12.4: Problems on Variance, Covariance, Linear Regression
Exercise \(\PageIndex{1}\)
(See Exercise 1 from " Problems on Distribution and Density Functions ", and Exercise 1 from " Problems on Mathematical Expectation ", 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 \(\text{Var} [X]\).
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npr07_01 Data are in T and pc EX = T*pc' EX = 2.7000 VX = (T.^2)*pc' - EX^2 VX = 1.5500 [X,PX] = csort(T,pc); % Alternate Ex = X*PX' Ex = 2.7000 Vx = (X.^2)*PX' - EX^2 Vx = 1.5500
Exercise \(\PageIndex{2}\)
(See Exercise 2 from " Problems on Distribution and Density Functions ", and Exercise 2 from " Problems on Mathematical Expectation ", 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
\(X = 3.5 I_{C_1} + 5.0 I_{C_2} + 3.5 I_{C_3} + 7.5 I_{C_4} + 5.0 I_{C_5} + 5.0 I_{C_6} + 3.5 I_{C_7} + 7.5 I_{C_8}\)
Determine \(\text{Var} [X]\).
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npr07_02 Data are in T, pc EX = T*pc'; VX = (T.^2)*pc' - EX^2 VX = 2.8525
Exercise \(\PageIndex{3}\)
(See Exercise 12 from " Problems on Random Variables and Probabilities ", Exercise 3 from " Problems on Mathematical Expectation ", 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]
Consider \(X = I_A + I_B + I_C + I_D\), which counts the number of these events which occur on a trial. Determine \(\text{Var} [X]\).
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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 VX = (X.^2)*PX' - (X*PX')^2 VX = 0.7309
Exercise \(\PageIndex{4}\)
(See Exercise 4 from " Problems on Mathematical Expectation "). In a thunderstorm in a national park there are 127 lightning strikes. Experience shows that the probability of each lightning strike starting a fire is about 0.0083. Determine \(\text{Var} [X]\).
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\(X\) ~ binomial (127, 0.0083). \(\text{Var} [X] = 127 \cdot 0.0083 \cdot (1-0.0083) = 1.0454\).
Exercise \(\PageIndex{5}\)
(See Exercise 5 from " Problems on Mathematical Expectation "). Two coins are flipped twenty times. Let \(X\) be the number of matches (both heads or both tails). Determine \(\text{Var} [X]\).
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\(X\) ~ binomial (20, 1/2). \(\text{Var}[X] = 20 \cdot (1/2)^2 = 5\).
Exercise \(\PageIndex{6}\)
(See Exercise 6 from " Problems on Mathematical Expectation "). 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
\(X = 50 \cdot 10 - 30 N\), where \(N\) is the number of winners
Determine \(\text{Var} [X]\).
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\(N\) ~ binomial (50, 0.2). \(\text{Var}[N] = 50 \cdot 0.2 \cdot 0.8 = 8\). \(\text{Var} [X] = 30^2\ \text{Var} [N] = 7200\).
Exercise \(\PageIndex{7}\)
(See Exercise 7 from " Problems on Mathematical Expectation "). The number of noise pulses arriving on a power circuit in an hour is a random quantity \(X\) having Poisson (7) distribution. Determine \(\text{Var} [X]\).
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\(X\) ~ Poisson (7). \(\text{Var} [X] = \mu = 7\).
Exercise \(\PageIndex{8}\)
(See Exercise 24 from " Problems on Distribution and Density Functions ", and Exercise 8 from " Problems on Mathematical Expectation "). The total operating time for the units in Exercise 24 from " Problems on Distribution and Density Functions " is a random variable \(T\) ~ gamma (20, 0.0002). Determine \(\text{Var} [T]\).
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\(T\) ~ gamma (20, 0.0002). \(\text{Var}[T] = 20/0.0002^2 = 500,000,000\).
