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8.3: Problems on Random Vectors and Joint Distributions

  • Page ID
    10882
    • Contributed by Paul Pfeiffer
    • Professor emeritus (Computational and Applied Mathematics) at Rice University

    Exercise \(\PageIndex{1}\)

    Two cards are selected at random, without replacement, from a standard deck. Let \(X\) be the number of aces and \(Y\) be the number of spades. Under the usual assumptions, determine the joint distribution and the marginals.

    Answer

    Let \(X\) be the number of aces and \(Y\) be the number of spades. Define the events \(AS_i\), \(A_i\), \(S_i\), and \(N_i\), \(i = 1, 2\) of drawing ace of spades, other ace, spade (other than the ace), and neither on the i selection. Let \(P(i, k) = P(X = i, Y = k)\).

    \(P(0, 0) = P(N_1N_2) = \dfrac{36}{52} \cdot \dfrac{35}{51} = \dfrac{1260}{2652}\)

    \(P(0, 1) = P(N_1S_2 \bigvee S_1N_2) = \dfrac{36}{52} \cdot \dfrac{12}{51} + \dfrac{12}{52} \cdot \dfrac{36}{51} = \dfrac{864}{2652}\)

    \(P(0, 2) = P(S_1 S_2) = \dfrac{12}{52} \cdot \dfrac{11}{51} = \dfrac{132}{2652}\)

    \(P(1, 0) = P(A_N_2 \bigvee N_1 S_2) = \dfrac{3}{52} \cdot \dfrac{36}{51} + \dfrac{36}{52} \cdot \dfrac{3}{51} = \dfrac{216}{2652}\)

    \(P(1, 1) = P(A_1S_2 \bigvee S_1A_2 \bigvee AS_1N_2 \bigvee N_1AS_2) = \dfrac{3}{52} \cdot \dfrac{12}{51} + \dfrac{12}{52} \cdot \dfrac{3}{51} + \dfrac{1}{52} \cdot \dfrac{36}{51} + \dfrac{36}{52} \cdot \dfrac{1}{51} = \dfrac{144}{2652}\)

    \(P(1, 2) = P(AS_1S_2 \bigvee S_1AS_2) = \dfrac{1}{52} \cdot \dfrac{12}{51} + \dfrac{12}{52} \cdot \dfrac{1}{51} = \dfrac{24}{2652}\)

    \(P(2, 0) = P(A_1A_2) = \dfrac{3}{52} \cdot \dfrac{2}{51} = \dfrac{6}{2652}\)

    \(P(2, 1) = P(AS_1A_2 \bigvee A_1AS_2) = \dfrac{1}{52} \cdot \dfrac{3}{51} + \dfrac{3}{52} \cdot \dfrac{1}{51} = \dfrac{6}{2652}\)

    \(P(2, 2) = P(\emptyset) = 0\)

    % 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

    Exercise \(\PageIndex{2}\)

    Two positions for campus jobs are open. Two sophomores, three juniors, and three seniors apply. It is decided to select two at random (each possible pair equally likely). Let \(X\) be the number of sophomores and \(Y\) be the number of juniors who are selected. Determine the joint distribution for the pair \(\{X, Y\}\) and from this determine the marginals for each.

    Answer

    Let \(A_i, B_i, C_i\) be the events of selecting a sophomore, junior, or senior, respectively, on the \(i\)th trial. Let \(X\) be the number of sophomores and \(Y\) be the number of juniors selected.

    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

    Exercise \(\PageIndex{3}\)

    A die is rolled. Let \(X\) be the number that turns up. A coin is flipped \(X\) times. Let \(Y\) be the number of heads that turn up. Determine the joint distribution for the pair \(\{X, Y\}\). Assume \(P(X = k) = 1/6\) for \(1 \le k \le 6\) and for each \(k\), \(P(Y = j|X = k)\) has the binomial (\(k\), 1/2) distribution. Arrange the joint matrix as on the plane, with values of \(Y\) increasing upward. Determine the marginal distribution for \(Y\). (For a MATLAB based way to determine the joint distribution see Example 14.1.7 from "Conditional Expectation, Regression")

    Answer

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

    % 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

    Exercise \(\PageIndex{4}\)

    As a variation of Exercise 8.3.3., Suppose a pair of dice is rolled instead of a single die. Determine the joint distribution for the pair \(\{X, Y\}\) and from this determine the marginal distribution for \(Y\).

