11.3: Problems on Mathematical Expectation
Exercise \(\PageIndex{1}\)
(See Exercise 1 from "Problems on Distribution and Density Functions", m-file npr07_01.m). The class \(\{C_j: 1 \le j \le 10\}\) is a partition. Random variable \(X\) has values {1, 3, 2, 3, 4, 2, 1, 3, 5, 2} on \(C_1\) through \(C_{10}\), respectively, with probabilities 0.08, 0.13, 0.06, 0.09, 0.14, 0.11, 0.12, 0.07, 0.11, 0.09. Determine \(E[X]\)
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% file npr07_01.m % Data for Exercise 1 from "Problems on Distribution and Density Functions" T = [1 3 2 3 4 2 1 3 5 2]; pc = 0.01*[8 13 6 9 14 11 12 7 11 9]; disp('Data are in T and pc') npr07_01 Data are in T and pc EX = T*pc' EX = 2.7000 [X,PX] csort(T,pc): % Alternate using X, PX ex = X*PX' ex = 2.7000
Exercise \(\PageIndex{2}\)
(See Exercise 2 from " Problems on Distribution and Density Functions ", m-file npr07_02.m ). A store has eight items for sale. The prices are $3.50, $5.00, $3.50, $7.50, $5.00, $5.00, $3.50, and $7.50, respectively. A customer comes in. She purchases one of the items with probabilities 0.10, 0.15, 0.15, 0.20, 0.10 0.05, 0.10 0.15. The random variable expressing the amount of her purchase may be written
\(X = 3.5I_{C_1} + 5.0 I_{C_2} + 3.5I_{C_3} + 7.5I_{C_4} + 5.0I_{C_5} + 5.0I_{C_6} + 3.5I_{C_7} + 7.5I_{C_8}\)
Determine the expection \(E[X]\) of the value of her purchase.
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% file npr07_02.m % Data for Exercise 2 from "Problems on Distribution and Density Functions" T = [3.5 5.0 3.5 7.5 5.0 5.0 3.5 7.5]; pc = 0.01*[10 15 15 20 10 5 10 15]; disp('Data are in T and pc') npr07_02 Data are in T and pc EX = T*pc' EX = 5.3500 [X,PX] csort(T,pc) ex = X*PX' ex = 5.3500
Exercise \(\PageIndex{3}\)
See Exercise 12 from " Problems on Random Variables and Probabilities ", and Exercise 3 from " Problems on Distribution and Density Functions ," m-file npr06_12.m ). The class \(\{A, B, C, D\}\) has minterm probabilities
\(pm = \) 0.001 * [5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302]
Determine the mathematical expection for the random variable \(X = I_A + I_B + I_C + I_D\), which counts the number of the events which occur on a trial.
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% file npr06_12.m % Data for Exercise 12 from "Problems on Random Variables and Probabilities" pm = 0.001*[5 7 6 8 9 14 22 33 21 32 50 75 86 129 201 302]; c = [1 1 1 1 0]; disp('Minterm probabilities in pm, coefficients in c') npr06_12 Minterm probabilities in pm, coefficients in c canonic Enter row vector of coefficients c Enter row vector of minterm probabilities pm Use row matrices X and PX for calculations call for XDBN to view the distribution EX = X*PX' EX = 2.9890 T = sum(mintable(4)); [x,px] = csort(T,pm); ex = x*px ex = 2.9890
Exercise \(\PageIndex{4}\)
(See Exercise 5 from " Problems on Distribution and Density Functions "). In a thunderstorm in a national park there are 127 lightning strikes. Experience shows that the probability of of a lightning strike starting a fire is about 0.0083. Determine the expected number of fires.
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\(X\) ~ binomial (127, 0.0083), \(E[X] = 127 \cdot 0.0083 = 1.0541\)
Exercise \(\PageIndex{5}\)
(See Exercise 8 from " Problems on Distribution and Density Functions "). Two coins are flipped twenty times. Let \(X\) be the number of matches (both heads or both tails). Determine \(E[X]\)
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\(X\) ~ binomial (20, 1/2). \(E[X] = 20 \cdot 0.5 = 10\)
Exercise \(\PageIndex{6}\)
(See Exercise 12 from " Problems on Distribution and Density Functions "). A residential College plans to raise money by selling “chances” on a board. Fifty chances are sold. A player pays $10 to play; he or she wins $30 with probability \(p = 0.2\). The profit to the College is
\(X = 50 \cdot 10 - 30N\), where \(N\) is the numbe of winners
Determine the expected profit \(E[X]\).
