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7 3 1,.,,.,, *1.,,..,.,,., ,,? *2.,.,..,,, *1,,.,. *2 A, B, N. A p, B q 1 p., A N m, B N n. A Q. Q = P! m+n 1 k k=m p m q k m. m
8 4 1,., : (P. Fermat) (B, Pascal) (J. Bernoulli) (de Moivre) (P.S. Laplace) ,,., ( ). n. A k, A P[A] = k/n., * ,,.. *4, : : n, 1 n 1. n, n 1 /n 1/6. *5 1/6 n, n 1/6..,,. 1.2,., *3, *4??,??. *5 1.
9 1.2 5, , ; X [0, 1],,, X [a, b] p. [ ] X [0, 1], (1.2.1) p = [a, b] [0, 1]. p = b a 1 = b a., (1.2.1)., X, [a, b]?, ( ). 1 C.., 3 q.,.?? [ 1],., x.! L C 2 3, L y 1/2. L, 2 (0, 1), (0, 1) 2 (0, 1/2), (0, 1/2).
10 6 1, q = 1/2. [ 2], O. O C x.! O L C 2 3, L x, π/3. L C 2 π/3., π/3 π.
11 1.3 7, q = 1/ , q = 1/2, q = 1/3., (1.3.1a) (1.3.1b) x, [ 1, 1], O, x [0, π],., 1.2.2,.,, (1.3.1a) (1.3.1b) 1, ,., (a) :, Ω. (b) :, F.,.
12 8 1 (c) : Ω, P. P. (Ω, F, P),., (i) (Ω, F, P),. F P, 2.1. (ii), Ω, (i), (iii) Ω, (c) ( Ω, P ),. 20, A. N. (iii) *6,. *6,.
13 9 2,,., (Ω, F), P,.. 2.1, (Ω, F, P) probability space.. (i) Ω n Ω = {w 1, w 2,, w n }, sample space *1, w k Ω A Ω sample. (ii) F Ω (= ) power sets. F = {, w 1,, w n, w 1 w 2,, w 1 w 2 w 3,, Ω} 2 n. (iii) P, A F P[A] ( ) probability measure. : (2.1.1a) (2.1.1b) 0 P[A] 1 : P[Ω] = 1, P[ ] = 0, A 1, A 2 F A 1 A 1 = P[A 1 A 2 ] = P[A 1 ] + P[A 2 ] (i), P,. *1,
14 10 2 (ii) (i), Ω., F. Ω,., F (Ω, P)., (Ω, P). (i) P[A c ] = 1 P[A], (ii) A B P[A] P[B] (iii) {A 1, A 2,, A m } : (2.1.2) i j A i, A j A i A j = P[A 1 A 2 A m ] = P[A 1 ] + P[A 2 ] + + P[A m ]. [ ] (i) Ω = A A c A A c =, (2.1.1a) (2.1.1b) 1 = P[Ω] = P[A A c ] = P[A] + P[A c ] P[A c ] = 1 P[A]. (ii) A B B = A ( A c B ). A ( A c B ) =, (2.1.1b) P[B] = P[A (A c B)] = P[A] + P[A c B]. (2.1.1a) P[A c B] 0 P[B] = P[A] + P[A c B] P[A]. (iii). m = 2 (2.1.1b) (iii). m = k (iii), m = k + 1. C A 1 A k. (2.1.2) C 1 A k+1 = (A 1 A k+1 ) (A 2 A k+1 ) (A k A k+1 ) = =, (2.1.1b) P[A 1 A k A k+1 ] = P[C A k+1 ] = P[C] + P[A k+1 ] m = k (iii) P[C] = P[A 1 ] + P[A 2 A k ]. (iii) m = k
15 (1 ). 1. w 1 { }, w 2 { } Ω = {w 1, w 2 }., p (0 < p < 1) P[w 1 ] = p, P[w 2 ] = 1 p (1 ). 1. w 1 {1 }, w 2 {2 },, w 6 {6 } 6, Ω = {w 1,, w 6 } *2.,, P P[w 1 ] = 1 6, P[w 2] = 1 6,, P[w 6] = (2 ). 2. w 1 {1 =, 2 = }, w 2 {1 =, 2 = }, w 3 {1 =, 2 = }, w 4 {1 =, 2 = }, 4, Ω = {w 1,, w 4 }, (2.2.1), p, 0 < p < 1,. P P[w 1 ] = p 2, P[w 2 ] = P[w 3 ] = p(1 p), P[w 4 ] = (1 p) 2 2.,,. 20 Kolmogorov, n, (= Kolmogorov ),.,. *2, F Ω F {, Ω, w 1,, w 6, w 1 w 2,, w 5 w 6, w 1 w 2 w 3, } 2 6 = 64.
