12.3 Problems on variance, covariance, linear regression (Page 5/5)

Applied probability Page 5 / 5

(See Exercise 18 from "Problems On Random Vectors and Joint Distributions", and Exercise 28 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = \frac{3}{23} (t + 2 u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq max {2 - t, t}$ .

E [X] = \frac{53}{46}, E [Y] = \frac{22}{23}, E [X^{2}] = \frac{9131}{5290}

E [X Y] = \frac{3}{23} \int_{0}^{1} \int_{0}^{2 - t} t u (t + 2 u) d u d t + \frac{3}{23} \int_{1}^{2} \int_{0}^{t} t u (t + 2 u) d u d t = \frac{251}{230}

Cov [X, Y] = - \frac{57}{5290}, Var [X] = \frac{4217}{10580}

a = Cov [X, Y] / Var [X] = - \frac{114}{4217} b = E [Y] - a E [X] = \frac{4165}{4217}

tuappr: [0 2] [0 2]200 200 (3/23)*(t + 2*u).*(u<=max(2-t,t))
VX = 0.3984 CV = -0.0108  a = -0.0272  b = 0.9909

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(See Exercise 21 from "Problems On Random Vectors and Joint Distributions", and Exercise 31 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = \frac{2}{13} (t + 2 u)$ , for $0 \leq t \leq 2$ , $0 \leq u \leq min {2 t, 3 - t}$ .

E [X] = \frac{16}{13}, E [Y] = \frac{11}{12}, E [X^{2}] = \frac{2847}{1690}

E [X Y] = \frac{2}{13} \int_{0}^{1} \int_{0}^{3 - t} t u (t + 2 u) d u d t + \frac{2}{13} \int_{1}^{2} \int_{0}^{2 t} t u (t + 2 u) d u d t = \frac{431}{390}

Cov [X, Y] = - \frac{3}{130} Var [X] = \frac{287}{1690}

a = Cov [X, Y] / Var [X] = - \frac{39}{297} b = E [Y] - a E [X] = \frac{3733}{3444}

tuappr: [0 2] [0 2]400 400 (2/13)*(t + 2*u).*(u<=min(2*t,3-t))
VX = 0.1698  CV = -0.0229  a = -0.1350  b = 1.0839

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(See Exercise 22 from "Problems On Random Vectors and Joint Distributions", and Exercise 32 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = I_{[0, 1]} (t) \frac{3}{8} (t^{2} + 2 u) + I_{(1, 2]} (t) \frac{9}{14} t^{2} u^{2}$ , for $0 \leq u \leq 1$ .

E [X] = \frac{243}{224}, E [Y] = \frac{11}{16}, E [X^{2}] = \frac{107}{70}

E [X Y] = \frac{3}{8} \int_{0}^{1} \int_{0}^{1} t u (t^{2} + 2 u) d u d t + \frac{9}{14} \int_{1}^{2} \int_{0}^{1} t^{3} u^{3} d u d t = \frac{347}{448}

Cov [X, Y] = \frac{103}{3584}, Var [X] = \frac{88243}{250880}

a = Cov [X, Y] / Var [X] = \frac{7210}{88243} b = E [Y] - a E [X] = \frac{105691}{176486}

tuappr: [0 2] [0 1]400 200 (3/8)*(t.^2 + 2*u).*(t<=1) + (9/14)*t.^2.*u.^2.*(t>1)
VX = 0.3517  CV = 0.0287 a = 0.0817  b = 0.5989

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The class ${X, Y, Z}$ of random variables is iid (independent, identically distributed) with common distribution

X = [- 5 - 1 3 4 7] P X = 0.01 * [15 20 30 25 10]

Let $W = 3 X - 4 Y + 2 Z$ . Determine $E [W]$ and $Var [W]$ . Do this using icalc, then repeat with icalc3 and compare results.

x = [-5 -1 3 4 7];px = 0.01*[15 20 30 25 10];EX = dot(x,px)                % Use of properties
EX =   1.6500VX = dot(x.^2,px) - EX^2
VX =  12.8275EW = (3 - 4+ 2)*EX
EW =   1.6500VW = (3^2 + 4^2 + 2^2)*VX
VW = 371.9975icalc                         % Iterated use of icalc
Enter row matrix of X-values  xEnter row matrix of Y-values  x
Enter X probabilities  pxEnter Y probabilities  px
 Use array operations on matrices X, Y, PX, PY, t, u, and PG = 3*t - 4*u;
[R,PR]= csort(G,P);
icalcEnter row matrix of X-values  R
Enter row matrix of Y-values  xEnter X probabilities  PR
Enter Y probabilities  pxUse array operations on matrices X, Y, PX, PY, t, u, and P
H = t + 2*u;[W,PW] = csort(H,P);EW = dot(W,PW)
EW =   1.6500VW = dot(W.^2,PW) - EW^2
VW = 371.9975icalc3                        % Use of icalc3
Enter row matrix of X-values  xEnter row matrix of Y-values  x
Enter row matrix of Z-values  xEnter X probabilities  px
Enter Y probabilities  pxEnter Z probabilities  px
Use array operations on matrices X, Y, Z,PX, PY, PZ, t, u, v, and P
S = 3*t - 4*u + 2*v;[w,pw] = csort(S,P);Ew = dot(w,pw)
Ew =   1.6500Vw = dot(w.^2,pw) - Ew^2
Vw = 371.9975