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\(N\) ~ binomial (50, 0.2). \(E[N] = 50 \cdot 0.2 = 10\). \(E[X] = 500 - 30E[N] = 200\).
Exercise \(\PageIndex{7}\)
(See Exercise 19 from " Problems on Distribution and Density Functions "). The number of noise pulses arriving on a power circuit in an hour is a random quantity having Poisson (7) distribution. What is the expected number of pulses in an hour?
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\(X\) ~ Poisson (7). \(E[X] = 7\).
Exercise \(\PageIndex{8}\)
(See Exercise 24 and Exercise 25 from " Problems on Distribution and Density Functions "). The total operating time for the units in Exercise 24 is a random variable \(T\) ~ gamma (20, 0.0002). What is the expected operating time?
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\(X\) ~ gamma (20, 0.0002). \(E[X] = 20/0.0002 = 100,000\).
Exercise \(\PageIndex{9}\)
(See Exercise 41 from " Problems on Distribution and Density Functions "). Random variable \(X\) has density function
\(f_X (t) = \begin{cases} (6/5) t^2 & \text{for } 0 \le t \le 1 \\ (6/5)(2 - t) & \text{for } 1 \le t \le 2 \end{cases} = I_{[0, 1]}(t) \dfrac{6}{5} t^2 + I_{(1, 2]} (t) \dfrac{6}{5} (2 - t)\).
What is the expected value \(E[X]\)?
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\(E[X] = \int t f_X(t)\ dt = \dfrac{6}{5} \int_{0}^{1} t^3 \ dt + \dfrac{6}{5} \int_{1}^{2} (2t - t^2)\ dt = \dfrac{11}{10}\)
Exercise \(\PageIndex{10}\)
Truncated exponential. Suppose \(X\) ~ exponential (\(\lambda\)) and \(Y = I_{[0, a]} (X) X + I_{a, \infty} (X) a\).
a. Use the fact that
\(\int_{0}^{\infty} te^{-\lambda t} \ dt = \dfrac{1}{\lambda ^2}\) and \(\int_{a}^{\infty} te^{-\lambda t}\ dt = \dfrac{1}{\lambda ^2} e^{-\lambda t} (1 + \lambda a)\)
to determine an expression for \(E[Y]\).
b. Use the approximation method, with \(\lambda = 1/50\), \(a = 30\). Approximate the exponential at 10,000 points for \(0 \le t \le 1000\). Compare the approximate result with the theoretical result of part (a).
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\(E[Y] = \int g(t) f_X (t)\ dt = \int_{0}^{a} t \lambda e^{-\lambda t} \ dt + aP(X > a) =\)
\(\dfrac{\lambda}{\lambda ^2} [1 - e^{-\lambda a} (1 + \lambda a)] + a e^{-\lambda a} = \dfrac{1}{\lambda} (1 - e^{-\lambda a})\)
tappr Enter matrix [a b] of x-range endpoints [0 1000] Enter number of x approximation points 10000 Enter density as a function of t (1/50)*exp(-t/50) Use row matrices X and PX as in the simple case G = X.*(X<=30) + 30*(X>30); EZ = G8PX' EZ = 22.5594 ez = 50*(1-exp(-30/50)) %Theoretical value ez = 22.5594
Exercise \(\PageIndex{11}\)
(See Exercise 1 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_01.m ). 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. Determine \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\), and \(E[XY]\).
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npr08_01 Data in Pn, P, X, Y jcalc Enter JOINT PROBABILITIES (as on the plane) P Enter row marix of VALUES of X X Enter row marix of VALUES of Y Y Use array operations on matrices X, Y, PX, PY, t, u, and P EX = X*PX' EX = 0.1538 ex = total(t.*P) % Alternate ex = 0.1538 EY = Y*PY' EY = 0.5000 EX2 = (X.^2)*PX' EX2 = 0.1629 EY2 = (Y.^2)*PY' EY2 = 0.6176 EXY = total(t.*u.*P) EXY = 0.0769
Exercise \(\PageIndex{12}\)
(See Exercise 2 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_02.m ). 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 \(\{X, Y\}\) and \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\), and \(E[XY]\).