16 (n ). (2.2.1) n. (2.2.2) w 1 {1 =, 2 =, 3 =,, n = }, w 2 {1 =, 2 =, 3 =,, n = }, w 3 {1 =, 2 =, 3 =,, n = },. w N {1 =, 2 =, 3 =,, n = }, N = 2 n, Ω = {w 1,, w N }. Ω, (2.2.2) ( ). 1, 0, (2.2.3) w Ω n w = (w (1), w (2),, w (n) ). w w (k) 1= 0., w 1 = 1 =, 2 =, 3 =,, n = w 1 = (1, 1, 1,, 1). (2.2.2) w 1 = (1, 1,, 1), w 2 = (0, 1, 1), w 3 = (1, 0, 1,, 1),, w N = (0, 0,, 0). (2.2.3) w Ω w w (1) + w (2) + + w (n) =, P (2.2.4) w Ω P[w] = p w (1 p) n w. 2.3 (Ω, P) P[B] 0 A, B, B, A conditional probability P[A/B], (2.3.1) P[A/B] P[A B] P[B] ( )
17 * A B independent (2.3.2) P[A B] = P[A] P[B]., A B A B. P[A/B] = P[A]. [ ]. (2.3.1) (2.3.2) P[A/B] = P[A B] P[B] = P[A] P[B] P[B]. (2.3.1) P[A B] P[B] = P[A/B] = P[A] = P[A].. P[B], (2.3.2). 3, (i) : 3 A, B, C, (2.3.3a) P[A B] = P[A] P[B], P[C A] = P[C] P[A], P[B C] = P[B] P[C], (2.3.3b) P[A B C] = P[A] P[B] P[C]. (ii) ( ) k A 1, A 2,, A k, A 1, A 2,, A k j A i1, A i2,, A ik, : P[A i1 A i2 A ik ] = P[A i1 ] P[A i2 ] P[A ik ] (i) A, B, C (2.3.3a) (2.3.3b). : 4 a, b, c, d, a 2, b 3, c 5, d 30., A, B, C A : B : C : 2, 3, 5, *3 T. Bayes. (2.3.1). P[A/B].
18 14 2. (2.3.3a), (2.3.3b). [ ] a, b, c, d 1/4, P[A B] = P[ d ] = 1 4 P[A] = P[ a d ] = 1 2, P[B] = P[ b d ] = 1 2, (2.3.3a).,.,, (2.3.3b). P[A B C] = P[ d ] = 1 4, P[A] P[B] P[C] = = 1 8 (ii), A, B, C (2.3.3b) (2.3.3a). ( ) 2.4, (2.3.1),,. P[A/H] H [P[H], P[A H] (2.3.1) (2.4.1) P[A H] = P[H] P[A/H] ( ). P[H] 0, P[A] 0 H, A, : (2.4.2) P[H/A] = P[H] P[A/H] P[H] P[A/H] + P[H c ] P[A/H c ]., P[H c ] = 0, P[H c ] P[A/H c ] = 0. [ ] (2.4.1). P[A H] + P[A H c ] = P[A], (2.4.2) = P[A H] P[A H] P[A H] + P[A H c = = (2.4.2). ] P[A]
19 ( 2). H 1,, H m H 1 H 2 H m = Ω; i j H i H j = k = 1,, m P[H k ] 0,. P[A] 0 A, : (2.4.3) P[H 1 /A] = P[H 1 ] P[A/H 1 ] P[H 1 ] P[A/H 1 ] + + P[H m ] P[A/H m ]. ( ) ( ) a 8 2, b a b?. [ ] A, B, W. A = a, W =. B = b, P[A/W ], P[B/W ]. (2.3.1) (2.4.4) P[A/W ] = P[A W ], P[B/W ] = P[B W ] P[W ] P[W ].,. 1/2, (2.4.1) P[A W ] = P[A] P[W/A] = = 8 20, P[B W ] = P[B] P[W/B] = = a b, W = ( A W ) ( B W ). ( A W ) ( B W ) =, (2.1.1b) P[W ] = P[(A W ) (B W )] = P[A W ] + P[B W ] = = (2.4.4), P[A/W ] = 8/20 13/20 = 8 5/ , P[B/W ] = 13 13/20 = , a ,.
