8.2 Problems on random vectors and joint distributions (Page 6/6)

Applied probability Page 6 / 6

$f_{X Y} (t, u) = \frac{24}{11} t u$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {1, 2 - t}$

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

Region is bounded by $t = 0, u = 0, u = 2, u = 2 - t$

f_{X} (t) = I_{[0, 1]} (t) \frac{24}{11} \int_{0}^{1} t u d u + I_{(1, 2]} (t) \frac{24}{11} \int_{0}^{2 - t} t u d u =

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

f_{Y} (u) = \frac{24}{11} \int_{0}^{2 - u} t u d t = \frac{12}{11} u {(u - 2)}^{2}, 0 \leq u \leq 1

P 1 = \frac{24}{11} \int_{0}^{1} \int_{0}^{1} t u d u d t = 6 / 11 P 2 = \frac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} t u d u d t = 5 / 11

P 3 = \frac{24}{11} \int_{0}^{1} \int_{t}^{1} t u d u d t = 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  400Enter 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 PM1 = (t<=1)&(u<=1);
P1 = total(M1.*P)P1 = 0.5447             % Theoretical = 6/11 = 0.5455P2 = total((t>1).*P)
P2 =  0.4553            % Theoretical = 5/11 = 0.4545P3 = total((t<u).*P)
P3 =  0.2705            % Theoretical = 3/11 = 0.2727

Got questions? Get instant answers now!

$f_{X Y} (t, u) = \frac{3}{23} (t + 2 u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq max {2 - t, t}$

P (X \geq 1, Y \geq 1), P (Y \leq 1), P (Y \leq X)

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

f_{X} (t) = I_{[0, 1]} (t) \frac{3}{23} \int_{0}^{2 - t} (t + 2 u) d u + I_{(1, 2]} (t) \frac{3}{23} \int_{0}^{t} (t + 2 u) d u = I_{[0, 1]} (t) \frac{6}{23} (2 - t) + I_{(1, 2]} (t) \frac{6}{23} t^{2}

f_{Y} (u) = I_{[0, 1]} (u) \frac{3}{23} \int_{0}^{2} (t + 2 u) d t + I_{(1, 2]} (u) [\frac{3}{23} \int_{0}^{2 - u} (t + 2 u) d t + \frac{3}{23} \int_{u}^{2} (t + 2 u) d t] =

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

P 1 = \frac{3}{23} \int_{1}^{2} \int_{1}^{t} (t + 2 u) d u d t = 13 / 46, P 2 = \frac{3}{23} \int_{0}^{2} \int_{0}^{1} (t + 2 u) d u d t = 12 / 23

P 3 = \frac{3}{23} \int_{0}^{2} \int_{0}^{t} (t + 2 u) d u d t = 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  200Enter 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 PM1 = (t>=1)&(u>=1);
P1 = total(M1.*P)P1 =  0.2841
13/46                 % Theoretical = 13/46 = 0.2826P2 = total((u<=1).*P)
P2 =  0.5190             % Theoretical = 12/23 = 0.5217P3 = total((u<=t).*P)
P3 =  0.6959             % Theoretical = 16/23 = 0.6957

Got questions? Get instant answers now!

$f_{X Y} (t, u) = \frac{12}{179} (3 t^{2} + u)$ , for $0 \leq t \leq 2$ , $0 \leq u \leq min {2, 3 - t}$

P (X \geq 1, Y \geq 1), P (X \leq 1, Y \leq 1), P (Y < X)

Region has two parts: (1) $0 \leq t \leq 1, 0 \leq u \leq 2$ (2) $1 < t \leq 2, 0 \leq u \leq 3 - t$

f_{X} (t) = I_{[0, 1]} (t) \frac{12}{179} \int_{0}^{2} (3 t^{2} + u) d u + I_{(1, 2]} (t) \frac{12}{179} \int_{0}^{3 - t} (3 t^{2} + u) d u =

I_{[0, 1]} (t) \frac{24}{179} (3 t^{2} + 1) + I_{(1, 2]} (t) \frac{6}{179} (9 - 6 t + 19 t^{2} - 6 t^{3})

f_{Y} (u) = I_{[0, 1]} (u) \frac{12}{179} \int_{0}^{2} (3 t^{2} + u) d t + I_{(1, 2]} (u) \frac{12}{179} \int_{0}^{3 - u} (3 t^{2} + u) d t =

I_{[0, 1]} (u) \frac{24}{179} (4 + u) + I_{(1, 2]} (u) \frac{12}{179} (27 - 24 u + 8 u^{2} - u^{3})

P 1 = \frac{12}{179} \int_{1}^{2} \int_{1}^{3 - t} (3 t^{2} + u) d u d t = 41 / 179 P 2 = \frac{12}{179} \int_{0}^{1} \int_{0}^{1} (3 t^{2} + u) d u d t = 18 / 179

P 3 = \frac{12}{179} \int_{0}^{3 / 2} \int_{0}^{t} (3 t^{2} + u) d u d t + \frac{12}{179} \int_{3 / 2}^{2} \int_{0}^{3 - t} (3 t^{2} + u) d u d t = 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  200Enter 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 Pfx = 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

Got questions? Get instant answers now!

