Cálculo de Probabilidades II Respuestas Tema 2 1. Si Xi ∼ Ber(p) y Y = n X xi i=1 entonces, MY (t) = E(E ty ) = E(etx1 etx2 ...etxn ) n Y = E(etxi ) i=1 t = (e p + (1 − p)n ) Y ∼ Bin(n, p) 2. Sea X ∼ χ2k Y Y = Z12 + Z22 + ... + Zn2 → MY (t) = E(ety ) n Y 2 = etk(zi ) i=1 = (1 − 2t)nk/2 → Y ∼ χ2P ni 3. Sea Y = aX + b → MY t = E(ety ) = E(et(aX+b) ) = etb MX (at) 4. Sea f (x; µ, σ) = 1 √ xσ 2π 2 /2σ 2 e−(lnx−µ) E(X) = eµ+σ 2 2 /2 V ar(X) = e2µ+2σ − e2µ+σ 1 2 5. Sea X ∼ N(0, 1), p.d. Z = X 2 ∼ χ2(1) Dem. ∞ Z e−x MZ (t) = 2 (t− 1 ) 2 0 = 1 t− 1 √ dx 2π 1 2 → Z = X 2 ∼ χ2(1) 6. Sean Xi ∼ Ga(α, 1) n X Y = Xi i=1 (a) Con n = 2 Z ∞ fX1 (x1 )fX2 (y − x1 )dx1 fX1 ,X2 (y) = Z−∞ y fX1 (x1 )fX2 (y − x1 )dx1 = 0 Z = 0 y xα−1 e−x1 (y − x1 )α−1 e−(y−x1 ) 1 dx1 Γ(α) Γ(α) Sea x1 = yt dx1 = ydt y sustituyendo en la integral anterior, −y α+α−1 Z fX1 ,X2 (y) = e y 1 ytα−1 (1 − t)α−1 0 −y 2α−1 = e y ∼ Ga(2α, 1) Γ(2α) (b) Sea Y ∼ Ga(α, n) y W = 2nY. P (W ≤ w) = P (2nY ≤ y) w = P (Y ≤ ) 2n w = FY ( ) 2n 2 1 dt Γ(α)Γ(α) w 1 fY ( ) 2n 2n wα−1 −w/2 1 = e ∼ χ22α Γ(α) 2α → fW (w) = (c) Sea X ∼ Ga(α, 1) y Y = X n X ≤ y) n = P (X ≤ ny) = FX (ny) P (Y ≤ y) = P ( → fY (y) = n ∗ fX (ny) 1 ny α−1 α = e y n ∼ Ga(α, n) Γ(α) 7. Sea Y = n X ai X i i=1 P.D. n n X X Y ∼ N( ai µ i , (ai σi )2 ) i=1 i=1 Sea u i = ai x i MU (t) = E(etai xi ) = exp(tai µi + Sea Y = n X (tai xi )2 ) 2 ui i=1 Pn MY (t) = E(e i=1 tui ) n Y = ui E(etui ) i=1 n n X X (tai σi )2 = exp( tai µi + ) 2 i=1 i=1 3 v u n n X uX → Y ∼ N( ai µi , t (ai σi )2 ) i=1 i=1 8. Sea fX (x) y Y = aX + b FY = P (Y ≤ y) = P (aX + b ≤ y) y−b = P (X ≤ ) a y−b d ) → fY = F X ( dy a 9. Sea Y = ex X ∼ N (µ, σ 2 ) P (Y ≤ y) = P (ex ≤ y) = P (X ≤ ln(y)) = FX (ln(y)) ln(y) − µ = P (Z ≤ ) σ ln(y) − µ = Φ( ) σ Derivando obtenemos ϕY (y) 10. Sean (a) Y =( x−b β ) a y X ∼ W eibull(b, a, β) x−b β ) ≤ y) a 1 = P (X ≤ ay β + b) Z ay β1 +b β x β−1 −(x/a)β = ( ) e dx, a a 0 = 1 − e−y P (Y ≤ y) = P (( → Y ∼ Exp(1) 4 Seaβ = 1 yb = 0 (b) Sean Y =( x−b β ) a 1 yX = ay β + b P (X ≤ x) = P (ay 1/β + b ≤ x) x−b β = P (Y ≤ ( ) ) a x−b β ) ) = 