20 (?).,.., 1 : 1. [ ] B, G., w 1 (B, B), w 2 (B, G), w 3 (G, B), w 4 (G, G), (2.4.5) Ω { w 1, w 2, w 3, w 4 }, P[wk ] = 1 4, k = 1,, 4. U = w 1 w 2 w 3, V = w 2 w 3 w 4. (2.4.5) P[ ] = P[V/U] = P[V U] P[U] = P[w 2 w 3 ] P[w 2 w 3 w 4 ] = 2/4 3/4 = ( ). 3 a, b, c,,. a b, c. (2.4.6a) (2.4.6b) a 1/3. b, c a 1/2.,. (2.4.7a) (2.4.7b) (2.4.7c) (2.4.7d) a, b, c 1/3, a, 1/2 b, c, b,,, (2.4.6b)? [ ] A,, D : A {a }, B {b }, C {c }, D { b } (2.4.6b) P[A/D] = 1 2..
21 ,, Ω (2.4.7a), (2.4.7b), (2.4.7c) (2.3.1) B C = Ω. P[A] = P[B] = P[C] = 1 3, P[D/A] = 1, P[D/C] = 1. 2 B C = Ω, (2.4.7d) P[D A] = P[D/A] P[A] = = 1 6, P[D C] = P[D/C] P[C] = = 1 3. P[D] = P[D A] + P[D B] + P[D C] = = 1 2 P[A/D] = P[D A] P[D] = 1/6 1/2 = 1 3, (2.4.6b). 1 %.,,., ( ).? [ ] Step 1.,. 2007, (2.4.8a) (2.4.8b) P[ / ] = 0.15, P[ / ] = 0.15,., A, B (2.4.9) A { }, B { }, (2.4.8a), (2.4.8b) (2.4.10) P[B c /A] = 0.15, P[B/A c ] = Step 2. (2.3.1) P[B/A] + P[B c /A] = = P[A] P[A] = 1. P[B A] P[A] + P[Bc A] P[A] = P[B A] + P[Bc A] P[A]
22 / (2.4.11) P[B/A] = 1 P[B c /A] = = Step 3., x (0 < x < 1): P[A] = x. Step 2, P[B]. (2.3.1), P[B A] 0.85 = P[B/A] = = P[A] 0.15 = P[B/A c ] = P[B Ac ] P[A c ] P[B A] x = P[B Ac ] 1 x P[B A] = 0.85 x. P[B A c ] = 0.15 (1 x). P[B] = P[B A] + P[B A c ] = 0.85 x (1 x) = x. Step 4. Q(x), Q(x) = P[A/B] =, : Q(x) = P[A/B] = P[A B] P[B] x, Q(x) : = 0.85 x x = 17x x. x 1% 2% 3% 4% 5% 10% 15% 20% 25% Q(x) 5.4% 10.4% 14.9% 19.1% 22.9% 38.6% 50% 58.6% 65.3%. : (i) x 4%, Q(x) 20%, /. 80 %. (ii) x 15%, Q(x) 50%, /., 50 %.
