8.3: Problemas en Vectores Aleatorios y Distribuciones Conjuntas

Ejercicio$\PageIndex{1}$

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

$X$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

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

$P(X = i, Y = k) = P(X = i) P(Y = k|X = i) = (1/6) P(Y = k|X = i)$.

% 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

% 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

% 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

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

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

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

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

1. Esbozar la región de definición y determinar analíticamente las funciones de densidad marginal$f_X$ y$f_Y$.
2. Utilice una aproximación discreta para trazar la densidad marginal$f_X$ y la función de distribución marginal$F_X$.
3. Calcular analíticamente las probabilidades indicadas.
4. 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

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:

1. $0 \le t \le 1$,$0 \le u \le 1 + t$
2. $1 < t \le 2$,$0 \le u \le 2$

$f_X(t) = I_{[0,1]} (t) \int_{0}^{1+t} f_{XY} (t, u) du + I_{(1, 2]} (t) \int_{0}^{2} f_{XY} (t, u) du =$

$I_{[0, 1]} (t) \dfrac{12}{227} (t^3 + 5t^2 + 4t) + I_{(1, 2]} (t) \dfrac{120}{227} t$

$f_Y(u) = I_{[0, 1]} (u) \int_{0}^{2} f_{XY} (t, u) dt + I_{(1, 2]} (u) \int_{u - 1}^{2} f_{XY} (t, u) dt =$

$I_{[0, 1]} (u) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (2u + 3) (3 + 2u - u^2)$

$= I_{[0, 1]} (u) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (9 + 12 u + u^2 - 2u^3)$

$P1 = \dfrac{12}{227} \int_{0}^{1/2} \int_{0}^{1 + t} (3t + 2tu) du dt = 139/3632$

$P2 = \dfrac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} (3t + 2tu) dudt + \dfrac{12}{227} \int_{1}^{3/2} \int_{1}^{2} (3t + 2tu) du dt = 68/227$

$P3 = \dfrac{12}{227} \int_{0}^{2} \int_{1}^{t} (3t + 2tu) dudt = 144/227$

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