$f_{X Y} (t, u) = \frac{12}{227} (3 t + 2 t u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {1 + t, 2}$

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

Region is in two parts:

$0 \leq t \leq 1, 0 \leq u \leq 1 + t$
(2) $1 < t \leq 2, 0 \leq u \leq 2$

f_{X} (t) = I_{[0, 1]} (t) \int_{0}^{1 + t} f_{X Y} (t, u) d u + I_{(1, 2]} (t) \int_{0}^{2} f_{X Y} (t, u) d u =

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

f_{Y} (u) = I_{[0, 1]} (u) \int_{0}^{2} f_{X Y} (t, u) d t + I_{(1, 2]} (u) \int_{u - 1}^{2} f_{X Y} (t, u) d t =

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

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

P 1 = \frac{12}{227} \int_{0}^{1 / 2} \int_{0}^{1 + t} (3 t + 2 t u) d u d t = 139 / 3632

P 2 = \frac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} (3 t + 2 t u) d u d t + \frac{12}{227} \int_{1}^{3 / 2} \int_{1}^{2} (3 t + 2 t u) d u d t = 68 / 227

P 3 = \frac{12}{227} \int_{0}^{2} \int_{1}^{t} (3 t + 2 t u) d u d t = 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  200Enter 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 PM1 = (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

Got questions? Get instant answers now!

$f_{X Y} (t, u) = \frac{2}{13} (t + 2 u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {2 t, 3 - t}$

P (X < 1), P (X \geq 1, Y \leq 1), P (Y \leq X / 2)

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

f_{X} (t) = I_{[0, 1]} (t) \frac{2}{13} \int_{0}^{2 t} (t + 2 u) d u + I_{(1, 2]} (t) \frac{2}{13} \int_{0}^{3 - t} (t + 2 u) d u = I_{[0, 1]} (t) \frac{12}{13} t^{2} + I_{(1, 2]} (t) \frac{6}{13} (3 - t)

f_{Y} (u) = I_{[0, 1]} (u) \frac{2}{13} \int_{u / 2}^{2} (t + 2 u) d t + I_{(1, 2]} (u) \frac{2}{13} \int_{u / 2}^{3 - u} (t + 2 u) d t =

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

P 1 = \int_{0}^{1} \int_{0}^{2 t} (t + 2 u) d u d t = 4 / 13 P 2 = \int_{1}^{2} \int_{0}^{1} (t + 2 u) d u d t = 5 / 13

P 3 = \int_{0}^{2} \int_{0}^{t / 2} (t + 2 u) d u d t = 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  400Enter 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 PP1 = total((t<1).*P)
P1 = 0.3076             % Theoretical = 4/13 = 0.3077M2 = (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

Got questions? Get instant answers now!

$f_{X Y} (t, u) = I_{[0, 1]} (t) \frac{3}{8} (t^{2} + 2 u) + I_{(1, 2]} (t) \frac{9}{14} t^{2} u^{2}$ for $0 \leq u \leq 1$ .

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

Region is rectangle bounded by $t = 0, t = 2, u = 0, u = 1$

f_{X Y} (t, u) = I_{[0, 1]} (t) \frac{3}{8} (t^{2} + 2 u) + I_{(1, 2]} (t) \frac{9}{14} t^{2} u^{2}, 0 \leq u \leq 1

f_{X} (t) = I_{[0, 1]} (t) \frac{3}{8} \int_{0}^{1} (t^{2} + 2 u) d u + I_{(1, 2]} (t) \frac{9}{14} \int_{0}^{1} t^{2} u^{2} d u = I_{[0, 1]} (t) \frac{3}{8} (t^{2} + 1) + I_{(1, 2]} (t) \frac{3}{14} t^{2}

f_{Y} (u) = \frac{3}{8} \int_{0}^{1} (t^{2} + 2 u) d t + \frac{9}{14} \int_{1}^{2} t^{2} u^{2} d t = \frac{1}{8} + \frac{3}{4} u + \frac{3}{2} u^{2} 0 \leq u \leq 1

P 1 = \frac{3}{8} \int_{1 / 2}^{1} \int_{0}^{1 / 2} (t^{2} + 2 u) d u d t + \frac{9}{14} \int_{1}^{3 / 2} \int_{0}^{1 / 2} t^{2} u^{2} d u d t = 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  400Enter 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 PM = (1/2<=t)&(t<=3/2)&(u<=1/2);
P = total(M.*P)P =  0.1228          % Theoretical = 55/448 = 0.1228

Got questions? Get instant answers now!

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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