11.2 Problems on mathematical expectation (Page 6/6)

Applied probability Page 6 / 6

The pair ${X, Y}$ has the joint distribution (in m-file npr08_07.m ):

P (X = t, 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

Let $Z = g (X, Y) = 3 X^{2} + 2 X Y - Y^{2}$ . Determine $E [Z]$ and $E [Z^{2}]$ .


npr08_07 Data are in X, Y, P
jcalc- - - - - - - - -
G = 3*t.^2 + 2*t.*u - u.^2;EG = total(G.*P)
EG =  5.2975ez2 = total(G.^2.*P)
EG2 = 1.0868e+03[Z,PZ] = csort(G,P);   % AlternateEZ = Z*PZ'
EZ =  5.2975EZ2 = (Z.^2)*PZ'
EZ2 = 1.0868e+03

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For the pair ${X, Y}$ in [link] , let

W = g 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 $E [W]$ and $E [W^{2}]$ .

H = t.*(t+u<=4) + 2*u.*(t+u>4);
EH = total(H.*P)EH =   4.7379
EH2 = total(H.^2.*P)EH2 = 61.4351
[W,PW]= csort(H,P);  % Alternate
EW = W*PW'EW =   4.7379
EW2 = (W.^2)*PW'EW2 = 61.4351

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For the distributions in Exercises 37-41 below

Determine analytically $E [Z]$ and $E [Z^{2}]$ .
Use a discrete approximation to calculate the same quantities.

$f_{X Y} (t, u) = \frac{3}{88} (2 t + 3 u^{2})$ for $0 \leq t \leq 2$ , $0 \leq u \leq 1 + t$ (see [link] ).

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

E [Z] = \frac{3}{88} \int_{0}^{1} \int_{0}^{1 + t} 4 t (2 t + 3 u^{2}) d u d t + \frac{3}{88} \int_{1}^{2} \int_{0}^{1 + t} (t + u) (2 t + 3 u^{2}) d u d t = \frac{5649}{1760}

E [Z^{2}] = \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 + u)}^{2} (2 t + 3 u^{2}) d u d t = \frac{4881}{440}

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);
EG = total(G.*P)EG =   3.2086
EG2 = total(G.^2.*P)EG2 = 11.0872

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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 [link] ).

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

E [Z] = \frac{12}{11} \int_{0}^{1} \int_{t}^{1} t^{2} u d u d t + \frac{24}{11} \int_{0}^{1} \int_{0}^{t} t u^{3} d u d t + \frac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} t u^{3} d u d t = \frac{16}{55}

E [Z^{2}] = \frac{6}{11} \int_{0}^{1} \int_{t}^{1} t^{3} u d u d t + \frac{24}{11} \int_{0}^{1} \int_{0}^{t} t u^{5} d u d t + \frac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} t u^{5} d u d t = \frac{39}{308}

tuappr:  [0 2] [0 1]400 200  (24/11)*t.*u.*(u<=min(1,2-t))
G = (1/2)*t.*(u>t) + u.^2.*(u<=t);
EZ = 0.2920  EZ2 = 0.1278

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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 [link] ).

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

E [Z] = \frac{3}{23} \int_{0}^{1} \int_{0}^{1} (t + u) (t + 2 u) d u d t + \frac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 2 u (t + 2 u) d u d t + \frac{3}{23} \int_{1}^{2} \int_{1}^{t} 2 u (t + 2 u) d u d t = \frac{175}{92}

E [Z^{2}] = \frac{3}{23} \int_{0}^{1} \int_{0}^{1} {(t + u)}^{2} (t + 2 u) d u d t + \frac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 4 u^{2} (t + 2 u) d u d t + \frac{3}{23} \int_{1}^{2} \int_{1}^{t} 4 u^{2} (t + 2 u) d u d t = \frac{2063}{460}

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)
EZ =  1.9048EZ2 = total(G.^2.*P)
EZ2 = 4.4963

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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 [link] ).

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

\frac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 2 u^{2} (3 t^{2} + u) d u d t = \frac{1422}{895}

E [Z^{2}] = \frac{12}{179} \int_{0}^{1} \int_{1}^{2} {(t + u)}^{2} (3 t^{2} + u) d u d t + \frac{12}{179} \int_{0}^{1} \int_{0}^{1} 4 u^{4} (3 t^{2} + u) d u d t +

\frac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 4 u^{4} (3 t^{2} + u) d u d t = \frac{28296}{6265}

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)
EZ =  1.5898EZ2 = total(G.^2.*P)
EZ2 = 4.5224

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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 [link] ).

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

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

\frac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} t u (3 t + 2 t u) d u d t + \frac{12}{227} \int_{1}^{2} \int_{2 - t}^{2} t u (3 t + 2 t u) d u d t = \frac{5774}{3405}

E [Z^{2}] = \frac{56673}{15890}

tuappr: [0 2] [0 2]400 400 (12/227)*(3*t + 2*t.*u).*(u<= min(1+t,2))
M = u<= min(1,2-t);
G = t.*M + t.*u.*(1 - M);EZ = total(G.*P)
EZ =  1.6955EZ2 = total(G.^2.*P)
EZ2 = 3.5659

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The class ${X, Y, Z}$ is independent. (See Exercise 16 from "Problems on Functions of Random Variables", m-file npr10_16.m )

$X = - 2 I_{A} + I_{B} + 3 I_{C}$ . Minterm probabilities are (in the usual order)

0.255 0.025 0.375 0.045 0.108 0.012 0.162 0.018

$Y = I_{D} + 3 I_{E} + I_{F} - 3$ . The class ${D, E, F}$ is independent with

P (D) = 0.32 P (E) = 0.56 P (F) = 0.40

Z has distribution

Value	-1.3	1.2	2.7	3.4	5.8
Probability	0.12	0.24	0.43	0.13	0.08

$W = X^{2} + 3 X Y^{2} - 3 Z$ . Determine $E [W]$ and $E [W^{2}]$ .


npr10_16 Data are in cx, pmx, cy, pmy, Z, PZ
[X,PX]= canonicf(cx,pmx);
[Y,PY]= canonicf(cy,pmy);
icalc3input: X, Y, Z, PX, PY, PZ
- - - - - - -Use array operations on matrices X, Y, Z,
PX, PY, PZ, t, u, v, and PG = t.^2 + 3*t.*u.^2 - 3*v;
[W,PW]= csort(G,P);
EW = W*PW'EW =   -1.8673
EW2 = (W.^2)*PW'EW2 = 426.8529

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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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