11.2 Problems on mathematical expectation (Page 4/6)

Applied probability Page 4 / 6

For the joint densities in Exercises 20-32 below

Determine analytically $E [X]$ , $E [Y]$ , $E [X^{2}]$ , $E [Y^{2}]$ , and $E [X Y]$ .
Use a discrete approximation for $E [X]$ , $E [Y]$ , $E [X^{2}]$ , $E [Y^{2}]$ , and $E [X Y]$ .

(See Exercise 10 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = 1$ for $0 \leq t \leq 1$ , $0 \leq u \leq 2 (1 - t)$ .

f_{X} (t) = 2 (1 - t), 0 \leq t \leq 1, f_{Y} (u) = 1 - u / 2, 0 \leq u \leq 2

E [X] = \int_{0}^{1} 2 t (1 - t) d t = 1 / 3, E [Y] = 2 / 3, E [X^{2}] = 1 / 6, E [Y^{2}] = 2 / 3

E [X Y] = \int_{0}^{1} \int_{0}^{2 (1 - t)} t u d u d t = 1 / 6

tuappr:  [0 1] [0 2]200 400   u<=2*(1-t)
EX = 0.3333   EY = 0.6667  EX2 = 0.1667  EY2 = 0.6667EXY = 0.1667 (use t, u, P)

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(See Exercise 11 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = 1 / 2$ on the square with vertices at $(1, 0), (2, 1), (1, 2), (0, 1)$ .

f_{X} (t) = f_{Y} (t) = I_{[0, 1]} (t) t + I_{(1, 2]} (t) (2 - t)

E [X] = E [Y] = \int_{0}^{1} t^{2} d t + \int_{1}^{t} (2 t - t^{2}) d t = 1, E [X^{2}] = E [Y^{2}] = 7 / 6

E [X Y] = (1 / 2) \int_{0}^{1} \int_{1 - t}^{1 + t} d u d t + (1 / 2) \int_{1}^{2} \int_{t - 1}^{3 - t} d u d t = 1

tuappr: [0 2] [0 2]200 200  0.5*(u<=min(t+1,3-t))&(u>= max(1-t,t-1))
EX = 1.0000  EY = 1.0002  EX2 = 1.1684  EY2 = 1.1687 EXY = 1.0002

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(See Exercise 12 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = 4 t (1 - u)$ for $0 \leq t \leq 1$ , $0 \leq u \leq 1$ .

f_{X} (t) = 2 t, 0 \leq t \leq 1, f_{Y} (u) = 2 (1 - u), 0 \leq u \leq 1

E [X] = 2 / 3, E [Y] = 1 / 3, E [X^{2}] = 1 / 2, E [Y^{2}] = 1 / 6 E [X Y] = 2 / 9

tuappr:  [0 1] [0 1]200 200 4*t.*(1-u)
EX = 0.6667  EY = 0.3333  EX2 = 0.5000  EY2 = 0.1667  EXY = 0.2222

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(See Exercise 13 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = \frac{1}{8} (t + u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq 2$ .

f_{X} (t) = f_{Y} (t) = \frac{1}{4} (t + 1), 0 \leq t \leq 2

E [X] = E [Y] = \frac{1}{4} \int_{0}^{2} (t^{2} + t) d t = \frac{7}{6}, E [X^{2}] = E [Y^{2}] = 5 / 3

E [X Y] = \frac{1}{8} \int_{0}^{2} \int_{0}^{2} (t^{2} u + t u^{2}) d u d t = \frac{4}{3}

tuappr:  [0 2] [0 2]200 200  (1/8)*(t+u)
EX = 1.1667  EY = 1.1667  EX2 = 1.6667  EY2 = 1.6667  EXY = 1.3333

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(See Exercise 14 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = 4 u e^{- 2 t}$ for $0 \leq t, 0 \leq u \leq 1$ .