1 − exp(( a β x − b β−1 x−b β ( ) exp(( ) ) a a a X ∼ W eibull(a, b, β) → fX = 11. Sean X ∼ CauchyEstandar W = 1 X 1 ≤ w) x 1 = P (X ≥ ) w 1 arctan w1 ) =1−( + 2 π P (W ≤ w) = P ( fW = 1 π(w2 + 1) W ∼ CauchyEstandar 12. Sean X ∼ U(-1, 1) W = |X| P (W ≤ w) = P (−w ≤ X ≤ w) = w → fW = 1, 0≤w≤1 SeaY = X 2 + 1 p p p P (Y ≤ y) = P (− y − 1 ≤ X ≤ y − 1) = y − 1 1 1 fY = ( √ ) 2 y−1 1 X +1 1 z−1 P (Z ≤ z) = P (X ≥ − 1) = z 2z SeaZ = 5 13. Sean X ∼ exp(1) Y = ln(X) y P (Y ≤ y) = P (X ≤ ey ) = 1 − e−e fY = exp(−ey + y) 14. Sean Y = eX X ∼ U (0, 1) P (Y ≤ y) = P (X ≤ ln(y)) = ln(y) 1 → fY = I[1,e] (y) y 15. Sea θ ∼ U( −π π , ) 2 2 R = Asen(θ) arcsen( ar ) r P (R ≤ r) = P (θ ≤ arcsen( )) = a π 1 1 → fR (r) = (p ) πA 1 − ( Ar )2 16. Sean X ∼ U (0, 1) Y = −ln(X) P (Y ≤ y) = P (X ≥ 1 1 )=1− y y e e → fY (y) = e−y → Y ∼ Exp(1) 17. Sea Long ∼ N (3.25, 0.052 ) L1 + L2 ∼ N (6.50, .005) 6.60 − 6.50 √ ) .005 = φ(1.4142) → P (L1 + L2 ≤ 6.60) = P (Z ≤ 18. Si X1 , X2 , X3 , ..., Xn son v.a.i.i.d. Yn = max(X1 , X2 , ..., Xn ) p.d.fYn (y) = n(FX (y))n−1 fX (y) 6 Dem. FY = P (Yn ≤ y) = P (max(X1 , X2 , ..., Xn ) ≤ y) n Y = P (Xi ≤ y) i=1 = (FX (y))n → fY (y) = n(FX (y))n−1 fX (y) 19. Si X ∼ U (0, 1) Y1 = minXi FY (y) = P (Y1 ≤ y) = P (minXi ≤ y) = 1 − (1 − FX (y))n 1 3 → P (Y1 ≤ ) = 1 − ( )n 4 4 20. Si X1 y X2 son v.a.i.i.d. N(0,1) y sea Y = (X2 − X1 )2 2 p.d. 1 1 Y ∼ Ga( , ) 2 2 MY (t) = E(eyt ) Z ∞Z ∞ (x2 − x1 )2 t 1 )exp(− (x21 x22 ))dx1 dx2 = exp( 2 2 0 0 1 1/2 1 = 1 2 para t < 2 −t 2 21. Si X1 , X2 v.a.i.i.d. N ormal(0, 1) Y1 = X12 + X22 Y2 = X2 1 1 −y1 /2 Iy22 ,∞ (y1 )I−∞,∞ (y2 ) f (y1 , y2 ) = p e 2 2π 2 y1 − y2 Y1 ∼ χ22 7 22. (a) Sea X1 , X2 v.a.i.i.d. Normal(0,1) 1 Y1 = (X12 + X22 ) 2 X2 Y2 = arctan( ) X1 2 y1 sec (y2 ) |J| = 1 + tan2 (y2 ) y1 −y12 /2 I< (y1 )I(0,π/2) (y2 ) e 2π → f (y1 , y2 ) = No son independientes. (b) Sea Xi ∼ U(0, 1) Y1 = X2 + X1 X1 Y2 − X2 y1 |J| = (1 − y2 )2 → f (y1 , y2 ) = y1 (1 − y2 )2 I(0,1) (y2 )I(0,1−y2 ) (y1 ) No son independientes. (c) Sea Xi ∼ Exp(1), i = 1, 2, 3 X1 X1 + X2 X 1 + X2 Y2 = X1 + X2 + X3 Y3 = X1 + X2 + X3 |J| = y2 y32 Y1 = → f (y1 , y2 , y3 ) = y2 y32 e−y3 I(0,1) (y1 , y2 , y3 ) Si son independientes. 