8.3: Problemas en Vectores Aleatorios y Distribuciones Conjuntas
- Page ID
- 150929
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Se seleccionan dos cartas al azar, sin reemplazo, de una baraja estándar. \(X\)Sea el número de ases y\(Y\) sea el número de espadas. Bajo los supuestos habituales, determinar la distribución conjunta y los marginales.
- Contestar
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\(X\)Sea el número de ases y\(Y\) sea el número de espadas. Definir los eventos\(AS_i\)\(A_i\),\(S_i\),, y\(N_i\),\(i = 1, 2\) de dibujar as de espadas, otro as, pala (que no sea el as), y ninguno en la selección i. Vamos\(P(i, k) = P(X = i, Y = k)\).
\(P(0, 0) = P(N_1N_2) = \dfrac{36}{52} \cdot \dfrac{35}{51} = \dfrac{1260}{2652}\)
\(P(0, 1) = P(N_1S_2 \bigvee S_1N_2) = \dfrac{36}{52} \cdot \dfrac{12}{51} + \dfrac{12}{52} \cdot \dfrac{36}{51} = \dfrac{864}{2652}\)
\(P(0, 2) = P(S_1 S_2) = \dfrac{12}{52} \cdot \dfrac{11}{51} = \dfrac{132}{2652}\)
\(P(1, 0) = P(A_N_2 \bigvee N_1 S_2) = \dfrac{3}{52} \cdot \dfrac{36}{51} + \dfrac{36}{52} \cdot \dfrac{3}{51} = \dfrac{216}{2652}\)
\(P(1, 1) = P(A_1S_2 \bigvee S_1A_2 \bigvee AS_1N_2 \bigvee N_1AS_2) = \dfrac{3}{52} \cdot \dfrac{12}{51} + \dfrac{12}{52} \cdot \dfrac{3}{51} + \dfrac{1}{52} \cdot \dfrac{36}{51} + \dfrac{36}{52} \cdot \dfrac{1}{51} = \dfrac{144}{2652}\)
\(P(1, 2) = P(AS_1S_2 \bigvee S_1AS_2) = \dfrac{1}{52} \cdot \dfrac{12}{51} + \dfrac{12}{52} \cdot \dfrac{1}{51} = \dfrac{24}{2652}\)
\(P(2, 0) = P(A_1A_2) = \dfrac{3}{52} \cdot \dfrac{2}{51} = \dfrac{6}{2652}\)
\(P(2, 1) = P(AS_1A_2 \bigvee A_1AS_2) = \dfrac{1}{52} \cdot \dfrac{3}{51} + \dfrac{3}{52} \cdot \dfrac{1}{51} = \dfrac{6}{2652}\)
\(P(2, 2) = P(\emptyset) = 0\)
% type npr08_01 % file npr08_01.m % Solution for Exercise 8.3.1. X = 0:2; Y = 0:2; Pn = [132 24 0; 864 144 6; 1260 216 6]; P = Pn/(52*51); disp('Data in Pn, P, X, Y') npr08_01 % Call for mfile Data in Pn, P, X, Y % Result PX = sum(P) PX = 0.8507 0.1448 0.0045 PY = fliplr(sum(P')) PY = 0.5588 0.3824 0.0588
Ejercicio\(\PageIndex{2}\)
Dos puestos para trabajos de campus están abiertos. Aplican dos estudiantes de segundo año, tres juniors y tres seniors. Se decide seleccionar dos al azar (cada par posible igualmente probable). \(X\)Sea el número de alumnos de segundo año y\(Y\) sea el número de juniors que sean seleccionados. Determinar la distribución conjunta para el par\(\{X, Y\}\) y a partir de esto determinar los marginales para cada uno.
- Contestar
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Dejen\(A_i, B_i, C_i\) ser los eventos de seleccionar a un segundo, junior, o senior, respectivamente, en el\(i\) th juicio. Dejar\(X\) ser el número de estudiantes de segundo año y\(Y\) ser el número de juniors seleccionados.
Set\(P(i, k) = P(X = i, Y = k)\)
\(P(0, 0) = P(C_1C_2) = \dfrac{3}{8} \cdot \dfrac{2}{7} = \dfrac{6}{56}\)
\(P(0, 1) = P(B_1C_2) + P(C_1B_2) = \dfrac{3}{8} \cdot \dfrac{3}{7} + \dfrac{3}{8} \cdot \dfrac{3}{7} = \dfrac{18}{56}\)
\(P(0, 2) = P(B_1B_2) = \dfrac{3}{8} \cdot \dfrac{2}{7} = \dfrac{6}{56}\)
\(P(1, 0) = P(A_1C_2) + P(C_1A_2) = \dfrac{2}{8} \cdot \dfrac{3}{7} + \dfrac{3}{8} \cdot \dfrac{2}{7} = \dfrac{12}{56}\)
\(P(1, 1) = P(A_1B_2) + P(B_1A_2) = \dfrac{2}{8} \cdot \dfrac{3}{7} + \dfrac{3}{8} \cdot \dfrac{2}{7} = \dfrac{12}{56}\)
\(P(2, 0) = P(A_1A_2) = \dfrac{2}{8} \cdot \dfrac{1}{7} = \dfrac{2}{56}\)
\(P(1, 2) = P(2, 1) = P(2, 2) = 0\)
\(PX =\)[30/56 24/56 2/56]\(PY =\) [20/56 30/56 6/56]
% file npr08_02.m % Solution for Exercise 8.3.2. X = 0:2; Y = 0:2; Pn = [6 0 0; 18 12 0; 6 12 2]; P = Pn/56; disp('Data are in X, Y,Pn, P') npr08_02 Data are in X, Y,Pn, P PX = sum(P) PX = 0.5357 0.4286 0.0357 PY = fliplr(sum(P')) PY = 0.3571 0.5357 0.1071
Ejercicio\(\PageIndex{3}\)
Se enrolla un dado. Deja\(X\) ser el número que aparece. Una moneda es volteada\(X\) veces. \(Y\)Sea el número de cabezas que aparecen. Determinar la distribución conjunta para el par\(\{X, Y\}\). Asumir\(P(X = k) = 1/6\) para\(1 \le k \le 6\) y para cada uno\(k\),\(P(Y = j|X = k)\) tiene la distribución binomial (\(k\), 1/2). Organizar la matriz conjunta como en el plano, con valores de\(Y\) aumento hacia arriba. Determinar la distribución marginal para\(Y\). (Para una forma basada en MATLAB de determinar la distribución conjunta, consulte el Ejemplo 14.1.7 de “Expectativa condicional, regresión”)
- Contestar
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\(P(X = i, Y = k) = P(X = i) P(Y = k|X = i) = (1/6) P(Y = k|X = i)\).