Exercise \(\PageIndex{9}\)
The class \(\{A, B, C, D, E, F\}\) is independent, with respective probabilities
0.43, 0.53, 0.46, 0.37, 0.45, 0.39. Let
\(X = 6 I_A + 13 I_B - 8I_C\), \(Y = -3I_D + 4 I_E + I_F - 7\)
a. Use properties of expectation and variance to obtain \(E[X]\), \(\text{Var} [X]\), \(E[Y]\), and \(\text{Var}[Y]\). Note that it is not necessary to obtain the distributions for \(X\) or \(Y\).
b. Let \(Z = 3Y - 2X\).
Determine \(E[Z]\), and \(\text{Var} [Z]\).
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cx = [6 13 -8 0]; cy = [-3 4 1 -7]; px = 0.01*[43 53 46 100]; py = 0.01*[37 45 39 100]; EX = dot(cx,px) EX = 5.7900 EY = dot(cy,py) EY = -5.9200 VX = sum(cx.^2.*px.*(1-px)) VX = 66.8191 VY = sum(cy.^2.*py.*(1-py)) VY = 6.2958 EZ = 3*EY - 2*EX EZ = -29.3400 VZ = 9*VY + 4*VX VZ = 323.9386
Exercise \(\PageIndex{10}\)
Consider \(X = -3.3 I_A - 1.7 I_B + 2.3 I_C + 7.6 I_D - 3.4\). The class \(\{A, B, C, D\}\) has minterm probabilities (data are in m-file npr12_10.m )
\(\text{pmx} =\) [0.0475 0.0725 0.0120 0.0180 0.1125 0.1675 0.0280 0.0420 \(\cdot\cdot\cdot\)
0.0480 0.0720 0.0130 0.0170 0.1120 0.1680 0.0270 0.0430]
a. Calculate \(E[X]\) and \(\text{Var} [X]\).
b. Let \(W = 2X^2 - 3X + 2\).
Calculate \(E[W]\) and \(\text{Var} [W]\)
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npr12_10 Data are in cx, cy, pmx and pmy canonic Enter row vector of coefficients cx Enter row vector of minterm probabilities pmx Use row matrices X and PX for calculations Call for XDBN to view the distribution EX = dot(X,PX) EX = -1.2200 VX = dot(X.^2,PX) - EX^2 VX = 18.0253 G = 2*X.^2 - 3*X + 2; [W,PW] = csort(G,PX); EW = dot(W,PW) EW = 44.6874 VW = dot(W.^2,PW) - EW^2 VW = 2.8659e+03
Exercise \(\PageIndex{11}\)
Consider a second random variable \(Y = 10 I_E + 17 I_F + 20 I_G - 10\) in addition to that in Exercise 12.4.10. The class \(\{E, F, G\}\) has minterm probabilities (in mfile npr12_10.m )
\(\text{pmy} =\) [0.06 0.14 0.09 0.21 0.06 0.14 0.09 0.21]
The pair \(\{X, Y\}\) is independent.
a. Calculate \(E[Y]\) and \(\text{Var} [Y]\).
b. Let \(Z = X^2 + 2XY - Y\).
Calculate \(E[Z]\) and \(\text{Var} [Z]\).
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(Continuation of Exercise 12.4.10)
[Y,PY] = canonicf(cy,pmy); EY = dot(Y,PY) EY = 19.2000 VY = dot(Y.^2,PY) - EY^2 VY = 178.3600 icalc Enter row matrix of X-values X Enter row matrix of Y-values Y Enter X probabilities PX Enter Y probabilities PY Use array operations on matrices X, Y, PX, PY, t, u, and P H = t.^2 + 2*t.*u - u; [Z,PZ] = csort(H,P); EZ = dot(Z,PZ) EZ = -46.5343 VZ = dot(Z.^2,PZ) - EZ^2 VZ = 3.7165e+04
Exercise \(\PageIndex{12}\)
Suppose the pair \(\{X, Y\}\) is independent, with \(X\) ~ gamma (3, 0.1) and
\(Y\) ~ Poisson (13). Let \(Z = 2X - 5Y\). Determine \(E[Z]\) and \(\text{Var} [Z]\).
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\(X\) ~ gamma (3, 0.1) implies \(E[X] = 30\) and \(\text{Var} [X] = 300.\) \(Y\) ~ Poisson (13) implies \(E[Y] = \text{Var} [Y] = 13\). Then
\(E[Z] = 2\cdot 30 - 5 \cdot 13 = -5\), \(\text{Var}[Z] = 4 \cdot 300 + 25 \cdot 13 = 1525\).