23. Sea Xi ∼ N (µi , σi2 ) MW (t) = E(etw ) n Y = MXi (t) i=1 = prodni=1 expµi t + σi2 t2 X X → Y ∼ N( µi , σi2 ) 8 24. (a) Sea X ∼ U (0, 2) Y ∼ U (0, 1) Z =X +Y 1 1 fZ = z I(0,1) (z) + I(1,2) (z) 2 2 (b) X, Y ∼ N (0, 1) Y X W =X → |J| = W Z= w −w2 (z2 +1) e I(−∞,∞) (w, z) Z2π∞ w −w2 (z2 +1) fZ (z) = e dw I(−∞,∞) (z) −∞ 2π 1 = I< (z) π(1 + z 2 ) f(Z,W ) = (c) X, Y tal que fT (t) = btb−1 I(0,1) (t)I(0,∞) (b) Z =YX W =Y → |J| = 1 W b2 b−1 z I(0,1) (z)I(z,1) (w)I(0,∞) (b) w → fZ = b2 z b−1 ln(Z) I(0,1) (z)I(0,∞) (b) f(Z,W ) = 25. Sea f (xi ) = 2xi I(0,1) (x) Z = min(Xi ) FZ (z) = 1 − (1 − x2 )3 → fZ = 6x(1 − x2 )2 1 mediana = √ 2 √ 1 1 P (Z > √ ) = 1 − Fz (1/ 2) = (1 − 1/2)3 = 8 2 9 26. (a) Sean X1 , X2 , ..., Xn v.a.i.i.d. con FX (x) y la funcion de densidad de la j-esima estadistica de orden Yj esta dada por: f (yj ) = n! fX (y)(FX (y))j−1 (1 − FX (y))n−j (j − 1)!(n − j)! Para variables continuas. (b) Sea Y1 = min(Xi ) FY1 = P (Y1 ≤ y1 ) P (min(Xi ) ≤ y1 ) = 1 − P (min(Xi ) > y1 ) n Y =1− P (Xi > y1 ) i=1 = 1 − (1 − FX (y1 ))n → fY1 (y1 ) = n(1 − FX (y1 ))n−1 fX (y1 ) (c) Sea Yn = max(Xi ) FYn = P (Yn ≤ yn ) P (max(Xi ) ≤ yn ) n Y = P (Xi ≤ yn ) i=1 = (FX (yn ))n → fYn (yn ) = n(FX (y1 ))n−1 fX (yn ) 27. (a) Sea X1 , X2 , ..., Xn una muestra aleatoria, en donde √ √ fXi (xi ) ∼ U (µ − 3σ, µ + 3σ) Obtener el Rango muestral: Z = Yn − Y1 W = Y1 → |J| = 1 w 1 z + w n−1 ( √ ) I(µ−√3σ,µ+√3σ) (w)(I(0,2√3σ) (z) f(Z,W ) (z, w) = n2 (1 − √ )n−1 12σ 2 3σ 2 3σ Z µ−√3σ 1 z + w n−1 w fZ (z) = n2 (1 − √ )n−1 ( √ ) dw I(0,2√3σ) (z) √ 12σ 2 3σ 2 3σ µ− 3σ 10 Obtener el rango medio muestral: Y1 − Yn 2 W = Yn → |J| = 2 Z= 2z − w n−1 1 w √ ) ( √ )n−1 I(µ−√3σ,µ) (z)I(µ−√3σ,2z−µ+√3σ) (w) 12σ 2 3σ 2 3σ √ √ + I(2z−µ− 3σ,µ+ 3σ) (w)I(µ,µ+√3σ) (z) fZ,W (z, w) =2n2 (1 − Z → fZ (z) = √ 2z−µ+ 3σ √ µ− 3σ Z + 2n2 (1 − √ µ+ 3σ √ 2z−µ− 3σ w 2z − w n−1 1 √ ) ( √ )n−1 dw 12σ 2 3σ 2 3σ w 2z − w n−1 1 √ ) ( √ )n−1 dw 12σ 2 3σ 2 3σ 2n2 (1 − I(µ−√3σ,µ) (z) I(µ,µ+√3σ) (z) (b) Sean