    Answer
    % 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

    Exercise \(\PageIndex{5}\)

    Suppose a pair of dice is rolled. Let \(X\) be the total number of spots which turn up. Roll the pair an additional \(X\) times. Let \(Y\) be the number of sevens that are thrown on the \(X\) rolls. Determine the joint distribution for the pair \(\{X, Y\}\) and from this determine the marginal distribution for \(Y\). What is the probability of three or more sevens?

    Answer
    % 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

    Exercise \(\PageIndex{6}\)

    The pair \(\{X, Y\}\) has the joint distribution (in 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]

    屏幕快照 2020-04-11 上午11.56.15.png

    Determine the marginal distribution and the corner values for \(F_{XY}\). Determine \(P(X + Y > 2)\) and \(P(X \ge Y)\).

    Answer
    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

    Exercise \(\PageIndex{7}\)

    The pair \(\{X, Y\}\) has the joint distribution (in 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

    Determine the marginal distributions and the corner values for \(F_{XY}\). Determine \(P(1 \le X \le 4, Y > 4)\) and \(P(|X - Y| \le 2)\).

    Answer
    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

    Exercise \(\PageIndex{8}\)

    The pair \(\{X, Y\}\) has the joint distribution (in 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

    Determine the marginal distributions. Determine \(F_{XY} (10, 6)\) and \(P(X > Y)\).

    Answer
    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

    Exercise \(\PageIndex{9}\)

    Data were kept on the effect of training time on the time to perform a job on a production line. \(X\) is the amount of training, in hours, and \(Y\) is the time to perform the task, in minutes. The data are as follows (in 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

    Determine the marginal distributions. Determine \(F_{XY}(2, 3)\) and \(P(Y/X \ge 1.25)\).

    Answer
    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
    

    For the joint densities in Exercises 10-22 below

    1. Sketch the region of definition and determine analytically the marginal density functions \(f_X\) and \(f_Y\).
    2. Use a discrete approximation to plot the marginal density \(f_X\) and the marginal distribution function \(F_X\).
    3. Calculate analytically the indicated probabilities.
    4. Determine by discrete approximation the indicated probabilities.

    Exercise \(\PageIndex{10}\)

    \(f_{XY}(t, u) = 1\) for \(0 \le t \le 1\), \(0 \le u \le 2(1 - t)\).

    \(P(X > 1/2, Y > 1), P(0 \le X \le 1/2, Y > 1/2), P(Y \le X)\)

    Answer

    Region is triangle with vertices (0, 0), (1, 0), (0, 2).

    \(f_{X} (t) = \int_{0}^{2(1-t)} du = 2(1 - t)\), \(0 \le t \le 1\)

    \(f_{Y} (u) = \int_{0}^{1 - u/2} dt = 1 - u/2\), \(0 \le u \le 2\)

    \(M1 = \{(t, u):t > 1/2, u> 1\}\) lies outside the trianlge \(P((X, Y) \in M1) = 0\)

    \(M2 = \{(t, u): 0 \le t \le 1/2, u > 1/2\}\) has area in the triangle = 1/2

    \(M3\) = the region in the triangle under \(u = t\), which has area 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

    Exercise \(\PageIndex{11}\)

    \(f_{XY} (t, u) = 1/2\) on the square with vertices at (1, 0), (2, 1), (1, 2), (0, 1).

    \(P(X > 1, Y > 1), P(X \le 1/2, 1 < Y), P(Y \le X)\)

    Answer

    The region is bounded by lines \(u = 1 + t\), \( u = 1 - t\), \(u = 3 - t\), and \(u = t - 1\)

    \(f_X (t) = I_{[0,1]} (t) 0.5 \int_{1 - t}^{1 + t} du + I_{(1, 2]} (t) 0.5 \int_{t - 1}^{3 - t} du = I_{(1, 2]} (t) (2 - t) = f_Y(t)\) by symmetry

    \(M1 = \{(t, u): t > 1, u > 1\}\) has area in the trangle = 1/2, so \(PM1 = 1/4\)

    \(M2 = \{(t, u): t \le 1/2, u > 1\}\) has area in the trangle = 1/8\), so \(PM2 = 1/16\)

    \(M3 = \{(t, u): u \le t\}\) has area in the trangle = 1, so \(PM3 = 1/2\)

    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

    Exercise \(\PageIndex{12}\)

    \(f_{XY} (t, u) = 4t(1 - u)\) for \(0 \le t \le 1\), \(0 \le u \le 1\).