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npr08_02 Data are in X, Y, Pn, P jcalc ----------------------- EX = X*PX' EX = 0.5000 EY = Y*PY' EY = 0.7500 EX2 = (X.^2)*PX' EX2 = 0.5714 EY2 = (Y.^2)*PY' EY2 = 0.9643 EXY = total(t.*u.*P) EXY = 0.2143
Exercise \(\PageIndex{13}\)
(See Exercise 3 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_03.m ). A die is rolled. Let X be the number of spots that turn 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 expected value \(E[Y]\)
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npr08_03 Data are in X, Y, P, PY jcalc ----------------------- EX = X*PX' EX = 3.5000 EY = Y*PY' EY = 1.7500 EX2 = (X.^2)*PX' EX2 = 15.1667 EY2 = (Y.^2)*PY' EY2 = 4.6667 EXY = total(t.*u.*P) EXY = 7.5833
Exercise \(\PageIndex{14}\)
(See Exercise 4 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_04.m ). As a variation of Exercise, suppose a pair of dice is rolled instead of a single die. Determine the joint distribution for \(\{X, Y\}\) and determine \(E[Y]\).
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npr08_04 Data are in X, Y, P jcalc ----------------------- EX = X*PX' EX = 7 EY = Y*PY' EY = 3.5000 EX2 = (X.^2)*PX' EX2 = 54.8333 EY2 = (Y.^2)*PY' EY2 = 15.4583
Exercise \(\PageIndex{15}\)
(See Exercise 5 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_05.m ). 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 \(\{X,Y\}\) and determine \(E[Y]\)
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npr08_05 Data are in X, Y, P, PY jcalc ----------------------- EX = X*PX' EX = 7.0000 EY = Y*PY' EY = 1.1667
Exercise \(\PageIndex{16}\)
(See Exercise 6 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_06.m ). The pair \(\{X,Y\}\) has the joint distribution:
\(X = \) [-2.3 -0.7 1.1 3.9 5.1] \(Y =\) [1.3 2.5 4.1 5.3]
\(P = \begin{bmatrix} 0.0483 & 0.0357 & 0.0420 & 0.0399 & 0.0441 \\ 0.0437 & 0.0323 & 0.0380 & 0.0361 & 0.0399 \\ 0.0713 & 0.0527 & 0.0620 & 0.0609 & 0.0551 \\ 0.0667 & 0.0493 & 0.0580 & 0.0651 & 0.0589 \end{bmatrix}\)
Determine \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\) and \(E[XY]\).
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npr08_06 Data are in X, Y, P jcalc --------------------- EX = X*PX' EX = 1.3696 EY = Y*PY' EY = 3.0344 EX2 = (X.^2)*PX' EX2 = 9.7644 EY2 = (Y.^2)*PY' EY2 = 11.4839 EXY = total(t.*u.*P) EXY = 4.1423
Exercise \(\PageIndex{17}\)
(See Exercise 7 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_07.m ). The pair \(\{X, Y\}\) has the joint distribution:
\(P(X = t, Y = u)\)
| 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 |
Determine \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\) and \(E[XY]\).
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npr08_07 Data are in X, Y, P jcalc --------------------- EX = X*PX' EX = 0.8590 EY = Y*PY' EY = 1.1455 EX2 = (X.^2)*PX' EX2 = 5.8495 EY2 = (Y.^2)*PY' EY2 = 19.6115 EXY = total(t.*u.*P) EXY = 3.6803
Exercise \(\PageIndex{18}\)
(See Exercise 8 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_08.m ). The pair \(\{X, Y\}\) has the joint distribution:
\(P(X = t, Y = u)\)
| 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 \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\) and \(E[XY]\).
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npr08_08 Data are in X, Y, P jcalc --------------------- EX = X*PX' EX = 10.1000 EY = Y*PY' EY = 3.0016 EX2 = (X.^2)*PX' EX2 = 133.0800 EY2 = (Y.^2)*PY' EY2 = 41.5564 EXY = total(t.*u.*P) EXY = 22.2890
Exercise \(\PageIndex{19}\)
(See Exercise 9 from " Problems On Random Vectors and Joint Distributions ", m-file npr08_09.m ). 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:
\(P(X = t, Y = u)\)
| 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 \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\) and \(E[XY]\).