Exercise \(\PageIndex{13}\)
The pair \(\{X, Y\}\) is jointly distributed with the following parameters:
\(E[X] = 3\), \(E[Y] = 4\), \(E[XY] = 15\), \(E[X^2] = 11\), \(\text{Var} [Y] = 5\)
Determine \(\text{Var} [3X - 2Y]\).
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EX = 3; EY = 4; EXY = 15; EX2 = 11; VY = 5; VX = EX2 - EX^2 VX = 2 CV = EXY - EX*EY CV = 3 VZ = 9*VX + 4*VY - 6*2*CV VZ = 2
Exercise \(\PageIndex{14}\)
The class \(\{A, B, C, D, E, F\}\) is independent, with respective probabilities
0.47, 0.33, 0.46, 0.27, 0.41, 0.37
Let
\(X = 8I_A + 11 I_B - 7I_C\), \(Y = -3I_D + 5I_E + I_F - 3\), and \(Z = 3Y - 2X\)
a. Use properties of expectation and variance to obtain \(E[X]\), \(\text{Var} [X]\), \(E[Y]\), and \(\text{Var}[Y]\).
b. Determine \(E[Z]\), and \(\text{Var} [Z]\).
c. Use appropriate m-programs to obtain \(E[X]\), \(\text{Var} [X]\), \(E[Y]\), \(\text{Var} [Y]\), \(E[Z]\), and \(\text{Var} [Z]\). Compare with results of parts (a) and (b).
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px = 0.01*[47 33 46 100]; py = 0.01*[27 41 37 100]; cx = [8 11 -7 0]; cy = [-3 5 1 -3]; ex = dot(cx,px) ex = 4.1700 ey = dot(cy,py) ey = -1.3900 vx = sum(cx.^2.*px.*(1 - px)) vx = 54.8671 vy = sum(cy.^2.*py.*(1-py)) vy = 8.0545 [X,PX] = canonicf(cx,minprob(px(1:3))); [Y,PY] = canonicf(cy,minprob(py(1:3))); icalc Enter row matrix of X-values X Enter row matrix of Y-values Y Enter X probabilities PX Enter Y probabilities PY Use array operations on matrices X, Y, PX, PY, t, u, and P EX = dot(X,PX) EX = 4.1700 EY = dot(Y,PY) EY = -1.3900 VX = dot(X.^2,PX) - EX^2 VX = 54.8671 VY = dot(Y.^2,PY) - EY^2 VY = 8.0545 EZ = 3*EY - 2*EX EZ = -12.5100 VZ = 9*VY + 4*VX VZ = 291.9589
Exercise \(\PageIndex{15}\)
For the Beta (\(r, s\)) distribution.
a. Determine \(E[X^n]\), where \(n\) is a positive integer.
b. Use the result of part (a) to determine \(E[X]\) and \(\text{Var} [X]\).
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\(E[X^n] = \dfrac{\Gamma (r + s)}{\Gamma (r) \Gamma (s)} \int_0^1 t^{r + n - 1} dt = \dfrac{\Gamma (r + s)}{\Gamma (r) \Gamma (s)} \cdot \dfrac{\Gamma (r + n) \Gamma (s)}{\Gamma (r + s + n)} =\)
\(\dfrac{\Gamma (r + n) \Gamma (r + s)}{\Gamma (r + s + n) \Gamma (r)}\)
Using \(\Gamma (x + 1) = x \Gamma (x)\) we have
\(E[X] = \dfrac{r}{r + s}\), \(E[X^2] = \dfrac{r(r + 1)}{(r + s) (r + s + 1)}\)
Some algebraic manipulations show that
\(\text{Var} [X] = E[X^2] - E^2[X] = \dfrac{rs} {(r + s)^2 (r + s + 1)}\)
Exercise \(\PageIndex{16}\)
The pair \(\{X, Y\}\) has joint distribution. Suppose
\(E[X] = 3\), \(E[X^2] = 11\), \(E[Y] = 10\), \(E[Y^2] = 101\), \(E[XY] = 30\)
Determine \(\text{Var} [15X - 2Y]\).