    \(P(1/2 < X < 3/4, Y > 1/2)\), \(P(X \le 1/2, Y > 1/2)\), \(P(Y \le X)\)

    Answer

    Region is the unit square,

    \(f_X (t) = \int_{0}^{1} 4t(1 - u) du = 2t\), \(0 \le t \le 1\)

    \(f_Y(u) = \int_{0}^{1} 4t(1 - u) dt = 2(1 - u)\), \(0 \le u \le 1\)

    \(P1 = \int_{1/2}^{3/4} \int_{1/2}^{1} 4t (1 - u) du dt = 5/64\) \(P2 = \int_{0}^{1/2} \int_{1/2}^{1} 4t(1 - u) dudt = 1/16\)

    \(P3 = \int_{0}^{1} \int_{0}^{t} 4t(1 - u) du dt = 5/6\)

    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

    Exercise \(\PageIndex{13}\)

    \(f_{XY} (t, u) = \dfrac{1}{8} (t + u)\) for \(0 \le t \le 2\), \(0 \le u \le 2\).

    \(P(X > 1/2, Y > 1/2), P(0 \le X \le 1, Y > 1), P(Y \le X)\)

    Answer

    Region is the square \(0 \le t \le 2\), \(0 \le u \le 2\)

    \(f_X (t) = \dfrac{1}{8} \int_{0}^{2} (t + u) = \dfrac{1}{4} ( t + 1) = f_Y(t)\), \(0 \le t \le 2\)

    \(P1 = \int_{1/2}^{2} \int_{1/2}^{2} (t + u) dudt = 45/64\) \(P2 = \int_{0}^{1} \int_{1}^{2} (t + u) du dt = 1/4\)

    \(P3 = \int_{0}^{2} \int_{0}^{1} (t + u) dudt = 1/2\)

    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

    Exercise \(\PageIndex{14}\)

    \(f_{XY}(t, u) = 4ue^{-2t}\) for \(0 \le t, 0 \le u \le 1\)

    \(P(X \le 1, Y > 1), P(X > 0, 1/2 < Y < 3/4), P(X < Y)\)

    Answer

    Region is strip by \(t = 0, u = 0, u = 1\)

    \(f_X(t) = 2e^{-2t}\), \(0 \le t\), \(f_Y(u) = 2u\), \(0 \le u \le 1\), \(f_{XY} = f_X f_Y\)

    \(P1 = 0\), \(P2 = \int_{0.5}^{\infty} 2e^{-2t} dt \int_{1/2}^{3/4} 2udu = e^{-1} 5/16\)

    \(P3 = 4 \int_{0}^{1} \int_{t}^{1} ue^{-2t} dudt = \dfrac{3}{2} e^{-2} + \dfrac{1}{2} = 0.7030\)

    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

    Exercise \(\PageIndex{15}\)

    \(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1 + t\).

    \(F_{XY} (1, 1)\), \(P(X \le 1, Y > 1)\), \(P(|X - Y| < 1)\)

    Answer

    Region bounded by \(t = 0\), \(t = 2\), \(u = 0\), \(u = 1 + t\)

    \(f_X (t) = \dfrac{3}{88} \int_{0}^{1 + t} (2t + 3u^2) du = \dfrac{3}{88}(1 + t)(1 + 4t + t^2) = \dfrac{3}{88} ( 1 + 5t + 5t^2 + t^3)\), \(0 \le t \le 2\)

    \(f_Y(u) = I_{[0,1]} (u) \dfrac{3}{88} \int_{0}^{2} (2t + 3u^2) dt + I_{(1, 3]} (u) \dfrac{3}{88} \int_{u - 1}^{2} (2t + 3u^2) dt = \)

    \(I_{[0,1]} (u) \dfrac{3}{88} (6u^2 + 4) + I_{(1,3]} (t) \dfrac{3}{88} (3 + 2u + 8u^2 - 3u^3)\)