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npr08_09 Data are in X, Y, P jcalc --------------------- EX = X*PX' EX = 1.9250 EY = Y*PY' EY = 2.8050 EX2 = (X.^2)*PX' EX2 = 4.0375 EY2 = (Y.^2)*PY' EXY = total(t.*u.*P) EY2 = 8.9850 EXY = 5.1410
For the joint densities in Exercise 20-32 below
a. Determine analytically \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\) and \(E[XY]\).
b. Use a discrete approximation for \(E[X]\), \(E[Y]\), \(E[X^2]\), \(E[Y^2]\) and \(E[XY]\).
Exercise \(\PageIndex{20}\)
(See Exercise 10 from " Problems On Random Vectors and Joint Distributions "). \(f_{XY}(t, u) = 1\) for \(0 \le t \le 1\). \(0 \le u \le 2(1-t)\).
\(f_X(t) = 2(1 -t)\), \(0 \le t \le 1\), \(f_Y(u) = 1 - u/2\), \(0 \le u \le 2\)
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\(E[X] = \int_{0}^{1} 2t(1 - t)\ dt = 1/3\), \(E[Y] = 2/3\), \(E[X^2] = 1/6\), \(E[Y^2] = 2/3\)
\(E[XY] = \int_{0}^{1} \int_{0}^{2(1-t)} tu\ dudt = 1/6\)
tuappr: [0 1] [0 2] 200 400 u<=2*(1-t) EX = 0.3333 EY = 0.6667 EX2 = 0.1667 EY2 = 0.6667 EXY = 0.1667 (use t, u, P)
Exercise \(\PageIndex{21}\)
(See Exercise 11 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = 1/2\) on the square with vertices at (1, 0), (2, 1) (1, 2), (0, 1).
\(f_{X} (t) = f_{Y} (t) = I_{[0, 1]} (t) t + I_{(1, 2]} (t) (2 - t)\)
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\(E[X] = E[Y] = \int_{0}^{1} t^2 \ dt + \int_{1}^{t} (2t - t^2) \ dt = 1\), \(E[X^2] = E[Y^2] = 7/6\)
\(E[XY] = (1/2) \int_{0}^{1} \int_{1 - t}^{1 + t} dt dt + (1/2) \int_{1}^{2} \int_{t - 1}^{3 - t} du dt = 1\)
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
Exercise \(\PageIndex{22}\)
(See Exercise 12 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = 4t (1 - u)\) for \(0 \le t \le 1\). \(0 \le u \le 1\)
\(f_X (t) = 2t\), \(0 \le t \le 1\), \(f_Y(u) = 2(1 - u)\), \(0 \le u \le 1\)
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\(E[X] = 2/3\), \(E[Y] = 1/3\), \(E[X^2] = 1/2\), \(E[Y^2] = 1/6\), \(E[XY] = 2/9\)
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
Exercise \(\PageIndex{23}\)
(See Exercise 13 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = \dfrac{1}{8} (t + u)\) for \(0 \le t \le 2\), \(0 \le u \le 2\)
\(f_{X} (t) = f_{Y} (t) = \dfrac{1}{4} (t + 1)\), \(0 \le t \le 2\)
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\(E[X] = E[Y] = \dfrac{1}[4} \int_{0}^{2} (t^2 + t) \ dt = \dfrac{7}{6}\), \(E[X^2] = E[Y^2] = 5/3\)
\(E[XY] = \dfrac{1}{8} \int_{0}^{2} \int_{0}^{2} (t^2u + tu^2) \ dudt = \dfrac{4}{3}\)
tuappr: [0 1] [0 1] 200 200 4*t.*(1-u) EX = 1.1667 EY = 1.1667 EX2 = 1.6667 EY2 = 1.6667 EXY = 1.3333
Exercise \(\PageIndex{24}\)
(See Exercise 14 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = 4ue^{-2t}\) for \(0 \le t, 0 \le u \le 1\)
\(f_X (t) = 2e^{-2t}\), \(0 \le t\), \(f_Y(u) = 2u\), \(0 \le u \le 1\)
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\(E[X] = \int_{0}^{\infty} 2te^{-2t} \ dt = \dfrac{1}{2}\), \(E[Y] = \dfrac{2}{3}\), \(E[X^2] = \dfrac{1}{2}\), \(E[Y^2] = \dfrac{1}{2}\), \(E[XY] = \dfrac{1}{3}\)
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
Exercise \(\PageIndex{25}\)
(See Exercise 15 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1 + t\).
\(f_X(t) = \dfrac{3}{88} (1 + t) (1 + 4t + t^2) = \dfrac{3}{88} (1 + 5t + 5t^2 + t^3)\), \(0 \le t \le 2\)
\(f_Y(t) = I_{[0, 1]} (u) \dfrac{3}{88} (6u^2 + 4) + I_{(1, 3]} (u) \dfrac{3}{88} (3 + 2u + 8u^2 - 3u^3)\)
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\(E[X] = \dfrac{313}{220}\), \(E[Y] = \dfrac{1429}{880}\), \(E[X^2] = \dfrac{49}{22}\), \(E[Y^2] = \dfrac{172}{55}\), \(E[XY] = \dfrac{2153}{880}\)
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
Exercise \(\PageIndex{26}\)
(See Exercise 16 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = 12t^2 u\) on the parallelogram with vertices
(-1, 0), (0, 0), (1, 1), (0, 1)
\(f_X(t) = I_{[-1, 0]} (t) 6t^2 (t + 1)^2 + I_{(0, 1]} (t) 6t^2 (1 - t^2)\), \(f_Y(u) 12u^3 - 12u^2 + 4u\), \(0 \le u \le 1\)
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\(E[X] = \dfrac{2}{5}\), \(E[Y] = \dfrac{11}{15}\), \(E[X^2] = \dfrac{2}{5}\), \(E[Y^2] = \dfrac{3}{5}\), \(E[XY] = \dfrac{2}{5}\)
tuappr: [-1 1] [0 1] 400 300 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
Exercise \(\PageIndex{27}\)
(See Exercise 17 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1, 2-t\}\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{11}t + I_{(1, 2]} (t) \dfrac{12}{11} t (2 - t)^2\), \(f_Y(u) = \dfrac{12}{11} u(u - 2)^2\), \(0 \le u \le 1\)
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\(E[X] = \dfrac{52}{55}\), \(E[Y] = \dfrac{32}{55}\), \(E[X^2] = \dfrac{57}{55}\), \(E[Y^2] = \dfrac{2}{5}\), \(E[XY] = \dfrac{28}{55}\)
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
Exercise \(\PageIndex{28}\)
(See Exercise 18 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{max } \{2 - t, t\}\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{6}{23} (2 - t) + I_{(1, 2]} (t) \dfrac{6}{23} t^2\), \(f_Y(u) = I_{[0, 1]} (u) \dfrac{6}{23} (2u + 1) + I_{(1, 2]} (u) \dfrac{3}{23} (4 + 6u - 4u^2)\)
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\(E[X] = \dfrac{53}{46}\), \(E[Y] = \dfrac{22}{23}\), \(E[X^2] = \dfrac{397}{230}\), \(E[Y^2] = \dfrac{261}{230}\), \(E[XY] = \dfrac{251}{230}\)
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
Exercise \(\PageIndex{29}\)
(See Exercise 19 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2, 3 - t\}\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{24}{179} (3t^2 + 1) + I_{(1, 2]} (t) \dfrac{6}{179} (9 - 6t + 19t^2 - 6t^3)\)
\(f_Y (u) = I_{[0, 1]} (t) \dfrac{24}{179} (4 + u) + I_{(1, 2]} (t) \dfrac{12}{179} (27 - 24u + 8u^2 - u^3)\)
- Answer
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\(E[X] = \dfrac{2313}{1790}\), \(E[Y] = \dfrac{778}{895}\), \(E[X^2] = \dfrac{1711}{895}\), \(E[Y^2] = \dfrac{916}{895}\), \(E[XY] = \dfrac{1811}{1790}\)
tuappr: [0 2] [0 2] 400 400 (12/179)*(3*t.^2 + u).*(u<=min(2,3-t)) EX = 1.2923 EY = 0.8695 EX2 = 1.9119 EY2 = 1.0239 EXY = 1.0122