% file npr08_03.m % Solution for Exercise 8.3.3. X = 1:6; Y = 0:6; P0 = zeros(6,7); % Initialize for i = 1:6 % Calculate rows of Y probabilities P0(i,1:i+1) = (1/6)*ibinom(i,1/2,0:i); end P = rot90(P0); % Rotate to orient as on the plane PY = fliplr(sum(P')); % Reverse to put in normal order disp('Answers are in X, Y, P, PY') npr08_03 % Call for solution m-file Answers are in X, Y, P, PY disp(P) 0 0 0 0 0 0.0026 0 0 0 0 0.0052 0.0156 0 0 0 0.0104 0.0260 0.0391 0 0 0.0208 0.0417 0.0521 0.0521 0 0.0417 0.0625 0.0625 0.0521 0.0391 0.0833 0.0833 0.0625 0.0417 0.0260 0.0156 0.0833 0.0417 0.0208 0.0104 0.0052 0.0026 disp(PY) 0.1641 0.3125 0.2578 0.1667 0.0755 0.0208 0.0026
Ejercicio\(\PageIndex{4}\)
Como variación del Ejercicio 8.3.3. , Supongamos que se tira un par de dados en lugar de un solo dado. Determinar la distribución conjunta para el par\(\{X, Y\}\) y a partir de esto determinar la distribución marginal para\(Y\).
- Contestar
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% file npr08_04.m % Solution for Exercise 8.3.4. X = 2:12; Y = 0:12; PX = (1/36)*[1 2 3 4 5 6 5 4 3 2 1]; P0 = zeros(11,13); for i = 1:11 P0(i,1:i+2) = PX(i)*ibinom(i+1,1/2,0:i+1); end P = rot90(P0); PY = fliplr(sum(P')); disp('Answers are in X, Y, PY, P') npr08_04 Answers are in X, Y, PY, P disp(P) Columns 1 through 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0005 0 0 0 0 0 0.0013 0.0043 0 0 0 0 0.0022 0.0091 0.0152 0 0 0 0.0035 0.0130 0.0273 0.0304 0 0 0.0052 0.0174 0.0326 0.0456 0.0380 0 0.0069 0.0208 0.0347 0.0434 0.0456 0.0304 0.0069 0.0208 0.0312 0.0347 0.0326 0.0273 0.0152 0.0139 0.0208 0.0208 0.0174 0.0130 0.0091 0.0043 0.0069 0.0069 0.0052 0.0035 0.0022 0.0013 0.0005 Columns 8 through 11 0 0 0 0.0000 0 0 0.0000 0.0001 0 0.0001 0.0003 0.0004 0.0002 0.0008 0.0015 0.0015 0.0020 0.0037 0.0045 0.0034 0.0078 0.0098 0.0090 0.0054 0.0182 0.0171 0.0125 0.0063 0.0273 0.0205 0.0125 0.0054 0.0273 0.0171 0.0090 0.0034 0.0182 0.0098 0.0045 0.0015 0.0078 0.0037 0.0015 0.0004 0.0020 0.0008 0.0003 0.0001 0.0002 0.0001 0.0000 0.0000 disp(PY) Columns 1 through 7 0.0269 0.1025 0.1823 0.2158 0.1954 0.1400 0.0806 Columns 8 through 13 0.0375 0.0140 0.0040 0.0008 0.0001 0.0000
Ejercicio\(\PageIndex{5}\)
Supongamos que se tira un par de dados. Dejar\(X\) ser el número total de manchas que aparecen. Enrolle el par una\(X\) vez más. \(Y\)Sea el número de sietes que se lanzan en los\(X\) rollos. Determinar la distribución conjunta para el par\(\{X, Y\}\) y a partir de esto determinar la distribución marginal para\(Y\). ¿Cuál es la probabilidad de tres o más sietes?
- Contestar
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% file npr08_05.m % Data and basic calculations for Exercise 8.3.5. PX = (1/36)*[1 2 3 4 5 6 5 4 3 2 1]; X = 2:12; Y = 0:12; P0 = zeros(11,13); for i = 1:11 P0(i,1:i+2) = PX(i)*ibinom(i+1,1/6,0:i+1); end P = rot90(P0); PY = fliplr(sum(P')); disp('Answers are in X, Y, P, PY') npr08_05 Answers are in X, Y, P, PY disp(PY) Columns 1 through 7 0.3072 0.3660 0.2152 0.0828 0.0230 0.0048 0.0008 Columns 8 through 13 0.0001 0.0000 0.0000 0.0000 0.0000 0.0000
Ejercicio\(\PageIndex{6}\)
El par\(\{X, Y\}\) tiene la distribución conjunta (en m-file npr08_06.m):
\(X =\)[-2.3 -0.7 1.1 3.9 5.1]\(Y =\) = [1.3 2.5 4.1 5.3]
Determinar la distribución marginal y los valores de esquina para\(F_{XY}\). Determinar\(P(X + Y > 2)\) y\(P(X \ge Y)\).