    \(F_{XY}(1, 1) = \int_{0}^{1} \int_{0}^{1} f_{XY} (t, u) dudt = 3/44\)

    \(P1 = \int_{0}^{1} \int_{1}^{1 + t} f_{XY} (t, u)dudt = 41/352\) \(P2 = \int_{0}^{1} \int_{1}^{1 + t} f_{XY} (t, u) dudt = 329/352\)

    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

    Exercise \(\PageIndex{16}\)

    \(f_{XY} (t, u) = 12t^2u\) on the parallelogram with vertices (-1, 0), (0, 0), (1, 1), (0, 1).

    \(P(X \le 1/2, Y > 0), P(X < 1/2, Y \le 1/2), P(Y \ge 1/2)\)

    Answer

    Region bounded by \(u = 0\), \(u = t\), \(u = 1\), \(u = t + 1\)

    \(f_X (t) = I_{[-1, 0]} (t) 12 \int_{0}^{t + 1} t^2 u du + I_{(0, 1]} (t) 12 \int_{t}^{1} t^2 u du = I_{[-1, 0]} (t) 6t^2 (t + 1)^2 + I_{(0, 1]}(t) 6t^2(1 - t^2)\)

    \(f_Y(u) = 12\int_{u - 1}^{t} t^2 udu + 12u^3 - 12u^2 + 4u\), \(0 \le u \le 1\)

    \(P1 = 1 - 12 \int_{1/2}^{1} \int_{t}^{1} t^2 ududt = 33/80\), \(P2 = 12 \int_{0}^{1/2} \int_{u - 1}^{u} t^2 udtdu = 3/16\)

    \(P3 = 1 - P2 = 13/16\)

    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

    Exercise \(\PageIndex{17}\)

    \(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \le \text{min}\ \{1, 2 - t\}\)

    \(P(X \le 1, Y \le 1), P(X > 1), P(X < Y)\)

    Answer

    Region is bounded by \(t = 0, u = 0, u = 2, u = 2 - t\)

    \(f_X (t) = I_{[0, 1]} (t) \dfrac{24}{11} \int_{0}^{1} tudu + I_{(1, 2]} (t) \dfrac{24}{11} \int_{0}^{2 - t} tudu =\)

    \(I_{[0, 1]} (t) \dfrac{12}{11} t + I_{(1, 2]} (t) \dfrac{12}{11} t(2 - t)^2\)

    \(f_Y (u) = \dfrac{24}{11} \int_{0}^{2 - u} tudt = \dfrac{12}{11} u(u - 2)^2\), \(0 \le u \le 1\)

    \(P1 = \dfrac{24}{11} \int_{0}^{1} \int_{0}^{1} tududt = 6/11\) \(P2 = \dfrac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} tududt = 5/11\)

    \(P3 = \dfrac{24}{11} \int_{0}^{1} \int_{t}^{1} tududt = 3/11\)

    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

    Exercise \(\PageIndex{18}\)

    \(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{max}\ \{2 - t, t\}\)

    \(P(X \ge 1, Y \ge 1), P(Y \le 1), P(Y \le X)\)

    Answer

    Region is bounded by \(t = 0, t = 2, u = 0, u = 2 - t\) \((0 \le t \le 1)\), \(u = t (1 < t \le 2)\)

    \(f_X(t) = I_{[0,1]} (t) \dfrac{3}{23} \int_{0}^{2 - t} (t + 2u) du + I_{(1, 2]} (t) \dfrac{3}{23} \int_{0}^{t} (t + 2u) du = I_{[0, 1]} (t) \dfrac{6}{23} (2 - t) + I_{(1, 2]} (t) \dfrac{6}{23}t^2\)

    \(f_Y(u) = I_{[0, 1]} (u) \dfrac{3}{23} \int_{0}^{2} (t + 2u) du + I_{(1, 2]} (u) [\dfrac{3}{23} \int_{0}^{2 - u} (t + 2u) dt + \dfrac{3}{23} \int_{u}^{2} (t + 2u) dt]=\)

    \(I_{[0,1]} (u) \dfrac{6}{23} (2u + 1) + I_{(1, 2]} (u) \dfrac{3}{23} (4 + 6u - 4u^2)\)