Exercise \(\PageIndex{30}\)
(See Exercise 20 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = \dfrac{12}{227} (3t + 2tu)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1 + t, 2\}\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{227} (t^3 + 5t^2 + 4t) + I_{(1, 2]} (t) \dfrac{120}{227} t\)
\(f_Y (u) = I_{[0, 1]} (t) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (2u + 3) (3 + 2u - u^2)\)
\( = I_{[0, 1]} (u) \dfrac{24}{227} (2u + 3) + I_{(1, 2]} (u) \dfrac{6}{227} (9 + 12u + u^2 - 2u^3)\)
- Answer
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\(E[X] = \dfrac{1567}{1135}\), \(E[Y] = \dfrac{2491}{2270}\), \(E[X^2] = \dfrac{476}{227}\), \(E[Y^2] = \dfrac{1716}{1135}\), \(E[XY] = \dfrac{5261}{3405}\)
tuappr: [0 2] [0 2] 400 400 (12/227)*(3*t + 2*t.*u).*(u<=min(1+t,2)) EX = 1.3805 EY = 1.0974 EX2 = 2.0967 EY2 = 1.5120 EXY = 1.5450
Exercise \(\PageIndex{31}\)
(See Exercise 21 from " Problems On Random Vectors and Joint Distribution "). \(f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2t, 3-t\}\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{13} t^2 + I_{(1, 2]} (t) \dfrac{6}{13} (3 - t)\)
\(f_Y(u) = I_{[0, 1]} (u) (\dfrac{4}{13} + \dfrac{8}{13} u - \dfrac{9}{52} u^2) + I_{(1, 2]} (u) (\dfrac{9}{13} + \dfrac{6}{13} u - \dfrac{51}{52} u^2)\)
- Answer
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\(E[X] = \dfrac{16}{13}\), \(E[Y] = \dfrac{11}{12}\), \(E[X^2] = \dfrac{219}{130}\), \(E[Y^2] = \dfrac{83}{78}\), \(E[XY] = \dfrac{431}{390}\)
tuappr: [0 2] [0 2] 400 400 (2/13)*(t + 2*u).*(u<=min(2*t,3-t)) EX = 1.2309 EY = 0.9169 EX2 = 1.6849 EY2 = 1.0647 EXY = 1.1056
Exercise \(\PageIndex{32}\)
(See Exercise 22 from " Problems On Random Vectors and Joint Distribution ").
\(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2 u^2\), for \(0 \le u \le 1\).
\(f_X(t) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 1) + I_{(1, 2]} (t) \dfrac{3}{14} t^2\), \(f_Y(u) = \dfrac{1}{8} + \dfrac{3}{4} u + \dfrac{3}{2} u^2\) (0 \le u \le 1\)
- Answer
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\(E[X] = \dfrac{243}{224}\), \(E[Y] = \dfrac{11}{16}\), \(E[X^2] = \dfrac{107}{70}\), \(E[Y^2] = \dfrac{127}{240}\), \(E[XY] = \dfrac{347}{448}\)
tuappr: [0 2] [0 1] 400 200 (3/8)*(t.^2 + 2*u).*(t<=1) + (9/14)*(t.^2.*u.^2).*(t > 1) EX = 1.0848 EY = 0.6875 EX2 = 1.5286 EY2 = 0.5292 EXY = 0.7745
Exercise \(\PageIndex{33}\)
The class \(\{X, Y, Z\}\) of random variables is iid(independent, identically distributed) with common distribution
\(X =\) [-5 -1 3 4 7] \(PX =\) 0.01 * [15 20 30 25 10]
Let \(W = 3X - 4Y + 2Z\). Determine \(E[W]\). Do this using icalc, then repeat with icalc3 and compare results.
- Answer
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Use \(x\) and \(px\) to prevent renaming.