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EX = 3; EX2 = 11; EY = 10; EY2 = 101; EXY = 30; VX = EX2 - EX^2 VX = 2 VY = EY2 - EY^2 VY = 1 CV = EXY - EX*EY CV = 0 VZ = 15^2*VX + 2^2*VY VZ = 454
Exercise \(\PageIndex{17}\)
The pair \(\{X, Y\}\) has joint distribution. Suppose
\(E[X] = 2\), \(E[X^2] = 5\), \(E[Y] = 1\), \(E[Y^2] = 2\), \(E[XY] = 1\)
Determine \(\text{Var} [3X + 2Y]\).
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EX = 2; EX2 = 5; EY = 1; EY2 = 2; EXY = 1; VX = EX2 - EX^2 VX = 1 VY = EY2 - EY^2 VY = 1 CV = EXY - EX*EY CV = -1 VZ = 9*VX + 4*VY + 2*6*CV VZ = 1
Exercise \(\PageIndex{18}\)
The pair \(\{X, Y\}\) is independent, with
\(E[X] = 2\), \(E[Y] = 1\), \(\text{Var} [X] = 6\), \(\text{Var} [Y] = 4\)
Let \(Z = 2X^2 + XY^2 - 3Y + 4\).
Determine \(E[Z]\).
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EX = 2; EY = 1; VX = 6; VY = 4; EX2 = VX + EX^2 EX2 = 10 EY2 = VY + EY^2 EY2 = 5 EZ = 2*EX2 + EX*EY2 - 3*EY + 4 EZ = 31
Exercise \(\PageIndex{19}\)
(See Exercise 9 from " Problems on Mathematical Expectation "). Random variable X has density function
\(f_X (t) = \begin{cases} (6/5) t^2 & \text{for } 0 \le t \le 1 \\ (6/5)(2 - t) & \text{for } 1 < t \le 2 \end{cases} = I_{[0, 1]} (t) \dfrac{6}{5} t^2 + I_{(1, 2]} (t) \dfrac{6}{5} (2 - t)\)
\(E[X] = 11/10\). Determine \(\text{Var} [X]\).
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\(E[X^2] = \int t^2 f_X (t)\ dt = \dfrac{6}{5} \int_0^1 t^4\ dt + \dfrac{6}{5} \int_1^2 (2t^2 - t^3)\ dt = \dfrac{67}{50}\)
\(\text{Var} [X] = E[X^2] - E^2[X] = \dfrac{13}{100}\)
For the distributions in Exercises 20-22
Determine \(\text{Var} [X]\), \(\text{Cov} [X, Y]\), and the regression line of \(Y\) on \(X\).
Exercise \(\PageIndex{20}\)
(See Exercise 7 from " Problems On Random Vectors and Joint Distributions ", and Exercise 17 from " Problems on Mathematical Expectation "). The pair \(\{X, Y\}\) has the joint distribution (in 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 |
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npr08_07 Data are in X, Y, P jcalc - - - - - - - - - - - EX = dot(X,PX); EY = dot(Y,PY); VX = dot(X.^2,PX) - EX^2 VX = 5.1116 CV = total(t.*u.*P) - EX*EY CV = 2.6963 a = CV/VX a = 0.5275 b = EY - a*EX b = 0.6924 % Regression line: u = at + b
Exercise \(\PageIndex{21}\)
(See Exercise 8 from " Problems On Random Vectors and Joint Distributions ", and Exercise 18 from " Problems on Mathematical Expectation "). The pair \(\{X, Y\}\) has the joint distribution (in file npr08_08.m ):
\(P(X = t, Y = u)\)
| 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 |
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npr08_08 Data are in X, Y, P jcalc - - - - - - - - - - - - EX = dot(X,PX); EY = dot(Y,PY); VX = dot(X.^2,PX) - EX^2 VX = 31.0700 CV = total(t.*u.*P) - EX*EY CV = -8.0272 a = CV/VX a = -0.2584 b = EY - a*EX b = 5.6110 % