概率统计第二章习题答案(全部)

3 1 1 − = , 10 5 10 1 3 2 P (x = 0) = − = , 2 10 10 1
1 1 = . 2 2 We come to conclusion that the probability density matrix of X is P (x = 2) = 1 − −5 −2
Solutions for Homework
Huaming King 2012 c 10 19 F
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§ vkc[u
§ Œ[w
‹<†ØgC?U˜e¿wŠ
§1
x
1
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3. If F (x) = −∞ Ce−t dt, x ∈ (−∞, ∞) is a distribution function, what should C be? Solution:Since F (x) is a distribution function, then
¥gS 4‡¦¤kŒU {kA4 8 . {n = 4 } ¿ › X 1 ˜ ng Ñü‡x¥˜‡ç¥§ • ˜‡¥ Ñx¥" l¥ ôÚþw§ k±eA«ŒU"
(x§ x§ ç§ x) (x§ ç§ x§ x) (ç§ x§ x§ x)
Ù¢ùn«ŒU´ù 5 §• ˜‡´x¥Ø7+§"c>n‡ ˜¥À˜‡5–ç 1 «À{" ¥kC3 ¤±´n«ü{" e¡·‚ïÄA4 x§ ç§ x) ù«5Ÿ kõ « {" ´• 8 ¥ ¥•{¥÷v(x§ 1 A1 A1 « {" kA1 A 4 3 4 2 1 1 1 ¤± ÷v(x§ ç§ x§ x), Ú(ç§ x§ x§ x) {•´A1 4 A3 A4 A2 «"
(x§ x§ ç§ ç§ x) (x§ ç§ x§ ç§ x) (ç§ x§ ç§ x§ x) 2
(ç§ ç§ x§ x§ x) (ç§ x§ x§ ç§ x) (x§ ç§ ç§ x§ x)
Ù¢ù8«ŒU´ù 5 §• ˜‡´x¥Ø7+§"c>o‡ ˜¥Àü‡5–ç 2 «À{" ¥kC4 ¤±´6«ü{" e¡·‚ïÄA5 ¥•{¥÷v(x§ x§ ç§ ç§ x) ù«5Ÿ kõ « {" 8¥ 1 A1 A1 A1 « {" ´• kA1 A 4 3 4 3 2 1 1 1 1 ÷v, Ô«ŒU|Ü {•ˆ´´A1 ¤± 4 A3 A4 A3 A2 «"
k=1 n
x1 x2 · · · 1 1 ··· n n
x4
1 n
.
n
xk × x2 k×
k=1
1 1 = (x1 + x2 + ... + xn ); n n 1 1 2 = (x2 + x2 2 + ... + xn ); n n 1
EX 2 =
5
DX = EX 2 − (EX )2 =
1 2 2 (x + x2 2 + ... + xn ) − n 1
3 3
4 5 3 5 2 5 1 5
3
;
3
;
3
;
3
.
,˜•¡§ N´•
P (X = k ) = P (X ≥ k ) − P (X = k + 2).
¤±
P (X = 1) = 1 − P (X = 3) = P (X = 5) = P (X = 7) = 4 5 3 5 2 5
3
4 5 −
3
= 3 5 2 5 1 5
3
61 ; 125 = 37 ; 125 19 ; 125 7 ; 125
3
3
−
3
=
3
− 1 5
3
= 1 . 125
P (X = 9) =
=
,
·‚•
E (X ) = 1 × 61 37 19 7 1 +3× +5× +7× +9× = 2.6. 125 125 125 125 125
14. Suppose that X∼ Find EX and DX. Solution: EX =
1 (x1 + x2 + ... + xn ) n
2
.
17. Suppose that X is a nonnegative random variable with density function p(x). Show that √ Y = X is a continuous random variable with density function py (y ) = 2yp(y 2 ) y > 0 0 y ≤ 0.
11. Let R be the return rate of a program. It is a random variable with distribution density matrix 1% 2% 3% 4% 5% 6% . 0.1 0.1 0.2 0.3 0.2 0.1 If one man invests 100 thousands Yuan in this programme, find the expectation and variance of his return. Solution: Note that ER = (1% × 0.1 + 2% × 0.1 + 3% × 0.2 + 4% × 0.3 + 5% × 0.2 + 6% × 0.1) = 0.037. Suppose his return is X. One has that EX = E (100(1 + R)) = 100 + 100ER = 103.7. To calculate DX note that ER2 = (1%)2 × 0.1 + (2%)2 × 0.1 + (3%)2 × 0.2 + (4%)2 × 0.3 + (5%)2 × 0.2 + (6%)2 × 0.1 = 0.00157 implying that DR = ER2 − (ER)2 = 0.00157 − 0.0372 = 0.000201. Then one follows that DX = D(100(1 + R)) = 1002 D(1 + R) = 1002 DR = 2.01
4
2 k 3
, k = 0, 1, 2, 3. Find
C
k=0
2 3
k
= C (1 +
2 4 8 + + ) = 1, 3 9 27
implying that C =
27 65 .