f_{X} (t) = 2 e^{- 2 t}, 0 \leq t, f_{Y} (u) = 2 u, 0 \leq u \leq 1

E [X] = \int_{0}^{\infty} 2 t e^{- 2 t} d t = \frac{1}{2}, E [Y] = \frac{2}{3}, E [X^{2}] = \frac{1}{2}, E [Y^{2}] = \frac{1}{2}, E [X Y] = \frac{1}{3}

tuappr: [0 6] [0 1]600 200  4*u.*exp(-2*t)
EX = 0.5000  EY = 0.6667  EX2 = 0.4998  EY2 = 0.5000  EXY = 0.3333

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(See Exercise 15 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = \frac{3}{88} (2 t + 3 u^{2})$ for $0 \leq t \leq 2$ , $0 \leq u \leq 1 + t$ .

f_{X} (t) = \frac{3}{88} (1 + t) (1 + 4 t + t^{2}) = \frac{3}{88} (1 + 5 t + 5 t^{2} + t^{3}), 0 \leq t \leq 2

f_{Y} (u) = I_{[0, 1]} (u) \frac{3}{88} (6 u^{2} + 4) + I_{(1, 3]} (u) \frac{3}{88} (3 + 2 u + 8 u^{2} - 3 u^{3})

E [X] = \frac{313}{220}, E [Y] = \frac{1429}{880}, E [X^{2}] = \frac{49}{22}, E [Y^{2}] = \frac{172}{55}, E [X Y] = \frac{2153}{880}

tuappr:  [0 2] [0 3]200 300 (3/88)*(2*t + 3*u.^2).*(u<1+t)
EX = 1.4229  EY = 1.6202  EX2 = 2.2277 EY2 = 3.1141  EXY = 2.4415

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(See Exercise 16 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = 12 t^{2} u$ on the parallelogram with vertices

(- 1, 0), (0, 0), (1, 1), (0, 1)

f_{X} (t) = I_{[- 1, 0]} (t) 6 t^{2} {(t + 1)}^{2} + I_{(0, 1]} (t) 6 t^{2} (1 - t^{2}), f_{Y} (u) = 12 u^{3} - 12 u^{2} + 4 u, 0 \leq u \leq 1

E [X] = \frac{2}{5}, E [Y] = \frac{11}{15}, E [X^{2}] = \frac{2}{5}, E [Y^{2}] = \frac{3}{5}, E [X Y] = \frac{2}{5}

tuappr: [-1 1] [0 1]400 200 12*t.^2.*u.*(u>= max(0,t)).*(u<= min(1+t,1))
EX = 0.4035  EY = 0.7342  EX2 = 0.4016  EY2 = 0.6009  EXY = 0.4021

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(See Exercise 17 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = \frac{24}{11} t u$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {1, 2 - t}$ .

f_{X} (t) = I_{[0, 1]} (t) \frac{12}{11} t + I_{(1, 2]} (t) \frac{12}{11} t {(2 - t)}^{2}, f_{Y} (u) = \frac{12}{11} u {(u - 2)}^{2}, 0 \leq u \leq 1

E [X] = \frac{52}{55}, E [Y] = \frac{32}{55}, E [X^{2}] = \frac{57}{55}, E [Y^{2}] = \frac{2}{5}, E [X Y] = \frac{28}{55}

tuappr:  [0 2] [0 1]400 200 (24/11)*t.*u.*(u<=min(1,2-t))
EX = 0.9458  EY = 0.5822  EX2 = 1.0368 EY2 = 0.4004   EXY = 0.5098

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(See Exercise 18 from "Problems On Random Vectors and Joint Distributions"). $f_{X Y} (t, u) = \frac{3}{23} (t + 2 u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq max {2 - t, t}$ .

f_{X} (t) = 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{6}{23} (2 u + 1) + I_{(1, 2]} (u) \frac{3}{23} (4 + 6 u - 4 u^{2})

E [X] = \frac{53}{46}, E [Y] = \frac{22}{23}, E [X^{2}] = \frac{397}{230}, E [Y^{2}] = \frac{261}{230}, E [X Y] = \frac{251}{230}

tuappr:  [0 2] [0 2]200 200 (3/23)*(t + 2*u).*(u<=max(2-t,t))
EX = 1.1518  EY = 0.9596  EX2 = 1.7251  EY2 = 1.1417  EXY = 1.0944

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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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