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$f_{X Y} (t, u) = \frac{3}{88} (2 t + 3 u^{2})$ for $0 \leq t \leq 2$ , $0 \leq u \leq 1 + t$ (see Exercise 25 and Exercise 37 from "Problems on Mathematical Expectation").

Z = I_{[0, 1]} (X) 4 X + I_{(1, 2]} (X) (X + Y)

E [X] = \frac{313}{220}, E [Z] = \frac{5649}{1760}, E [Z^{2}] = \frac{4881}{440}

Determine $Var [Z]$ and $Cov [X, Z]$ . Check with discrete approximation.

E [X Z] = \frac{3}{88} \int_{0}^{1} \int_{0}^{1 + t} 4 t^{2} (2 t + 3 u^{2}) d u d t + \frac{3}{88} \int_{1}^{2} \int_{0}^{1 + t} t (t + u) (2 t + 3 u^{2}) d u d t = \frac{16931}{3520}

Var [Z] = E [Z^{2}] - E^{2} [Z] = \frac{2451039}{3097600} Cov [X, Z] = E [X Z] - E [X] E [Z] = \frac{94273}{387200}

tuappr: [0 2] [0 3]200 300 (3/88)*(2*t+3*u.^2).*(u<=1+t)
G = 4*t.*(t<=1) + (t+u).*(t>1);
EZ = total(G.*P)EZ = 3.2110
EX = dot(X,PX)EX = 1.4220
CV = total(G.*t.*P) - EX*EZCV = 0.2445                       % Theoretical 0.2435
VZ = total(G.^2.*P) - EZ^2VZ = 0.7934                       % Theoretical 0.7913

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$f_{X Y} (t, u) = \frac{24}{11} t u$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {1, 2 - t}$ (see Exercise 27 and Exercise 38 from "Problems on Mathematical Expectation").

Z = I_{M} (X, Y) \frac{1}{2} X + I_{M^{c}} (X, Y) Y^{2}, M = {(t, u) : u > t}

E [X] = \frac{52}{55}, E [Z] = \frac{16}{55}, E [Z^{2}] = \frac{39}{308}

Determine $Var [Z]$ and $Cov [X, Z]$ . Check with discrete approximation.

E [X Z] = \frac{24}{11} \int_{0}^{1} \int_{t}^{1} t (t / 2) t u d u d t + \frac{24}{11} \int_{0}^{1} \int_{0}^{t} t u^{2} t u d u d t + \frac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} t t u^{2} t u d u d t = \frac{211}{770}

Var [Z] = E [Z^{2}] - E^{2} [Z] = \frac{3557}{84700} Cov [Z, X] = E [X Z] - E [X] E [Z] = - \frac{43}{42350}

tuappr:  [0 2] [0 1]400 200 (24/11)*t.*u.*(u<=min(1,2-t))
G = (t/2).*(u>t) + u.^2.*(u<=t);
VZ = total(G.^2.*P) - EZ^2VZ =   0.0425
CV = total(t.*G.*P) - EZ*dot(X,PX)CV = -9.2940e-04

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$f_{X Y} (t, u) = \frac{3}{23} (t + 2 u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq max {2 - t, t}$ (see Exercise 28 and Exercise 39 from "Problems on Mathematical Expectation").

Z = I_{M} (X, Y) (X + Y) + I_{M^{c}} (X, Y) 2 Y, M = {(t, u) : max (t, u) \leq 1}

E [X] = \frac{53}{46}, E [Z] = \frac{175}{92}, E [Z^{2}] = \frac{2063}{460}

Determine $Var [Z]$ and $Cov [Z]$ . Check with discrete approximation.

E [Z X] = \frac{3}{23} \int_{0}^{1} \int_{0}^{1} t (t + u) (t + 2 u) d u d t + \frac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 2 t u (t + 2 u) d u d t +

\frac{3}{23} \int_{1}^{2} \int_{1}^{t} 2 t u (t + 2 u) d u d t = \frac{1009}{460}

Var [Z] = E [Z^{2}] - E^{2} [Z] = \frac{36671}{42320} Cov [Z, X] = E [Z X] - E [Z] E [X] = \frac{39}{21160}

tuappr:  [0 2] [0 2]400 400 (3/23)*(t+2*u).*(u<=max(2-t,t))
M = max(t,u)<=1;
G = (t+u).*M + 2*u.*(1-M);EZ = total(G.*P);
EX = dot(X,PX);CV = total(t.*G.*P) - EX*EZ
CV =  0.0017

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$f_{X Y} (t, u) = \frac{12}{179} (3 t^{2} + u)$ , for $0 \leq t \leq 2$ , $0 \leq u \leq min {2, 3 - t}$ (see Exercise 29 and Exercise 40 from "Problems on Mathematical Expectation").