So P (X = k ) = 27 65
2 3 27 65
2 3
k
(2)P (1 ≤ X ≤ 2) = P (X = 1) + P (X = 2) =
2 10 1 10
0
2 10
2
5 10
.
5. •¥k4‡ç¥4‡x¥§ ؘ£/ ¦§ 3‡x¥•Ž§ ¦ Ѧoê ©Ù" Solutionµ ò8‡¥?Ò§ {1, 2, 3, 4, 5, 6, 7, 8} Ù¥{1, 2, 3, 4} “Lx¥§ {5, 6, 7, 8} “Lç¥" Ñoê•ne¡©O¦P (n = 3), P (n = 4), P (n = 5), P (n = 6), P (n = 7) VÇ" (1) P (n = 3) =? l8‡¥¥Ø˜£!P ¥gS 3‡¦¤kŒU {kA3 8 . ¯‡{n = 3} ¤áL« 3 l{1, 2, 3, 4} Ñ3‡¥ŒU {§ kA4 «"¤± P (n = 3) = (2) P (n = 4) =? l8‡¥¥Ø˜£!P 1 4×3×3 A3 4 = . = 3 8×7×6 14 A8
13. l1, 3, 5, 7, 9 k˜£/ n‡ê§ ±X L«Ù¥• êi§ Á¦EX. Solution: ng Ñ5 êi©O´X1 , X2 , X3 . KX = min[X1 , X2 , X3 ]. u´·‚k P (X ≥ k ) = P (min[X1 , X2 , X3 ] ≥ k ) = P ({X1 ≥ k } ∩ {X2 ≥ k } ∩ {X3 ≥ k }).
27 65
+
2 3
2 2 3 2 2 3
=
6 13 . 6 13 .
Fra Baidu bibliotek
(3)P (0 < X < 2.5) = P (X = 1) + P (X = 2) =
+
=
10. Suppose that random variable X has the following density matrix −1 0
P (n = 4) = (3) P (n = 5) =? l8‡¥¥Ø˜£!P
1 1 1 3A1 3×4×3×4×2 6 4 A3 A4 A2 = = 8 × 7 × 6 × 5 35 A4 8
nog U"
¥gS 5‡¦¤kŒU {kA5 8 . {n = 5 } ¿ › X 1 ˜ Ñü‡x¥ü‡ç¥§• ˜‡¥ Ñx¥"l¥ ôÚþw§k±eA«Œ
1 8 1 4
2
3 8
3
1 4
.
3
Find EX, EX 2 E (−2X + 1). Solution: 1 1 3 1 11 EX = −1 × + 0 × + 2 × + 3 × = ; 8 4 8 4 8 1 1 3 1 31 EX 2 = (−1)2 × + 02 × + 22 × + 32 × = ; 8 4 8 4 8 11 7 E (−2X + 1) = −2EX + 1 = −2 × +1=− . 8 4
P (n = 5) = (4) P (n = 6) = 2 7. 3 . (5) P (n = 7) = 14
1 1 1 1 9 6 A1 6×4×3×4×3×2 4 A3 A4 A3 A2 = = 5 8×7×6×5×4 35 A8
¤±
Ño¥ê
©Ù—ÝÝ
3
1 14
•
4
6 35
5
9 35
6
2 7
7
3 14
∞ −∞
Ce−|t| dt = 2C
0
∞
Ce−t dt = 2C = 1
which implying that 1 C= . 2
4. Suppose that random variable X has the following distribution function, 0, x < −5 −5 ≤ x < 2 1 5 3 F (x) = 10 −2 ≤ x < 0 1 0≤x<2 2 1 x≥2 Find the probability distribution of X. Solution: It is easy to see that F (x) is a purely jumping function. Therefore X is a discrete type random variable. Since P (X = x) = F (x) − F (x−), one follows that P (x = −5) = P (x = −2) = 1 1 −0= , 5 5
Proof: Let Fy (x) be the distribution function of Y. Then √ Fy (x) = P (Y ≤ x) = P ( X ≤ x)
x2
= P (X ≤ x ) =
0 x
2
p(t)dt (let t = y 2 )
(1)
=
0
2yp(y 2 )dy.
Then we conclude that Y is a continuous random variable with density function py (y ) = 2yp(y 2 ) y > 0.
du´k˜£ ¥§ ¤±zg ¥´Õá § 8 VÇ uVÇ ¦È, …X1 , X2 , X3 ä kƒÓ ©Ù"¤±
P (X ≥ k ) = P ({X1 ≥ k })P ({X2 ≥ k })P ({X3 ≥ k }) = [P (X1 ≥ k )]3 . 4
u´·‚k
P (X ≥ 1) = [P (X1 ≥ 1)]3 = 13 = 1; P (X ≥ 3) = [P (X1 ≥ 3)]3 = P (X ≥ 5) = [P (X1 ≥ 5)]3 = P (X ≥ 7) = [P (X1 ≥ 7)] = P (X ≥ 9) = [P (X1 ≥ 9)] =
.
7. Suppose that random variable X has distribution P (X = k ) = C (1) the value of C ; (2) P (1 ≤ X ≤ 2); (3) P (0 < X < 2.5). Solution: (1) One knows that