23 2.5 19,. ( Spam ) (Ham )., ( ).,. H, A H = { }, A = { },. (2.4.12) P[A/H] = 0.3, P[A/H c ] = 0.01, P[H c ] P[H] = 2. P[H/A]. [ ] P[H] P[A/H] P[H/A] = P[H] P[H/A] + P[H c ] P[A/H c ] P[A/H] = P[H/A] + ( P[H c ] / P[H] ) P[A/H c ]. P[H/A] = , (2.4.12), (Ω, P), A = {w 1, w 3, w 5 }, B 4 = {w 4, w 5, w 6 }.. (i) P[A B], (ii) P[A B], (iii) P[A] + P[B] P[A B] (Ω, P) (i) A w 1,, w 6. (ii) P[w 2 /A]. (iii) A w 2? (Ω, P) (i) A 1 w 1,, w 4.
24 20 2 (ii) B 2 w 1,, w 4. (iii) A B? ,. : 2, 3, 5 5 6, 10, , A, B, C A : B : C : 2, 3, 5,. (i) (2.3.3b). (ii) (2.3.3a) , 4. 10,. A { }, B {2 }, C {3 },,. (i) P[A B], (ii) P[B], (iii) P[B C], (iv) P[C] ( ). C,, (E 1 ) (E 2 ) (E 3 ) P[5 /E 1 ] = 0.2, 10 P[10 /E 3 ] = 0.1.,. P[E 1 ] = 0.2, P[E 2 ] = 0.3, P[E 3 ] = 0.5. C 10, P[E 3 /10 ],..
25 21 3,.,,.,. 3.1 (Ω, P) w Ω X : Ω R random variable X(w), Y (w), (3.1.1) X(w) a 1, a 2,, a n, Y (w) b 1, b 2,, b m. (i) (3.1.1) X(w), (3.1.2) p 1 P[X(w) = a 1 ], p 2 P[X(w) = a 2 ],, p n P[X(w) = a n ], (3.1.2) X(w) distribution of X(w). (ii), X(w) Y (w) independent : (3.1.3) 1 j n, 1 k m P[X(w) = a j, Y (w) = b k ] = P[X(w) = a j ] P[Y (w) = b k ]., : (3.1.2), : k = 1,, n p k 0; n p k = 1. k=1
26 (Ω, P) X(w) : { 1 (3.1.4) X(w) 1, P[X(w) = 1] = P[w 1 ] = p, P[X(w) = 1] = P[w 2 ] = 1 p (i) (Ω, P) X 1 (w), X 2 (w) : (3.1.5) { 1 1 X 1 (w) 1 1 { 1 2 X 2 (w) 1 2., P[X 1 (w) = 1] = P[w 1 w 3 ] = P[w 1 ] + P[w 3 ] = p 2 + p(1 p) = p, P[X 1 (w) = 1] = P[w 2 w 4 ] = P[w 2 ] + P[w 4 ] = p(1 p) + (1 p) 2 = 1 p, P[X 2 (w) = 1] = P[w 1 w 2 ] = P[w 1 ] + P[w 2 ] = p 2 + p(1 p) = p, P[X 2 (w) = 1] = P[w 3 w 4 ] = P[w 3 ] + P[w 4 ] = p(1 p) + (1 p) 2 = 1 p. (ii) P[X 1 (w) = 1, X 2 (w) = 1], P[X 1 (w) = 1, X 2 (w) = 1], P[X 1 (w) = 1, X 2 (w) = 1], P[X 1 (w) = 1, X 2 (w) = 1],, X 1 (w) X 2 (w) , X 1 (w) X 2 (w) (Ω, P) (3.1.6) Z(w) 100. (i) Z(w) (3.1.5) X 1 (w), X 2 (w). (ii). (a) P[Z(w) = 200], (b) P[Z(w) = 100], (c) P[Z(w) = 0].