- Contestar
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npr08_06 Data are in X, Y, P jcalc Enter JOINT PROBABILITIES (as on the plane) P Enter row matrix of VALUES of X X Enter row matrix of VALUES of Y Y Use array operations on matrices X, Y, PX, PY, t, u, and P disp([X;PX]') -2.3000 0.2300 -0.7000 0.1700 1.1000 0.2000 3.9000 0.2020 5.1000 0.1980 disp([Y;PY]') 1.3000 0.2980 2.5000 0.3020 4.1000 0.1900 5.3000 0.2100 jddbn Enter joint probability matrix (as on the plane) P To view joint distribution function, call for FXY disp(FXY) 0.2300 0.4000 0.6000 0.8020 1.0000 0.1817 0.3160 0.4740 0.6361 0.7900 0.1380 0.2400 0.3600 0.4860 0.6000 0.0667 0.1160 0.1740 0.2391 0.2980 P1 = total((t+u>2).*P) P1 = 0.7163 P2 = total((t>=u).*P) P2 = 0.2799
Ejercicio\(\PageIndex{7}\)
El par\(\{X, Y\}\) tiene la distribución conjunta (en m-file npr08_07.m):
\(P(X = i, Y = u)\)
t = | -3.1 | -0.5 | 1.2 | 2.4 | 3.7 | 4.9 |
u = 7.5 | 0.0090 | 0.0396 | 0.0594 | 0.0216 | 0.0440 | 0.0203 |
4.1 | 0.0495 | 0 | 0.1089 | 0.0528 | 0.0363 | 0.0231 |
-2.0 | 0.0405 | 0.1320 | 0.0891 | 0.0324 | 0.0297 | 0.0189 |
-3.8 | 0.0510 | 0.0484 | 0.0726 | 0.0132 | 0 | 0.0077 |
Determinar las distribuciones marginales y los valores de esquina para\(F_{XY}\). Determinar\(P(1 \le X \le 4, Y > 4)\) y\(P(|X - Y| \le 2)\).
- Contestar
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npr08_07 Data are in X, Y, P jcalc Enter JOINT PROBABILITIES (as on the plane) P Enter row matrix of VALUES of X X Enter row matrix of VALUES of Y Y Use array operations on matrices X, Y, PX, PY, t, u, and P disp([X;PX]') -3.1000 0.1500 -0.5000 0.2200 1.2000 0.3300 2.4000 0.1200 3.7000 0.1100 4.9000 0.0700 disp([Y;PY]') -3.8000 0.1929 -2.0000 0.3426 4.1000 0.2706 7.5000 0.1939 jddbn Enter joint probability matrix (as on the plane) P To view joint distribution function, call for FXY disp(FXY) 0.1500 0.3700 0.7000 0.8200 0.9300 1.0000 0.1410 0.3214 0.5920 0.6904 0.7564 0.8061 0.0915 0.2719 0.4336 0.4792 0.5089 0.5355 0.0510 0.0994 0.1720 0.1852 0.1852 0.1929 M = (1<=t)&(t<=4)&(u>4); P1 = total(M.*P) P1 = 0.3230 P2 = total((abs(t-u)<=2).*P) P2 = 0.3357
Ejercicio\(\PageIndex{8}\)
El par\(\{X, Y\}\) tiene la distribución conjunta (en m-file npr08_08.m):
\(P(X = t, Y = u)\)
t = | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17 | 19 |
u = 12 | 0.0156 | 0.0191 | 0.0081 | 0.0035 | 0.0091 | 0.0070 | 0.0098 | 0.0056 | 0.0091 | 0.0049 |
10 | 0.0064 | 0.0204 | 0.0108 | 0.0040 | 0.0054 | 0.0080 | 0.0112 | 0.0064 | 0.0104 | 0.0056 |
9 | 0.0196 | 0.0256 | 0.0126 | 0.0060 | 0.0156 | 0.0120 | 0.0168 | 0.0096 | 0.0056 | 0.0084 |
5 | 0.0112 | 0.0182 | 0.0108 | 0.0070 | 0.0182 | 0.0140 | 0.0196 | 0.0012 | 0.0182 | 0.0038 |
3 | 0.0060 | 0.0260 | 0.0162 | 0.0050 | 0.0160 | 0.0200 | 0.0280 | 0.0060 | 0.0160 | 0.0040 |
-1 | 0.0096 | 0.0056 | 0.0072 | 0.0060 | 0.0256 | 0.0120 | 0.0268 | 0.0096 | 0.0256 | 0.0084 |
-3 | 0.0044 | 0.0134 | 0.0180 | 0.0140 | 0.0234 | 0.0180 | 0.0252 | 0.0244 | 0.0234 | 0.0126 |
-5 | 0.0072 | 0.0017 | 0.0063 | 0.0045 | 0.0167 | 0.0090 | 0.0026 | 0.0172 | 0.0217 | 0.0223 |
Determinar las distribuciones marginales. Determinar\(F_{XY} (10, 6)\) y\(P(X > Y)\).