    \(P1 = \dfrac{3}{23} \int_{1}^{2} \int_{1}^{t} (t + 2u) du dt = 13/46\), \(P2 = \dfrac{3}{23} \int_{0}^{2} \int_{0}^{1} (t + 2u) du dt = 12/23\)

    \(P3 = \dfrac{3}{23} \int_{0}^{2} \int_{0}^{t} (t + 2u) dudt = 16/23\)

    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

    Exercise \(\PageIndex{19}\)

    \(f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1 + t, 2\}\)

    \(P(X \ge 1, Y \ge 1), P(X \le 1, Y \le 1), P(Y < X)\)

    Answer

    Region has two parts: (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

    Exercise \(\PageIndex{20}\)

    \(f_{XY} (t, u) = \dfrac{12}{227} (3t + 2tu)\) for \(0 \le t \le 2\), \(0 \le u \le \text{min} \{1 + t, 2\}\)

    \(P(X \le 1/2, Y \le 3/2), P(X \le 1.5, Y > 1), P(Y < X)\)

    Answer

    Region is in two parts:

    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

    Exercise \(\PageIndex{21}\)

    \(f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{min}\ \{2t, 3 - t\}\)

    \(P(X < 1), P(X \ge 1, Y \le 1), P(Y \le X/2)\)

    Answer

    Region bounded by \(t = 2, u = 2t\) \((0 \le t \le 1)\), \(3 - t\) \((1 \le t \le 2)\)

    \(f_X(t) = I_{[0, 1]} (t) \dfrac{2}{13} \int_{0}^{2t} (t + 2u) du + I_{(1, 2]} (t) \dfrac{2}{13} \int_{0}^{3 - t} (t + 2u) du = I_{[0, 1]} (t) \dfrac{12}{13} t^2 + I_{(1, 2]} (t) \dfrac{6}{13} (3 - t)\)

    \(f_Y (u) = I_{[0, 1]} (u) \dfrac{2}{13} \int_{u/2}^{2} (t + 2u) dt + I_{(1, 2]} (u) \dfrac{2}{13} \int_{u/2}^{3 - u} (t + 2u) dt =\)

    \(I_{[0, 1]} (u) (\dfrac{4}{13} + \dfrac{8}{13}u - \dfrac{9}{52} u^2) + I_{(1, 2]} (u) (\dfrac{9}{13} + \dfrac{6}{13} u - \dfrac{21}{52} u^2)\)

    \(P1 = \int_{0}^{1} \int_{0}^{2t} (t + 2u) dudt = 4/13\) \(P2 = \int_{1}^{2} \int_{0}^{1} (t + 2u)dudt = 5/13\)

    \(P3 = \int_{0}^{2} \int_{0}^{u/2} (t + 2u) dudt = 4/13\)

    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

    Exercise \(\PageIndex{22}\)

    \(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2u^2\) for \(0 \le u \le 1\).

    \(P(1/2 \le X \le 3/2, Y \le 1/2)\)

    Answer

    Region is rectangle bounded by \(t = 0\), \(t = 2\), \(u = 0\), \(u = 1\)

    \(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2 u^2\), \(0 \le u \le 1\)

    \(f_X (t) = I_{[0, 1]} (t) \dfrac{3}{8} \int_{0}^{1} (t^2 + 2u) du + I_{(1, 2]} (t) \dfrac{9}{14} \int_{0}^{1} t^2 u^2 du = I_{[0,1]} (t) \dfrac{3}{8} (t^2 + 1) + I_{(1, 2]} (t) \dfrac{3}{14} t^2\)

    \(f_Y(u) = \dfrac{3}{8} \int_{0}^{1} (t^2 + 2u0 dt + \dfrac{9}{14} \int_{1}^{2} t^2 u^2 dt = \dfrac{1}{8} + \dfrac{3}{4} u + \dfrac{3}{2} u^2\) \(0 \le u \le 1\)

    \(P1 = \dfrac{3}{8} \int_{1/2}^{1} \int_{0}^{1/2} (t^2 + 2u) dudt + \dfrac{9}{14} \int_{1}^{3/2} \int_{0}^{1/2} t^2 u^2 dudt = 55/448\)

    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
    
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