27 , n (Ω, P), X k (w), k = 1, 2,, n,. { 1 k (3.1.7) X k (w) 1 k P[X k (w) = ±1] : (2.2.3)., X k (w) P[X k (w) = 1] = P[{w Ω : w (k) = 1}] = P[w] = p P[X k (w) = 1] = P[{w Ω : w (k) = 0}] = w Ω,w (k) =1 w Ω,w (k) =0 P[w] = 1 p. X 1, X 2,, X n ( ). (3.1.7) X k (w), k = 1, 2,, n, { (3.1.8) S k (w) 0 k = 0 X 1 (w) + + X k (w) 1 k n (n ). random walk,, (3.1.8) S k (w), k = 0, 1,, n,, : (3.1.9) P[S k (w) = j] = k C (k+j)/2 p (k+j)/2 (1 p) (k j)/2, k j k k + j (2 ). ( (3.1.9), 2 binomial distribution (i) a 1,, a n X(w), E[X(w)] a 1 P[X(w) = a 1 ] + a 2 P[X(w) = a 2 ] + + a n P[X(w) = a n ] X(w) mean *1,. *1, expectation.
28 X(w). (3.1.4) E[X(w)] = 1 P[X(w) = 1] + ( 1) P[X(w) = 1] = 1 p + ( 1) (1 p) = 2p G : G : 1, 100, G *2? [ ] (Ω, P), G w = w k Z(w) = k 100, k = 1, 2,, 6. 5 Kolmogorov n. G n, n E[Z(w)]. G E[Z(w)] = 100 P[Z(w) = 100] P[Z(w) = 600] = = 2100 = 350 6, /.,., X(w), Y (w), a, b. (i) E [ a X(w) + b Y (w)] = a E[X(w)] + b E[Y (w)]. (ii) a, E[a] = a. (iii) X(w) Y (w) E[X(w) Y (w)] = E[X(w)] E[Y (w)] ,., n n ,.? [ ] N. A:, =N. *2.
29 3.2 25, B: (i) N = k r, k r G 1, G 2,, G r. (ii) G 1 k, k, S 1. (iii) S 1 G 1, 1. S 1 G 1, 1 + k. (iv) G 2,, G r (ii)+(iii). Step 1. :,.. Case 1:, A =N B =k, A > B, B. Case 2:, A =N B =r + r k = r + N, A < B, A. Step 2. q, 0 < q < 1,, Case 1 2. G 1 Y 1 (w), { 1 S1 Y 1 (w) = 1 + k S 1, S 1, k S 1 = (1 q) k. S 1 = 1 (1 q) k. Y 1 (w) k, E[Y 1 (w)] = 1 (1 q) k + (1 + k) {1 (1 q) k } = 1 + k k(1 q) k. Step 3. G 2 Y 2 (w),, G r Y r (w). Y 1 (w),, Y r (w), R(q, k) B
30 26 3 R(q, k) = E[Y 1 (w)] + + E[Y r (w)] = r {1 + k k(1 q) k } = r + r k r k (1 q) k = N k + N N(1 q)k = N ( 1 k + 1 (1 q)k). B A R(q, k) = N = { (1 q)k} k, 1. Step 4. q = 1/100, (3.2.1) Q(k) (1 q)k k, k Q(k) , Q(11) = : q = 1/100, 2 k 30 B A. k = 11, B A 20%. Step 5.. [ ] (3.2.1) Q(k) 0 = d dk Q(k) = 1 k 2 (1 q)k log(1 q) k Q(k), k., q, / (1 q) k 1 k q, log(1 q) q
31 d dk Q(k), q2 0 = dq(k) dk = 1 k 2 (1 kq) ( q) 1 k 2 + q k = 1 q Q(k) = 1 k + 1 (1 q)k 1 k + 1 (1 k q) = 1 k + k q, Q(1/ q) 2 q. q = 1/100, k 1/ p. (i) k F (k, p). (ii) Y (w). [ ] Step 1. (3.2.2) q 1 p, 1 : P[Y (w) = 1] = p, 2 : P[Y (w) = 2] = q p,. k : P[Y (w) = k] = q k 1 p,., F (k, p) = p + qp + q 2 p + + q k 1 p = p 1 qk 1 q = 1 (1 p)k., p F (k, p) : p F (3, p) F (5, p) F (7, p) F (10, p) : 10,. 10%.. Step 2. Y (w). (3.2.2), : E[Y (w)] = 1 p + 2 q p + 3 q 2 p + + k q k 1 p + = p ( 1 + 2q + 3q kq k 1 + ) p = (1 q) 2 = 1 p. ( I II ): 1 + 2q + 3q kq k 1 = 1 (1 q) 2.