x = [-5 -1 3 4 7]; px = 0.01*[15 20 30 25 10]; icalc Enter row matrix of X-values x Enter row matrix of Y-values x Enter X probabilities px Enter Y probabilities px Use array operations on matrices X, Y, PX, PY, t, u, and P G = 3*t - 4*u [R,PR] = csort(G,P); icalc Enter row matrix of X-values R Enter row matrix of Y-values x Enter X probabilities PR Enter Y probabilities px Use array operations on matrices X, Y, PX, PY, t, u, and P H = t + 2*u; EH = total(H.*P) EH = 1.6500 [W,PW] = csort(H,P); % Alternate EW = W*PW' EW = 1.6500 icalc3 % Solution with icalc3 Enter row matrix of X-values x Enter row matrix of Y-values x Enter row matrix of Z-values x Enter X probabilities px Enter Y probabilities px Enter Z probabilities px Use array operations on matrices X, Y, Z, PX, PY, PZ, t, u, v, and P K = 3*t - 4*u + 2*v; EK = total(K.*P) EK = 1.6500
Exercise \(\PageIndex{34}\)
(See Exercise 5 from " Problems on Functions of Random Variables ") The cultural committee of a student organization has arranged a special deal for tickets to a concert. The agreement is that the organization will purchase ten tickets at $20 each (regardless of the number of individual buyers). Additional tickets are available according to the following schedule:
11-20, $18 each; 21-30 $16 each; 31-50, $15 each; 51-100, $13 each
If the number of purchasers is a random variable \(X\), the total cost (in dollars) is a random quantity \(Z = g(X)\) described by
\(g(X) = 200 + 18I_{M1} (X) (X - 10) + (16 - 18) I_{M2} (X) (X - 20) +\)
\((15 - 16) I_{M3} (X) (X - 30) + (13 - 15) I_{M4} (X) (X - 50)\)
where \(M1 = [10, \infty)\), \(M2 = [20, \infty)\), \(M3 = [30, \infty)\), \(M4 = [50, \infty)\)
Suppose \(X\) ~ Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine \(E[Z]\) and \(E[Z^2]\).
- Answer
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X = 0:150; PX = ipoisson(75, X); G = 200 + 18*(X - 10).*(X>=10) + (16 - 18)*(X - 20).*(X>=20) + ... (15 - 16)*(X - 30).*(X>=30) + (13 - 15)*(X>=50); [Z,PZ] = csort(G,PX); EZ = Z*PZ' EZ = 1.1650e+03 EZ2 = (Z.^2)*PZ' EZ2 = 1/3699e+06
Exercise \(\PageIndex{35}\)
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 \(Z = g(X, Y) = 3X^2 + 2XY - Y^2)\). Determine \(E[Z]\) and \(E[Z^2]\).
- Answer
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npr08_07 Data are in X, Y, P jcalc ------------------ G = 3*t.^2 + 2*t.*u - u.^2; EG = total(G.*P) EG = 5.2975 ez2 = total(G.^2.*P) EG2 = 1.0868e+03 [Z,PZ] = csort(G,P); % Alternate EZ = Z*PZ' EZ = 5.2975 EZ2 = (Z.^2)*PZ' EZ2 = 1.0868e+03
Exercise \(\PageIndex{36}\)
For the pair \(\{X, Y\}\) in Exercise 11.3.35, let
\(W = g(X, Y) = \begin{cases} X & \text{for } X + Y \le 4 \\ 2Y & \text{for } X+Y > 4 \end{cases} = I_M (X, Y) X + I_{M^c} (X, Y)2Y\)
Determine \(E[W]\) and \(E[W^2]\).
- Answer
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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
For the distribution in Exercises 37-41 below
a. Determine analytically \(E[Z]\) and \(E[Z^2]\)
b. Use a discrete approximation to calculate the same quantities.
Exercise \(\PageIndex{37}\)
\(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1+t\) (see Exercise 25).
\(Z = I_{[0, 1]} (X)4X + I_{(1,2]} (X)(X+Y)\)
- Answer
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\(E[Z] = \dfrac{3}{88} \int_{0}^{1} \int_{0}^{1 + t} 4t (2t + 3u^2)\ dudt + \dfrac{3}{88} \int_{1}^{2} \int_{0}^{1 + t} (t + u) (2t + 3u^2)\ dudt = \dfrac{5649}{1760}\)
\(E[Z^2] = \dfrac{3}{88} \int_{0}^{1} \int_{0}^{1 + t} (4t)^2 (2t + 3u^2)\ dudt + \dfrac{3}{88} \int_{1}^{2} \int_{0}^{1 + t} (t + u)^2 (2t + 3u^2)\ dudt = \dfrac{4881}{440}\) -
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
Exercise \(\PageIndex{38}\)
\(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1, 2 - t\}\) (see Exercise 27)