- Contestar
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npr08_08 Data are in X, Y, P jcalc - - - - - - - - - Use array operations on matrices X, Y, PX, PY, t, u, and P disp([X;PX]') 1.0000 0.0800 3.0000 0.1300 5.0000 0.0900 7.0000 0.0500 9.0000 0.1300 11.0000 0.1000 13.0000 0.1400 15.0000 0.0800 17.0000 0.1300 19.0000 0.0700 disp([Y;PY]') -5.0000 0.1092 -3.0000 0.1768 -1.0000 0.1364 3.0000 0.1432 5.0000 0.1222 9.0000 0.1318 10.0000 0.0886 12.0000 0.0918 F = total(((t<=10)&(u<=6)).*P) F = 0.2982 P = total((t>u).*P) P = 0.7390
Ejercicio\(\PageIndex{9}\)
Se conservaron datos sobre el efecto del tiempo de capacitación en el tiempo para realizar un trabajo en una línea de producción. \(X\)es la cantidad de entrenamiento, en horas, y\(Y\) es el tiempo para realizar la tarea, en minutos. Los datos son los siguientes (en m-file npr08_09.m):
\(P(X = t, Y = u)\)
t = | 1 | 1.5 | 2 | 2.5 | 3 |
u = 5 | 0.039 | 0.011 | 0.005 | 0.001 | 0.001 |
4 | 0.065 | 0.070 | 0.050 | 0.015 | 0.010 |
3 | 0.031 | 0.061 | 0.137 | 0.051 | 0.033 |
2 | 0.012 | 0.049 | 0.163 | 0.058 | 0.039 |
1 | 0.003 | 0.009 | 0.045 | 0.025 | 0.017 |
Determinar las distribuciones marginales. Determinar\(F_{XY}(2, 3)\) y\(P(Y/X \ge 1.25)\).
- Contestar
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npr08_09 Data are in X, Y, P jcalc - - - - - - - - - - - - Use array operations on matrices X, Y, PX, PY, t, u, and P disp([X;PX]') 1.0000 0.1500 1.5000 0.2000 2.0000 0.4000 2.5000 0.1500 3.0000 0.1000 disp([Y;PY]') 1.0000 0.0990 2.0000 0.3210 3.0000 0.3130 4.0000 0.2100 5.0000 0.0570 F = total(((t<=2)&(u<=3)).*P) F = 0.5100 P = total((u./t>=1.25).*P) P = 0.5570
Para las densidades articulares en los Ejercicios 10-22 a continuación
- Esbozar la región de definición y determinar analíticamente las funciones de densidad marginal\(f_X\) y\(f_Y\).
- Utilice una aproximación discreta para trazar la densidad marginal\(f_X\) y la función de distribución marginal\(F_X\).
- Calcular analíticamente las probabilidades indicadas.
- Determinar por aproximación discreta las probabilidades indicadas.
Ejercicio\(\PageIndex{10}\)
\(f_{XY}(t, u) = 1\)para\(0 \le t \le 1\),\(0 \le u \le 2(1 - t)\).
\(P(X > 1/2, Y > 1), P(0 \le X \le 1/2, Y > 1/2), P(Y \le X)\)
- Contestar
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Región es triángulo con vértices (0, 0), (1, 0), (0, 2).
\(f_{X} (t) = \int_{0}^{2(1-t)} du = 2(1 - t)\),\(0 \le t \le 1\)
\(f_{Y} (u) = \int_{0}^{1 - u/2} dt = 1 - u/2\),\(0 \le u \le 2\)
\(M1 = \{(t, u):t > 1/2, u> 1\}\)yace fuera del trianlge\(P((X, Y) \in M1) = 0\)
\(M2 = \{(t, u): 0 \le t \le 1/2, u > 1/2\}\)tiene área en el triángulo = 1/2
\(M3\)= la región en el triángulo bajo\(u = t\), que tiene área 1/3
tuappr Enter matrix [a b] of X-range endpoints [0 1] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 200 Enter number of Y approximation points 400 Enter expression for joint density (t<=1)&(u<=2*(1-t)) Use array operations on X, Y, PX, PY, t, u, and P fx = PX/dx; FX = cumsum(PX); plot(X,fx,X,FX) % Figure not reproduced M1 = (t>0.5)&(u>1); P1 = total(M1.*P) P1 = 0 % Theoretical = 0 M2 = (t<=0.5)&(u>0.5); P2 = total(M2.*P) P2 = 0.5000 % Theoretical = 1/2 P3 = total((u<=t).*P) P3 = 0.3350 % Theoretical = 1/3
Ejercicio\(\PageIndex{11}\)
\(f_{XY} (t, u) = 1/2\)en el cuadrado con vértices en (1, 0), (2, 1), (1, 2), (0, 1).
\(P(X > 1, Y > 1), P(X \le 1/2, 1 < Y), P(Y \le X)\)
- Contestar
-
La región está delimitada por líneas\(u = 1 + t\),\( u = 1 - t\),\(u = 3 - t\), y\(u = t - 1\)
\(f_X (t) = I_{[0,1]} (t) 0.5 \int_{1 - t}^{1 + t} du + I_{(1, 2]} (t) 0.5 \int_{t - 1}^{3 - t} du = I_{(1, 2]} (t) (2 - t) = f_Y(t)\)por simetría
\(M1 = \{(t, u): t > 1, u > 1\}\)tiene área en el trangle = 1/2, entonces\(PM1 = 1/4\)
\(M2 = \{(t, u): t \le 1/2, u > 1\}\)tiene área en el trangle = 1/8\), entonces\(PM2 = 1/16\)
\(M3 = \{(t, u): u \le t\}\)tiene área en el trangle = 1, entonces\(PM3 = 1/2\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 200 Enter number of Y approximation points 200 Enter expression for joint density 0.5*(u<=min(1+t,3-t))& ... (u>=max(1-t,t-1)) Use array operations on X, Y, PX, PY, t, u, and P fx = PX/dx; FX = cumsum(PX); plot(X,fx,X,FX) % Plot not shown M1 = (t>1)&(u>1); PM1 = total(M1.*P) PM1 = 0.2501 % Theoretical = 1/4 M2 = (t<=1/2)&(u>1); PM2 = total(M2.*P) PM2 = 0.0631 % Theoretical = 1/16 = 0.0625 M3 = u<=t; PM3 = total(M3.*P) PM3 = 0.5023 % Theoretical = 1/2
Ejercicio\(\PageIndex{12}\)
\(f_{XY} (t, u) = 4t(1 - u)\)para\(0 \le t \le 1\),\(0 \le u \le 1\).