32 ,,. Z(w), 4. Z(w). [ ] (i) 1, a. (ii), a p = 3/4. 3/4. Y 1 (w). (iii) 3, 2/4. Y 2 (w). (iv) 4, 1/4. Y 3 (w). (v) 1 + Y 1 (w) + Y 2 (w) + Y 3 (w), , Y k (w), E[Z(w)] = E[1 + Y 1 (w) + + Y 3 (w)] = 1 + E[Y 1 (w)] + + E[Y 3 (w)] = / / /4 = 25 3.
33 29 4, X(w), m E[X(w)] a 1,, a n X(w), V[X(w)] E[ ( X(w) m ) 2 ] = n k=0 ( ak m ) 2 P[X(w) = ak ] X(w) variance X(w) Y (w). P[X(w) = 1] = 1 = P[X(w) = 1], 2 P[Y (w) = 100] = 1 = P[Y (w) = 100]. 2 E[X(w)] = = 0, E[Y (w)] = = 0, 0. V[X(w)] = E[ ( X(w) 0 ) 2 ] = ( 1)2 1 2 = 1 V[Y (w)] = E[ ( Y (w) 0 ) 2 ] = (100) ( 100)2 1 2 = X(w) 1, Y (w) 100, *1. *1,,.
34 ( ).. 2 A: 2,. B: 1,.. [ ] q, A X(w), B Y (w). Step 1. X(w), Y (w). E[X(w)] = 2 (1 q) + 0 q = 2(1 q), E[Y (w)] = 2 (1 q) (1 q) q + 1 q (1 q) + 0 q 2 = 2(1 q).. Step 2., , (iii) :. V[X(w)] = 2 2 (1 q) ( 2(1 q) ) 2 = 4q(1 q), V[Y (w)] = 2 2 (1 q) (1 q) q q (1 q) ( 2(1 q) ) 2 = 2q(1 q) = V[X(w)] 2, B, A 1/2, : (i) 1949,, A 4 AC 18. A. (ii) 1958,,, (iii) 1961, USA. 73., 1,, (=,.) ,. 2 X, Y. X = { 50 10, Y =
35 (i) X, Y 30. X Y? (ii) X Y?,., X(w), Y (w), c, c. (i) V[cX(w) + c ] = c 2 V[X(w)]. (ii) V[c] = 0. V[X(w)] = 0 X(w). (iii) V[X(w)] = E[ ( X(w) ) 2 ] ( E[X(w)] ) 2. (iv) X(w) Y (w) V[X(w) + Y (w)] = V[X(w)] + V[Y (w)] (3.1.5) X 1 (w) X 2 (w) X, Y, P[X = 2] = P[X = 2] = 1 2, P[Y = 3] = P[Y = 3] = 1 6, P[Y = 1] = P[Y = 1] = Y k (w), k = 1,, 4, ( ). [ ] E[Y (w)] = 1 p , (iii), V[Y (w)] = E[ ( Y (w) ) 2 ( ) 2 ] E[Y (w)] ( ) ( 1 2 = 1 2 p q p q 2 p + + k 2 q k 1 p + p) = p ( q q k 2 q k 1 + ) 1 p 2 = 4.1.6, (iv) p(1 + q) (1 q) 3 1 p 2 = q p 2 = 1 (1 p p 1). V[Z(w)] = V[1 + Y 1 (w) + + Y 4 (w)] = /4 ( 1 3/4 1) + 1 2/4 ( 1 2/4 1) + 1 1/4 ( 1 1/4 1) = 19.