Z = I_{M} (X, Y) (X + Y) + I_{M^{c}} (X, Y) 2 Y^{2}, M = {(t, u) : t \leq 1, u \geq 1}

E [X] = \frac{2313}{1790}, E [Z] = \frac{1422}{895}, E [Z^{2}] = \frac{28296}{6265}

Determine $Var [Z]$ and $Cov [X, Z]$ . Check with discrete approximation.

E [Z X] = \frac{12}{179} \int_{0}^{1} \int_{1}^{2} t (t + u) (3 t^{2} + u) d u d t + \frac{12}{179} \int_{0}^{1} \int_{0}^{1} 2 t u^{2} (3 t^{2} + u) d u d t +

\frac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 2 t u^{2} (3 t^{2} + u) d u d t = \frac{24029}{12530}

Var [Z] = E [Z^{2}] - E^{2} [Z] = \frac{11170332}{5607175} Cov [Z, X] = E [Z X] - E [Z] E [X] = - \frac{1517647}{11214350}

tuappr:  [0 2] [0 2]400 400 (12/179)*(3*t.^2 + u).*(u<= min(2,3-t))
M = (t<=1)&(u>=1);
G = (t + u).*M + 2*u.^2.*(1 - M);EZ = total(G.*P);
EX = dot(X,PX);CV = total(t.*G.*P) - EZ*EX
CV = -0.1347

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$f_{X Y} (t, u) = \frac{12}{227} (3 t + 2 t u)$ , for $0 \leq t \leq 2$ , $0 \leq u \leq min {1 + t, 2}$ (see Exercise 30 and Exercise 41 from "Problems on Mathematical Expectation").

Z = I_{M} (X, Y) X + I_{M^{c}} (X, Y) X Y, M = {(t, u) : u \leq min (1, 2 - t)}

E [X] = \frac{1567}{1135}, E [Z] = \frac{5774}{3405}, E [Z^{2}] = \frac{56673}{15890}

Determine $Var [Z]$ and $Cov [X, Z]$ . Check with discrete approximation.

E [Z X] = \frac{12}{227} \int_{0}^{1} \int_{0}^{1} t^{2} (3 t + 2 t u) d u d t + \frac{12}{227} \int_{1}^{2} \int_{0}^{2 - t} t^{2} (3 t + 2 t u) d u d t +

\frac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} t^{2} u (3 t + 2 t u) d u d t + \frac{12}{227} \int_{1}^{2} \int_{2 - t}^{2} t^{2} u (3 t + 2 t u) d u d t = \frac{20338}{7945}

Var [Z] = E [Z^{2}] - E^{2} [Z] = \frac{112167631}{162316350} Cov [Z, X] = E [Z X] - E [Z] E [X] = \frac{5915884}{27052725}

tuappr: [0 2] [0 2]400 400 (12/227)*(3*t + 2*t.*u).*(u<= min(1+t,2))
EX = dot(X,PX);M = u<= min(1,2-t);
G = t.*M + t.*u.*(1 - M);EZ = total(G.*P);
EZX = total(t.*G.*P)EZX =  2.5597
CV = EZX - EX*EZCV =   0.2188
VZ = total(G.^2.*P) - EZ^2VZ =   0.6907

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(See [link] , and Exercises 9 and 10 from "Problems on Functions of Random Variables"). For the pair ${X, Y}$ in [link] , let

Z = g (X, Y) = 3 X^{2} + 2 X Y - Y^{2}

W = h (X, Y) = \{\begin{matrix} X & for X + Y \leq 4 \\ 2 Y & for X + Y > 4 \end{matrix}) = I_{M} (X, Y) X + I_{M^{c}} (X, Y) 2 Y

Determine the joint distribution for the pair ${Z, W}$ and determine the regression line of W on Z .


npr08_07 Data are in X, Y, P
jointzwEnter joint prob for (X,Y) P
Enter values for X XEnter values for Y Y
Enter expression for g(t,u) 3*t.^2 + 2*t.*u - u.^2Enter expression for h(t,u) t.*(t+u<=4) + 2*u.*(t+u>4)
Use array operations on Z, W, PZ, PW, v, w, PZWEZ = dot(Z,PZ)
EZ =    5.2975EW = dot(W,PW)
EW =    4.7379VZ = dot(Z.^2,PZ) - EZ^2
VZ =    1.0588e+03CZW = total(v.*w.*PZW) - EZ*EW
CZW = -12.1697a = CZW/VZ
a =   -0.0115b = EW - a*EZ
b =    4.7988                 % Regression line: w = av + b

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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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