\(P(1/2 < X < 3/4, Y > 1/2)\),\(P(X \le 1/2, Y > 1/2)\),\(P(Y \le X)\)
- Contestar
-
Región es la unidad cuadrada,
\(f_X (t) = \int_{0}^{1} 4t(1 - u) du = 2t\),\(0 \le t \le 1\)
\(f_Y(u) = \int_{0}^{1} 4t(1 - u) dt = 2(1 - u)\),\(0 \le u \le 1\)
\(P1 = \int_{1/2}^{3/4} \int_{1/2}^{1} 4t (1 - u) du dt = 5/64\)\(P2 = \int_{0}^{1/2} \int_{1/2}^{1} 4t(1 - u) dudt = 1/16\)
\(P3 = \int_{0}^{1} \int_{0}^{t} 4t(1 - u) du dt = 5/6\)
tuappr Enter matrix [a b] of X-range endpoints [0 1] Enter matrix [c d] of Y-range endpoints [0 1] Enter number of X approximation points 200 Enter number of Y approximation points 200 Enter expression for joint density 4*t.*(1 - u) Use array operations on X, Y, PX, PY, t, u, and P fx = PX/dx; FX = cumsum(PX); plot(X,fx,X,FX) % Plot not shown M1 = (1/2<t)&(t<3/4)&(u>1/2); P1 = total(M1.*P) P1 = 0.0781 % Theoretical = 5/64 = 0.0781 M2 = (t<=1/2)&(u>1/2); P2 = total(M2.*P) P2 = 0.0625 % Theoretical = 1/16 = 0.0625 M3 = (u<=t); P3 = total(M3.*P) P3 = 0.8350 % Theoretical = 5/6 = 0.8333
Ejercicio\(\PageIndex{13}\)
\(f_{XY} (t, u) = \dfrac{1}{8} (t + u)\)para\(0 \le t \le 2\),\(0 \le u \le 2\).
\(P(X > 1/2, Y > 1/2), P(0 \le X \le 1, Y > 1), P(Y \le X)\)
- Contestar
-
Región es la plaza\(0 \le t \le 2\),\(0 \le u \le 2\)
\(f_X (t) = \dfrac{1}{8} \int_{0}^{2} (t + u) = \dfrac{1}{4} ( t + 1) = f_Y(t)\),\(0 \le t \le 2\)
\(P1 = \int_{1/2}^{2} \int_{1/2}^{2} (t + u) dudt = 45/64\)\(P2 = \int_{0}^{1} \int_{1}^{2} (t + u) du dt = 1/4\)
\(P3 = \int_{0}^{2} \int_{0}^{1} (t + u) dudt = 1/2\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 200 Enter number of Y approximation points 200 Enter expression for joint density (1/8)*(t+u) Use array operations on X, Y, PX, PY, t, u, and P fx = PX/dx; FX = cumsum(PX); plot(X,fx,X,FX) M1 = (t>1/2)&(u>1/2); P1 = total(M1.*P) P1 = 0.7031 % Theoretical = 45/64 = 0.7031 M2 = (t<=1)&(u>1); P2 = total(M2.*P) P2 = 0.2500 % Theoretical = 1/4 M3 = u<=t; P3 = total(M3.*P) P3 = 0.5025 % Theoretical = 1/2
Ejercicio\(\PageIndex{14}\)
\(f_{XY}(t, u) = 4ue^{-2t}\)para\(0 \le t, 0 \le u \le 1\)
\(P(X \le 1, Y > 1), P(X > 0, 1/2 < Y < 3/4), P(X < Y)\)
- Contestar
-
La región es despojada\(t = 0, u = 0, u = 1\)
\(f_X(t) = 2e^{-2t}\),\(0 \le t\),\(f_Y(u) = 2u\),\(0 \le u \le 1\),\(f_{XY} = f_X f_Y\)
\(P1 = 0\),\(P2 = \int_{0.5}^{\infty} 2e^{-2t} dt \int_{1/2}^{3/4} 2udu = e^{-1} 5/16\)
\(P3 = 4 \int_{0}^{1} \int_{t}^{1} ue^{-2t} dudt = \dfrac{3}{2} e^{-2} + \dfrac{1}{2} = 0.7030\)
tuappr Enter matrix [a b] of X-range endpoints [0 3] Enter matrix [c d] of Y-range endpoints [0 1] Enter number of X approximation points 400 Enter number of Y approximation points 200 Enter expression for joint density 4*u.*exp(-2*t) Use array operations on X, Y, PX, PY, t, u, and P M2 = (t > 0.5)&(u > 0.5)&(u<3/4); p2 = total(M2.*P) p2 = 0.1139 % Theoretical = (5/16)exp(-1) = 0.1150 p3 = total((t<u).*P) p3 = 0.7047 % Theoretical = 0.7030
Ejercicio\(\PageIndex{15}\)
\(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\)para\(0 \le t \le 2\),\(0 \le u \le 1 + t\).