36 32 4 Z(w), 25/3 4. : ( I II ): q q k 2 q k 1 + = 1 + q (1 q) ( ). X 1 (w), X 2 (w),, X n (w). X k (w), E[X k (w)] = m k, V[X k ] = c k, k = 1, 2,, n. E[X 1 (w) + X 2 (w) + + X n (w)] = m 1 + m 2 + m n, V[X 1 + X X n ] = c 1 + c c n ( ) ( ). :., j, 2 j., Y,, X(w) P[Y (w) = j] = 1, j = 1, 2,. 2j (4.1.1) E[X(w)] = 2 j P[Y (w) = j] = 1 = j=1., Q? j=1 [ ] Step 1. k X k. X 1, X 2,, (??), (?? ).. Step 2. N., X (N) (w) N E[X (N) (w)] = 2 j P[Y (w) = j] = 1 = N j=1 j=1
37 4.2 33, ( ) (??), N.. X (N) (w) V[X (N) (w)] ( 2] V[X (N) (w)] = E[ X (w)) (N) N 2 = N j=1 ( 2 j ) j = 2N 1 N 2., N V[X (N) (w)] N, X(w) Y (w), (3.1.3). X(w) Y (w) E[X(w) Y (w)] = E[X(w)] E[Y (w)]., X(w) Y (w), m X, m Y X(w), Y (w), (4.2.1) Cov[X, Y ] E [( X(w) m X ) ( Y (w) my )], ρ[x, Y ] Cov[X, Y ] V[X] V[Y ]. Cov[X, Y ] Covariance, ρ[x, Y ] correlation coefficient X(w), Y (w), Z(w), a, b. (i) Cov[X, X] = V[X]. (ii) Cov[aX + by, Z] = a Cov[X, Z] + b Cov[Y, Z]. (iii) V[X(w) + Y (w)] = V[X(w)] + 2 Cov[X, Y ] + V[Y (w)]. (iv) X(w) Y (w), Cov[X, Y ] = (i) X(w), Y (w), ρ[x, Y ] 1. (ii) X(w) Y (w), ρ[x, Y ] = 0. (, ρ[x, Y ] = 0 *2.) (iii), a, b, Y (w) = ax(w) + b, X(w) E[X(w)] V[X] = ρ[x, Y ] = 1., ρ[x, Y ] = 1 Y (w) E[Y (w)] V[Y ]. *
38 ( ). X(w), Y (w) P[X(w) = 1] = 1 4, P[X(w) = 0] = 1 2, P[X(w) = 1] = 1 4, Y (w) ( X(w) ) 2. P[X(w) = 1] P[Y (w) = 1] = = 1 8, P[X(w) = 1, Y (w) = 1] = P[X(w) = 1] = 1 4, P[X(w) = 1] P[Y (w) = 1] P[X(w) = 1, Y (w) = 1], X(w) Y (w), cf. (3.1.3)., E[X(w)] = 0, E[Y (w)] = 1/2, Cov[X, Y ] = E [ X(w) (Y (w) 1 )] ( ) 3] 1 = E[ X(w) 2 2 E[X(w)] = ( 1) = 0. V[X(w)] = 1/2, V[Y (w)] = 1/4, ρ[x, Y ] = 0 1/2 1/4 = 0, X(w) Y (w), A X(w) B Y (w)., V[X(w)] = 1, V[Y (w)] = 2, Cov[X, Y ] = 0.4. A x, B 1 x, x? [ ] A x, B 1 x F (x) , F (x) = V[xX(w) + (1 x) Y (w)] = V[x X(w)] + 2Cov[x X, (1 x) Y ] + V[(1 x) Y (w)] = x 2 V[X(w)] + 2x(1 x) Cov[X, Y ] + (1 x) 2 V[Y (w)] = x x(1 x) + 2 (1 x) 2 = 2.2 x x + 2 = 2.2 (x ) , x = 1.6/2.2 = 8/ X(w), Y (w) E[X(w)] = 0, V[X(w)] = 1, E[Y (w)] = 1, V[Y (w)] = 1, Cov[X, Y ] = 1.,.
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