx = Z 1
cx(1− x) dx (2 pts)
Hence c = 6 (3 pts)
E(x) = Z ∞
−∞
xf (x) dx = Z 1
0
x(6x(1− x)) dx (2 pts)
= 6 Z 1
0
x2− x3dx = 6 µx3
3 − x4 4
¶1 0
= 6 µ1
3− 1 4
¶
= 1
2 (3 pts) V(x)
µ
= Z ∞
−∞
(x− E(x))2f (x) dx
¶
= E(x2)− (E(x))2 (2 pts)
= Z ∞
−∞
x2f (x) dx− µ1
2
¶2
= 6 Z 1
0
x3− x4dx− µ1
2
¶2
= 6 µx4
4 −x5 5
¶1 0
− µ1
2
¶2
= 6 µ1
4− 1 5
¶
− µ1
2
¶2
= 1
20 (3 pts)
7. (15%) 若X , Y 為獨立之隨機變數並且有指數型機率密度函數
fX(t) = λe−λt = fY(t) 對t ≥ 0 , 而fX(t) = 0 = fY(t) 對t < 0 。
求 Z = X + Y 之機率密度函數 fZ(t) 及E(Z) , Var(Z) 。 (各5%) Sol:
(a) fZ(t) = Z ∞
−∞
fX(x)fY(t− x) dx (2 pts) The region of integration is 0≤ x ≤ t
Since 0≤ y ≤ ∞, −∞ ≤ −y ≤ 0, so 0 ≤ x = t − y ≤ t fZ(t) =
Z t 0
λe−λxλe−λ(t−x)dx = Z t
0
λ2e−λtdx = λ2te−λt (3 pts) (b) X, Y have the same distribution, so
E(Z) = E(X) + E(Y ) = 2E(X) = 2 Z ∞
0
λte−λtdt
= lim
b→∞2 Z b
0
λte−λtdt (2 pts)
= lim
b→∞−2te−λt¯¯¯b
0
+ 2 lim
b→∞
Z b
0
e−λtdt = 2
λ (3 pts)
(c) Since X, Y are i.i.d, Cov(X, Y ) = 0 and
Var(Z) = Var(X) + Var(Y ) + 2Cov(X, Y )
= Var(X) + Var(Y ) = 2Var(X)
= 2[E(X2)− (E(X))2] (2 pts)
E(X2) = Z ∞
0
t2λe−λtdt = lim
b→∞
Z b 0
t2λe−λtdt
= lim
b→∞−e−λtt2¯¯¯b
0
+ 2 lim
b→∞
Z b
0
te−λt = 2 λ2 So Var(X) = 1
λ2, and Var(Z) = 2
λ2 (3 pts)
8. (10%) 某公司之電話通數平均每小時30通 。 求在6 分鐘內至少有兩通電話之機率 。(假設此隨機現 象遵守 Poisson過程 。)
Sol:
Let X be the random variable of the times of phone calls during 6 seconds. Observe that the Poisson density is
fX(x) =
λxe−λ
x! x = 0, 1, 2, ...
0 otherwise
(3 pts)
and λ = 6· 30
60 = 3 (3 pts). Hence, we have
P(X ≥ 2) = 1 − fX(0)− fX(1) = 1− e−3− 3e−3 = 1− 4e−3 (4 pts)