\(F_{XY} (1, 1)\),\(P(X \le 1, Y > 1)\),\(P(|X - Y| < 1)\)
- Contestar
-
Región delimitada por\(t = 0\)\(t = 2\),\(u = 0\),\(u = 1 + t\)
\(f_X (t) = \dfrac{3}{88} \int_{0}^{1 + t} (2t + 3u^2) du = \dfrac{3}{88}(1 + t)(1 + 4t + t^2) = \dfrac{3}{88} ( 1 + 5t + 5t^2 + t^3)\),\(0 \le t \le 2\)
\(f_Y(u) = I_{[0,1]} (u) \dfrac{3}{88} \int_{0}^{2} (2t + 3u^2) dt + I_{(1, 3]} (u) \dfrac{3}{88} \int_{u - 1}^{2} (2t + 3u^2) dt = \)
\(I_{[0,1]} (u) \dfrac{3}{88} (6u^2 + 4) + I_{(1,3]} (t) \dfrac{3}{88} (3 + 2u + 8u^2 - 3u^3)\)
\(F_{XY}(1, 1) = \int_{0}^{1} \int_{0}^{1} f_{XY} (t, u) dudt = 3/44\)
\(P1 = \int_{0}^{1} \int_{1}^{1 + t} f_{XY} (t, u)dudt = 41/352\)\(P2 = \int_{0}^{1} \int_{1}^{1 + t} f_{XY} (t, u) dudt = 329/352\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 3] Enter number of X approximation points 200 Enter number of Y approximation points 300 Enter expression for joint density (3/88)*(2*t+3*u.^2).*(u<=1+t) Use array operations on X, Y, PX, PY, t, u, and P fx = PX/dx; FX = cumsum(PX); plot(X,fx,X,FX) MF = (t<=1)&(u<=1); F = total(MF.*P) F = 0.0681 % Theoretical = 3/44 = 0.0682 M1 = (t<=1)&(u>1); P1 = total(M1.*P) P1 = 0.1172 % Theoretical = 41/352 = 0.1165 M2 = abs(t-u)<1; P2 = total(M2.*P) P2 = 0.9297 % Theoretical = 329/352 = 0.9347
Ejercicio\(\PageIndex{16}\)
\(f_{XY} (t, u) = 12t^2u\)en el paralelogramo con vértices (-1, 0), (0, 0), (1, 1), (0, 1).
\(P(X \le 1/2, Y > 0), P(X < 1/2, Y \le 1/2), P(Y \ge 1/2)\)
- Contestar
-
Región delimitada por\(u = 0\)\(u = t\),\(u = 1\),\(u = t + 1\)
\(f_X (t) = I_{[-1, 0]} (t) 12 \int_{0}^{t + 1} t^2 u du + I_{(0, 1]} (t) 12 \int_{t}^{1} t^2 u du = I_{[-1, 0]} (t) 6t^2 (t + 1)^2 + I_{(0, 1]}(t) 6t^2(1 - t^2)\)
\(f_Y(u) = 12\int_{u - 1}^{t} t^2 udu + 12u^3 - 12u^2 + 4u\),\(0 \le u \le 1\)
\(P1 = 1 - 12 \int_{1/2}^{1} \int_{t}^{1} t^2 ududt = 33/80\),\(P2 = 12 \int_{0}^{1/2} \int_{u - 1}^{u} t^2 udtdu = 3/16\)
\(P3 = 1 - P2 = 13/16\)
tuappr Enter matrix [a b] of X-range endpoints [-1 1] Enter matrix [c d] of Y-range endpoints [0 1] Enter number of X approximation points 400 Enter number of Y approximation points 200 Enter expression for joint density 12*u.*t.^2.*((u<=t+1)&(u>=t)) Use array operations on X, Y, PX, PY, t, u, and P p1 = total((t<=1/2).*P) p1 = 0.4098 % Theoretical = 33/80 = 0.4125 M2 = (t<1/2)&(u<=1/2); p2 = total(M2.*P) p2 = 0.1856 % Theoretical = 3/16 = 0.1875 P3 = total((u>=1/2).*P) P3 = 0.8144 % Theoretical = 13/16 = 0.8125
Ejercicio\(\PageIndex{17}\)
\(f_{XY} (t, u) = \dfrac{24}{11} tu\)para\(0 \le t \le 2\),\(0 \le u \le \text{min}\ \{1, 2 - t\}\)
\(P(X \le 1, Y \le 1), P(X > 1), P(X < Y)\)
- Contestar
-
La región está delimitada por\(t = 0, u = 0, u = 2, u = 2 - t\)
\(f_X (t) = I_{[0, 1]} (t) \dfrac{24}{11} \int_{0}^{1} tudu + I_{(1, 2]} (t) \dfrac{24}{11} \int_{0}^{2 - t} tudu =\)
\(I_{[0, 1]} (t) \dfrac{12}{11} t + I_{(1, 2]} (t) \dfrac{12}{11} t(2 - t)^2\)
\(f_Y (u) = \dfrac{24}{11} \int_{0}^{2 - u} tudt = \dfrac{12}{11} u(u - 2)^2\),\(0 \le u \le 1\)
\(P1 = \dfrac{24}{11} \int_{0}^{1} \int_{0}^{1} tududt = 6/11\)\(P2 = \dfrac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} tududt = 5/11\)
\(P3 = \dfrac{24}{11} \int_{0}^{1} \int_{t}^{1} tududt = 3/11\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 1] Enter number of X approximation points 400 Enter number of Y approximation points 200 Enter expression for joint density (24/11)*t.*u.*(u<=2-t) Use array operations on X, Y, PX, PY, t, u, and P M1 = (t<=1)&(u<=1); P1 = total(M1.*P) P1 = 0.5447 % Theoretical = 6/11 = 0.5455 P2 = total((t>1).*P) P2 = 0.4553 % Theoretical = 5/11 = 0.4545 P3 = total((t<u).*P) P3 = 0.2705 % Theoretical = 3/11 = 0.2727