\(Z = I_M(X, Y) \dfrac{1}{2}X + I_{M^c} (X, Y) Y^2\), \(M = \{(t, u) : u > t\}\)
- Answer
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\(E[Z] = \dfrac{12}{11} \int_{0}^{1} \int_{t}^{1} t^2u\ dudt + \dfrac{24}{11} \int_{0}^{1} \int_{0}^{t} tu^3\ dudt + \dfrac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} tu^3\ dudt = \dfrac{16}{55}\)
\(E[Z^2] = \dfrac{6}{11} \int_{0}^{1} \int_{t}^{1} t^3u\ dudt + \dfrac{24}{11} \int_{0}^{1} \int_{0}^{t} tu^5\ dudt + \dfrac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} tu^5\ dudt = \dfrac{39}{308}\)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
Exercise \(\PageIndex{39}\)
\(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{max } \{2 - t, t\}\) (see Exercise 28)
\(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y)2Y\), \(M = \{(t, u): \text{max } (t, u) \le 1\}\)
- Answer
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\(E[Z] = \dfrac{3}{23} \int_{0}^{1} \int_{0}^{1} (t + u) (t + 2u) \ dudt + \dfrac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 2u (1 + 2u)\ dudt + \dfrac{3}{23} \int_{1}^{2} \int_{1}^{t} 2u (t + 2u)\ dudt = \dfrac{175}{92}\)
\(E[Z^2] = \dfrac{3}{23} \int_{0}^{1} \int_{0}^{1} (t + u)^2 (t + 2u) \ dudt + \dfrac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 4u^2 (1 + 2u)\ dudt + \dfrac{3}{23} \int_{1}^{2} \int_{1}^{t} 4u^2 (t + 2u)\ dudt = \)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 = 1.9048 EZ2 = total(G.^2.*P) EZ2 = 4.4963
Exercise \(\PageIndex{40}\)
\(f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2, 3-t\}\) (see Exercise 19)
\(Z = I_M (X,Y) (X + Y) + I_{M^c} (X, Y) 2Y^2\), \(M = \{(t, u): t \le 1, u \ge 1\}\)
- Answer
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\(E[Z] = \dfrac{12}{179} \int_{0}^{1} \int_{1}^{2} (t + u) (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{0}^{1} \int_{0}^{1} 2u^2 (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 2u^2 (3t^2 + u)\ dudt = \dfrac{1422}{895}\)
\(E[Z^2] = \dfrac{12}{179} \int_{0}^{1} \int_{1}^{2} (t + u)^2 (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{0}^{1} \int_{0}^{1} 4u^4 (3t^2 + u)\ dudt + \dfrac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 4u^4 (3t^2 + u)\ dudt = \dfrac{28296}{6265}\)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 = 1.5898 EZ2 = total(G.^2.*P) EZ2 = 4.5224
Exercise \(\PageIndex{41}\)
\(f_{XY} (t, u) = \dfrac{12}{227} (2t + 2tu)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1 + t, 2\}\) (see Exercise 30).
\(Z = I_M (X, Y) X + I_{M^c} (X, Y) XY\), \(M = \{(t, u): u \le \text{min } (1, 2 - t)\}\)
- Answer
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\(E[Z] = \dfrac{12}{227} \int_{0}^{1} \int_{0}^{1} t (3t + 2tu) \ dudt + \dfrac{12}{227} \int_{1}^{2} \int_{0}^{2 - t} t(3t + 2tu)\ dudt +\)
\(\dfrac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} tu(3t + 2tu)\ dudt + \dfrac{12}{227} \int_{1}^{2} \int_{2 - t}^{2} tu (3t + 2tu)\ dudt = \dfrac{5774}{3405}\)
\(E[Z^2] = \dfrac{56673}{15890}\) -
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 = 1.6955 EZ2 = total(G.^2.*P) EZ2 = 3.5659
Exercise \(\PageIndex{42}\)
The class \(\{X, Y, Z\}\) is independent. (See Exercise 16 from "Problems on Functions of Random Variables", m-file npr10_16.m)
\(X = -2I_A + I_B + 3I_C\). Minterm probabilities are (in the usual order)
0.255 0.025 0.375 0.045 0.108 0.012 0.162 0.018
\(Y = I_D + 3I_E + I_F - 3\). 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 |
\(W = X^2 + 3XY^2 - 3Z\). Determine \(E[W]\) and \(E[W^2]\).
- Answer
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npr10_16 Data are in cx, pmx, cy, pmy, Z, PZ [X,PX] = canonicf(cx,pmx); [Y,PY] - canonicf(cy,pmy); icalc3 input: X, Y, Z, PX, PY, PZ ------------- Use array operations on matrices X, Y, Z. PX, PY, PZ, t, u, v, and P G = 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