X1 , X2 , ..., Xn ∼ Exp(θ). Obtener el rando muestral: Z = Yn − Y1 W = Y1 → |J| = 1 fZ,W (z, w) = θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz I(0,∞) (z)I(z,∞) (w) Z ∞ fZ (z) = θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz dw I(0,∞) (z) z Obtener el Rango medio muestral: Y1 − Yn 2 W = Yn → |J| = 2 Z= fZ,W (z, w) = 2θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz I(0,∞) (z)I(0,2z) (w) Z 2z fZ (z) = 2θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz dw I(0,∞) (z) 0 28. Sea L ∼ N (µ, 1) (a) P (Ŷ > 8) = 1 − φ(4.5) 11 (b) P (6.2 ≤ Ŷ ≤ 6.8) = 2φ(0.9) − 1 (c) P (µ − 0.3 ≤ Ŷ ≤ µ + 0.3) = 0.95 √ → 2φ(0.3 n) − 1 = 0.95 → n = 42.6844 29. Sea N = 10, σ = 1 P (a ≤ S 2 ≤ b) = 0.90 n 1 X n−1 n−1 (xi − x̄)2 ∼ Ga( , ) n − 1 i=1 2 2σ 2 Z Z ( 92 )9/2 a 7 − 7x ( 29 )9/2 b 7 − 7x x 2 e 2 dx − x 2 e 2 dx → 0.90 = Γ( 92 ) 0 Γ( 29 ) 0 en donde S 2 = 30. Sea N1 = 6 N2 = 10 ambas con la misma varianza de poblacional. P( S12 ≤ b) = 0.90 S22 sabemos que S12 ∼ F (5, 9) S22 → b = 2.611 31. Sea P (X ≤ Q1 ) = 0.25, X ∼ N (0, 1) Qx1 = 1 − φ(0.68) Qx2 = 0 Qx3 = φ(0.68) Sea Y ∼ t − student(10) Qy1 = 1 − FX (.700) Qy2 = 0.129 Qy3 = 0.700 W ∼ χ2 (20) 12 Qw1 = 15425 Qw2 = 19337 Qw3 = 23828 32. Sean X ∼ U (0, 1) Y = ln( x ) 1−x x ≤ y) 1−x ey = P (x ≤ ) 1 + ey P (Y ≤ y) = P (ln( → fY (y) = ey (1 + ey )2 33. Sea Xi ∼ χ22 X 1 − X2 2 Z = X1 → |J| = 2 W = → fZ,W (z, w) = e−z−w I(0,∞) (w)I(2w,∞) (z) → fW (w) = e−3w I(0,∞) (w) 34. Sean X ∼ N (µ, σ 2 ) Y = x−µ ∼ N (0, 1) W = Y 2 σ 1 1 1 → fW (w) = √ w 2 e− 2w 2π 1 21 ( ) 1 1 = 2 1 w 2 e− 2w Γ( 2 ) √ 1 I(0,∞) (w) como π = Γ( ) 2 I(0,∞) (w) 1 1 → W ∼ Ga( , ) = χ21 2 2 35. Sea X ∼ Ga(α, 1) xi yi = Yk+1 Yk+1 = k+1 X i=1 13 xi sea x1 = y1 ∗ Yk+1 , x2 = y2 ∗ Yk+1 , ..., xk = yk ∗ Yk+1 Y xk+1 = (1 − k X yi ) ∗ yk+1 i=1 k k → |J| = |yk+1 − 2(yk+1 ∗ 1 k+1 → fY (y) = ( ) Γ(α) Sea ∞ Z k k ∗ −2(yk+1 (yk+1 X X yi )| X 2(α−1) yi ))∗e−yk+1 yk+1 (1− yi ) α−1 0 k Y i=1 k Y X X 1 k+1 α−1 γ=( yiα−1 ) (1 − 2 yi )(1 − yi ) Γ(α) i=1 Z ∞ 2(α−1) k yk+1 →γ e−yk+1 dyk+1 yk+1 = γ ∗ Γ(2α + k − 1) 0 k X X Y 1 k+1 α−1 → fY (y) = ( ) (1 − 2 yi )(1 − yi ) yiα−1 ∗ Γ(2α + k − 1) Γα i=1 Con K = 1 se tiene una Beta 36. (a) Sean SeaX1 , X2 , ..., Xn v.a.i.N (µ, σ) n 1 X 2 YS = (xi − x̄)2 n − 1 i=1 →M = X (xi − µ)2 (t) = M X (xi − x̄)2 (t) ∗ M (nx̄ − nµ)2 Si son independientes, P M (xi − µ)2 (t) →M (xi − x̄) (t) = M (nx̄ − nµ)2 P X xi − x̄ M ( xiσ−µ )2 (t) 2 →M ( ) (t) = σ )2 M ( nx̄−nµ σ X 2 En donde X xi − µ ( )2 (t) ∼ χ2n σ nx̄ − nµ 2 M( ) ∼ χ21 σ M 14 yiα−1 dyk+1 1 n ( 1 2−t ) 2 X xi − x̄ 2 →M ( ) (t) = 2 1 1 σ ( 1 2−t ) 2 2 = (1 2 1 2 −t ) n−1 2 Por Teorema de unicidad. X xi − x̄ (n − 1)S 2 ( )2 = σ σ2 ∼ χ2n−1 (b) Sabemos que si X ∼ N (0, 1) Y ∼ χ2n entonces X T = q ∼ t(n) Y n √ → x̄−µ ∼ N (0, 1) √σ n(x̄ − µ) n q 2 ∼ t(n − 1) =q χn−1 SX (n−1)S 2 ∼ σ 2 (n−1) n−1 37. Sea n = 3 (a) 1 P (Y1 > median) median = √ 2 Z P (Y1 > median) = √1 2 6(1 − y 2 )2 y1 dy1 0 = 0.125 (b) Sean Y1 Y2 Y2 Z2 = Y3 Z3 = Y3 → |J| = z2 z32 → f (z1 , z2 , z3 ) = z2 z32 fY1 (z1 z2 z3 )fY2 (z2 z3 )fY3 (z3 ) Z1 = Son independientes 15 38. Sea 12 x(x + y)I(0,1) (x)I(0,1) (y) 7 U = min(X, Y ) V = min(X, Y ) fX,Y (x, y) = FU,V = P (U ≤ u, V ≤ v) = P (V ≤ v) − P (U > u, V ≤ v) v Z Z P (V ≤ v) = =v 0 0 4 Z v Z v 12 x(x + y)dxdy 7 v 12 x(x + y)dydx u u 7 6 4 4 = v 4 + u4 − u2 v 2 − v 3 u − u3 v 7 7 7 → P (U > u, V ≤ v) = 1 fU,V (u, v) = (4v 3 u + 4u3 v − 7u4 + 6v 2 u2 ) I(0 < u < v < 1) 7 39. Si X1 , X2 , X3 , ..., Xn es una muestra aleatoria. E(Xi ) = µ V ar(Xi ) = σ 2 n X σ2 1 xi ) = V (x̄) = V ( n n i=1 n 1 X E(S ) = E( (xi − x̄)2 ) n − 1 i=1 2 n 1 X = ( V ar(xi ) − nV ar(x̄)) n − 1 i=1 → E(S 2 ) = σ 2 40. Sea Y = a + bx Cov(X, Y ) σY σX bV (X) = bσX σX =1 ρX,Y = 16 41. Si X1 , X2 , X3 , ..., Xn es una muestra con V(Xi ) = σ 2 . P.D. Cov(xi − x̄, x̄) = 0. Cov(xi − x̄, x̄) = Cov(xi , x̄) − V ar(x̄) σ2 σ2 = − n n =0 42. Sea X1 , X2 , X3 , ..., Xn una sucesion de v.a.i.i.d. con media µ y varianza σ 2 . (a) N m ∼ P o(γ) E(SN ) = E(E(N |SN )) nµ = γ V (SN ) = V (E(N |SN )) + E(V (N |SN )) n = 2 (nσ 2 + µ) γ (b) 1 1−p N ∼ Geo( ) γ = γ p E(SN ) = E(E(N |SN )) 1 = nµ γ V (SN ) = V (E(N |SN )) + E(V (N |SN )) 1 1 1 = (nσ )2 + (1 − )nµ γ γ γ 43. Sea X una v.a. con segundo momento finito. E((X − a)2 ) = E(X 2 ) − 2aE(X) + a2 + 2aE(X) − a2 = E(X 2 ) − E 2 (X) = V ar(X) 44. Sean Z ∼ N (0, 1) Y = a + bz + cz 2 Cov(Z, Y ) σZ σY b =√ 2 b + 2c2 → ρY,Z = 17 45. V( k X k k X X αi xi ) = E[( αi xi )2 ] − E 2 [ α i xi ] i=1 i=1 k X i=1 αi xi − = E[ i=1 k X = E[ i=1 k X = E[ n X αi µi ] i=1 αi (xi − µi )2 ] αi2 V ar(xi ) + 2 X i=1 αi αj cov(xi , xj )] i<j 46. Sean Z ∼ N (0, 1) Y = a + bz + cz 2 ρ= Cov(Z, Y ) σY σZ Cov(Z, Y ) = E(ZY ) − E(Z)E(Y ) = E(ZY ) = E(az + bz 2 + cz 3 ) = b + cE(Z 3 ) = b por independencia → σY = √ V ar(Z) = V ar(a + bz + cz 2 ) = b2 + 2c2 b2 + 2c2 b ρX,Y = √ 2 b + 2c2 47. Sea X ∼ N(0, 1) y sea I otra v.a.i. de X tal que P(I = 1) = P(I = 0) = 12 . Y sea tambien Y = X si I = 1 y Y = -X si I = 0. Y ∼ N (0, 1) c.p. 21 fY = Y ∼ N (0, 1) c.p. 12 1 −x2 e 2 ∼ N (0, 1) 2π Cov(X, Y ) = 0 → fY = 18 48. Sean f (Y |X) = Bin(n, x) X ∼ U (0, 1) n fX,Y (x, y) = (y )xy (1 − x)n−y n fX,Y (x, y) = (y ) Z I(0,1) (x)I0,1,2,...,n (y) 1 xy (1 − x)n−y dx I0,1,2,...,n (y) 0 Γ(y + 1)Γ(n − y + 1) = (y ) Γ(n + 2) n I0,1,2,...,n (y) * Kernel de una beta. 49. Sea ECM = E[(y − h(x))2 |X] P.D. minf ECM (f ) = E(Y |X = x) minfX E[(y − h(x))2 |X] = minc E[(y − c)2 |X] = minc [E(y 2 |X) − 2E(yc|X = x) + E(c2 |X = x)] → −2E(Y |X) + 2c = 0 → fX (x) = E(Y |X) 50. Sean u= f (y2 |y1 ) = y1 − µ1 σ1 v= y 2 − µ2 σ2 1√ −1 2 exp[ 1−ρ 2 (u − 2πσ1 σ2 1−ρ2 1 e−u2 /2 2πσ1 =√ 2ρuv + v 2 )] −1 v − ρu 2 1 p exp[ ( p )] 2 2πσ2 1 − ρ2 1 − ρ2 σ −1 y2 − (µ2 + ρ σ12 (y1 − µ1 )) p → f (y2 |y1 ) = √ exp[ ] 2 2πσ2 1 − ρ2 σ2 1 − ρ2 p σ2 → f (y2 |y1 ) ∼ N (µ2 + ρ (y1 − µ1 ), (σ2 1 − ρ2 )2 ) σ1 1 p 51. Sea X ∼ N (0, 1) 1 φX (t) = √ 2π Z 2 /2 = e−t ∞ eixt e−x Z−∞ ∞ −∞ −t2 /2 =e 19 2 /2 dx 1 2 √ e−(x−it) /2 dx 2π En general, sea W = µ + σx, W ∼ N (µ, σ 2 ) φW (t) = E(ei(µ∗σx)t ) = eiµt E(eitσx ) = eiµt φX (σt) = eiµt e−σ 52. fp|n = 1 Z 2 t2 /2 fp fn|p fn n+m ( n )pn (1 − p)m dp fn = 0 n+m =( n ) Γ(n + 1)Γ(m + 1) Γ(n + m + 2) fp|n = pn (1 − p)m Γ(n+1)Γ(m+1 Γ(n+m+2) 53. fX,Y = e−(x+y) I[0,∞) (x)I[0,∞) (y) Sea X Y W =Y ↔ X = WZ Z= ↔Y =W J= Z W 1 0 → |J| = W f(W,Z) = w ∗ f(X,Y ) (wz, w) = w ∗ e−w(z+1) I[0,∞) (w)I[0,∞) (z) Z fZ = ∞ we−w(z+1) dwI[0,∞) (z) 0 = 1 I[0,∞) (z) z+1 20 54. Sea X ∼ P o(θ), Y ∼ P o(λ), Z = X + Y ∼ P o(θ + λ). p.d. fX|Z ∼ Bin(Z, θ ) θ+λ fX,Z = f(X,Y ) (x, z − x) θx −(θ) λz−x −λ e I[0,1,2,..,z] (x)I[0,1,2,..) (y) = e x! (z − x)! fZ = fX,Z = fZ (λ + θ)z −(λ+θ) e I[0,1,2,..) (z) z! θx −(θ) λz−x −λ e e I(0,1,2,..,z) (x)I(0,1,2,..) (y) x! (z−x)! z (λ+θ) −(λ+θ) e I(0,1,2,..) (z) z! z! θx λz−x (λ + θ)−z I(0,1,2,..,z) (x) (x − z)!x! z θ x λ z−x = ( ) ( ) I[0,1,2,..,z) (x) x θ+λ θ+λ = f(X|Z) ∼ Bin(z, θ ) θ+λ 55. Sean X1 , X2 , X3 ,.., Xn v.a.i.i.d P (Xk = 1) = P (Xk = −1) = 1 2 N ∼ Geo(α) P (N = n) = α(1 − α)n I1,2,3,4,... (n) Y = n X xi xi ∼ Geo(α) i=1 → Y ∼ BinN eg(n, α) r+y−1 r → fY = α (1 − α)y I1,2,3,4,... (y) y 21 56. Sea (a) fS = k(s2 + 1) I(1,4) (s) (b) fT = ct I(0,600) (t) (c) Z 4 (s2 + 1)ds 1=k →k= 1 1 24 (d) Z 6 00tdt 1=c →c= 0 1 180000 (e) 100 c.p.0.11112, fX = 200 c.p.0.33336, 300 c.p.0.5556. (f) 1.5 c.p.0.1388, fY = 2.5 c.p.0.3055, 3.5 c.p.0.5555 (g) fY = x = 100 x = 100 x = 100 x = 200 x = 200 x = 200 x = 300 x = 300 x = 300 y y y y y y y y y = 1.5 = 2.5 = 3.5 = 1.5 = 2.5 = 3.5 = 1.5 = 2.5 = 3.5 c.p. 0.022224, c.p. 0.044448, c.p. 0.044448, c.p. 0.181963, c.p. 0.060654, c.p. 0.090981, c.p. 0.06486, c.p. 0.20039, c.p. 0.42007 57. Sean X1 , X2 , X3 , ...,Xn v.a.i.i.d. fXi = 2Xi I(0,1) (x) W = Yn − Y1 Z = Yn → |J| = 1 fW,Z = 4n2 (1 − (z − w)2 )n−1 (z − w)z 2n−1 I(0,1) (w)I(w,1) (z) Z 1 fW = 4n2 (1 − (z − w)2 )n−1 (z − w)z 2n−1 dz I(0,1) (x) w 22 Z 1 Z 1 4n2 (1 − (z − w)2 )n−1 (z − w)z 2n−1 dzdw E(W ) = w w 58. Sean 1 4 fX = I(−2,−1,1,2) (x) Y sea Y = X2 X/Y -2 -1 (a) 1 2 1 4 0 0.25 0.25 0 0.25 0 0 0.25 (b) CovXY = E(XY ) − E(X)E(Y ) X X XX yf (y) xf (x) xyf (x, y) − = x x y y =0 (c) ρXY = 0 (d) Pero no son independientes porque Y siempre va a depender de X 59. Sea X ∼ U (0, 1) Y Y = log P (Y ≤ log X 1−X ey X ) = P (X ≤ ) 1−X 1 + ey → fY = (1 + ey )ey − e2y (1 + ey )−2 I(−∞,∞) (Y ) 23