Ejercicio\(\PageIndex{18}\)
\(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\)para\(0 \le t \le 2\),\(0 \le u \le \text{max}\ \{2 - t, t\}\)
\(P(X \ge 1, Y \ge 1), P(Y \le 1), P(Y \le X)\)
- Contestar
-
La región está delimitada por\(t = 0, t = 2, u = 0, u = 2 - t\)\((0 \le t \le 1)\),\(u = t (1 < t \le 2)\)
\(f_X(t) = I_{[0,1]} (t) \dfrac{3}{23} \int_{0}^{2 - t} (t + 2u) du + I_{(1, 2]} (t) \dfrac{3}{23} \int_{0}^{t} (t + 2u) du = I_{[0, 1]} (t) \dfrac{6}{23} (2 - t) + I_{(1, 2]} (t) \dfrac{6}{23}t^2\)
\(f_Y(u) = I_{[0, 1]} (u) \dfrac{3}{23} \int_{0}^{2} (t + 2u) du + I_{(1, 2]} (u) [\dfrac{3}{23} \int_{0}^{2 - u} (t + 2u) dt + \dfrac{3}{23} \int_{u}^{2} (t + 2u) dt]=\)
\(I_{[0,1]} (u) \dfrac{6}{23} (2u + 1) + I_{(1, 2]} (u) \dfrac{3}{23} (4 + 6u - 4u^2)\)
\(P1 = \dfrac{3}{23} \int_{1}^{2} \int_{1}^{t} (t + 2u) du dt = 13/46\),\(P2 = \dfrac{3}{23} \int_{0}^{2} \int_{0}^{1} (t + 2u) du dt = 12/23\)
\(P3 = \dfrac{3}{23} \int_{0}^{2} \int_{0}^{t} (t + 2u) dudt = 16/23\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 200 Enter number of Y approximation points 200 Enter expression for joint density (3/23)*(t+2*u).*(u<=max(2-t,t)) Use array operations on X, Y, PX, PY, t, u, and P M1 = (t>=1)&(u>=1); P1 = total(M1.*P) P1 = 0.2841 13/46 % Theoretical = 13/46 = 0.2826 P2 = total((u<=1).*P) P2 = 0.5190 % Theoretical = 12/23 = 0.5217 P3 = total((u<=t).*P) P3 = 0.6959 % Theoretical = 16/23 = 0.6957
Ejercicio\(\PageIndex{19}\)
\(f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)\), para\(0 \le t \le 2\),\(0 \le u \le \text{min } \{1 + t, 2\}\)
\(P(X \ge 1, Y \ge 1), P(X \le 1, Y \le 1), P(Y < X)\)
- Contestar
-
La región tiene dos partes: (1)\(0 \le t \le 1, 0 \le u \le 2\) (2)\(1 < t \le 2, 0 \le u \le 3 - t\)
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{179} \int_{0}^{2} (3t^2 + u) du + I_{(1, 2]} (t) \dfrac{12}{179} \int_{0}^{3 - t} (3t^2 + u) du =\)
\(I_{[0, 1]} (t) \dfrac{24}{179} (3t^2 + 1) + I_{(1, 2]} (t) \dfrac{6}{179} (9 - 6t + 19t^2 - 6t^3)\)
\(f_Y(u) = I_{[0, 1]} (u) \dfrac{12}{179} \int_{0}^{2}(3t^2 + u) dt + I_{(1, 2]} (u) \dfrac{12}{179} \int_{0}^{3 - u} (3t^2 + u) dt =\)
\(I_{[0, 1]} (u) \dfrac{24}{179} (4 + u) + I_{(1, 2]} (u) \dfrac{12}{179} (27 - 24u + 8u^2 - u^3)\)
\(P1 = \dfrac{12}{179} \int{1}^{2} \int_{1}^{3 - t} (3t^2 + u) du dt = 41/179\)\(P2 = \dfrac{12}{179} \int_{0}^{1} \int_{0}^{1} (3t^2 + u) dudt = 18/179\)
\(P3 = \dfrac{12}{179} \int_{0}^{3/2} \int_{0}^{t} (3t^2 + u) dudt + \dfrac{12}{179} \int_{3/2}^{2} \int_{0}^{3 - t} (3t^2 + u) dudt = 1001/1432\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 200 Enter number of Y approximation points 200 Enter expression for joint density (12/179)*(3*t.^2+u).* ... (u<=min(2,3-t)) Use array operations on X, Y, PX, PY, t, u, and P fx = PX/dx; FX = cumsum(PX); plot(X,fx,X,FX) M1 = (t>=1)&(u>=1); P1 = total(M1.*P) P1 = 2312 % Theoretical = 41/179 = 0.2291 M2 = (t<=1)&(u<=1); P2 = total(M2.*P) P2 = 0.1003 % Theoretical = 18/179 = 0.1006 M3 = u<=min(t,3-t); P3 = total(M3.*P) P3 = 0.7003 % Theoretical = 1001/1432 = 0.6990
Ejercicio\(\PageIndex{20}\)
\(f_{XY} (t, u) = \dfrac{12}{227} (3t + 2tu)\)para\(0 \le t \le 2\),\(0 \le u \le \text{min} \{1 + t, 2\}\)
\(P(X \le 1/2, Y \le 3/2), P(X \le 1.5, Y > 1), P(Y < X)\)
- Contestar
-
La región se divide en dos partes:
- \(0 \le t \le 1\),\(0 \le u \le 1 + t\)
- \(1 < t \le 2\),\(0 \le u \le 2\)
\(f_X(t) = I_{[0,1]} (t) \int_{0}^{1+t} f_{XY} (t, u) du + I_{(1, 2]} (t) \int_{0}^{2} f_{XY} (t, u) du =\)
\(I_{[0, 1]} (t) \dfrac{12}{227} (t^3 + 5t^2 + 4t) + I_{(1, 2]} (t) \dfrac{120}{227} t\)
\(f_Y(u) = I_{[0, 1]} (u) \int_{0}^{2} f_{XY} (t, u) dt + I_{(1, 2]} (u) \int_{u - 1}^{2} f_{XY} (t, u) dt = \)
\(I_{[0, 1]} (u) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (2u + 3) (3 + 2u - u^2)\)
\(= I_{[0, 1]} (u) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (9 + 12 u + u^2 - 2u^3)\)
\(P1 = \dfrac{12}{227} \int_{0}^{1/2} \int_{0}^{1 + t} (3t + 2tu) du dt = 139/3632\)
\(P2 = \dfrac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} (3t + 2tu) dudt + \dfrac{12}{227} \int_{1}^{3/2} \int_{1}^{2} (3t + 2tu) du dt = 68/227\)
\(P3 = \dfrac{12}{227} \int_{0}^{2} \int_{1}^{t} (3t + 2tu) dudt = 144/227\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 200 Enter number of Y approximation points 200 Enter expression for joint density (12/227)*(3*t+2*t.*u).* ... (u<=min(1+t,2)) Use array operations on X, Y, PX, PY, t, u, and P M1 = (t<=1/2)&(u<=3/2); P1 = total(M1.*P) P1 = 0.0384 % Theoretical = 139/3632 = 0.0383 M2 = (t<=3/2)&(u>1); P2 = total(M2.*P) P2 = 0.3001 % Theoretical = 68/227 = 0.2996 M3 = u<t; P3 = total(M3.*P) P3 = 0.6308 % Theoretical = 144/227 = 0.6344
Ejercicio\(\PageIndex{21}\)
\(f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)\)para\(0 \le t \le 2\),\(0 \le u \le \text{min}\ \{2t, 3 - t\}\)
\(P(X < 1), P(X \ge 1, Y \le 1), P(Y \le X/2)\)
- Contestar
-
Región delimitada por\(t = 2, u = 2t\)\((0 \le t \le 1)\),\(3 - t\)\((1 \le t \le 2)\)
\(f_X(t) = I_{[0, 1]} (t) \dfrac{2}{13} \int_{0}^{2t} (t + 2u) du + I_{(1, 2]} (t) \dfrac{2}{13} \int_{0}^{3 - t} (t + 2u) du = I_{[0, 1]} (t) \dfrac{12}{13} t^2 + I_{(1, 2]} (t) \dfrac{6}{13} (3 - t)\)
\(f_Y (u) = I_{[0, 1]} (u) \dfrac{2}{13} \int_{u/2}^{2} (t + 2u) dt + I_{(1, 2]} (u) \dfrac{2}{13} \int_{u/2}^{3 - u} (t + 2u) dt =\)
\(I_{[0, 1]} (u) (\dfrac{4}{13} + \dfrac{8}{13}u - \dfrac{9}{52} u^2) + I_{(1, 2]} (u) (\dfrac{9}{13} + \dfrac{6}{13} u - \dfrac{21}{52} u^2)\)
\(P1 = \int_{0}^{1} \int_{0}^{2t} (t + 2u) dudt = 4/13\)\(P2 = \int_{1}^{2} \int_{0}^{1} (t + 2u)dudt = 5/13\)
\(P3 = \int_{0}^{2} \int_{0}^{u/2} (t + 2u) dudt = 4/13\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 2] Enter number of X approximation points 400 Enter number of Y approximation points 400 Enter expression for joint density (2/13)*(t+2*u).*(u<=min(2*t,3-t)) Use array operations on X, Y, PX, PY, t, u, and P P1 = total((t<1).*P) P1 = 0.3076 % Theoretical = 4/13 = 0.3077 M2 = (t>=1)&(u<=1); P2 = total(M2.*P) P2 = 0.3844 % Theoretical = 5/13 = 0.3846 P3 = total((u<=t/2).*P) P3 = 0.3076 % Theoretical = 4/13 = 0.3077
Ejercicio\(\PageIndex{22}\)
\(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2u^2\)para\(0 \le u \le 1\).
\(P(1/2 \le X \le 3/2, Y \le 1/2)\)
- Contestar
-
Región es rectángulo delimitado por\(t = 0\),\(t = 2\),\(u = 0\),\(u = 1\)
\(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2 u^2\),\(0 \le u \le 1\)
\(f_X (t) = I_{[0, 1]} (t) \dfrac{3}{8} \int_{0}^{1} (t^2 + 2u) du + I_{(1, 2]} (t) \dfrac{9}{14} \int_{0}^{1} t^2 u^2 du = I_{[0,1]} (t) \dfrac{3}{8} (t^2 + 1) + I_{(1, 2]} (t) \dfrac{3}{14} t^2\)
\(f_Y(u) = \dfrac{3}{8} \int_{0}^{1} (t^2 + 2u0 dt + \dfrac{9}{14} \int_{1}^{2} t^2 u^2 dt = \dfrac{1}{8} + \dfrac{3}{4} u + \dfrac{3}{2} u^2\)\(0 \le u \le 1\)
\(P1 = \dfrac{3}{8} \int_{1/2}^{1} \int_{0}^{1/2} (t^2 + 2u) dudt + \dfrac{9}{14} \int_{1}^{3/2} \int_{0}^{1/2} t^2 u^2 dudt = 55/448\)
tuappr Enter matrix [a b] of X-range endpoints [0 2] Enter matrix [c d] of Y-range endpoints [0 1] Enter number of X approximation points 400 Enter number of Y approximation points 200 Enter expression for joint density (3/8)*(t.^2+2*u).*(t<=1) ... + (9/14)*(t.^2.*u.^2).*(t > 1) Use array operations on X, Y, PX, PY, t, u, and P M = (1/2<=t)&(t<=3/2)&(u<=1/2); P = total(M.*P) P = 0.1228 % Theoretical = 55/448 = 0.1228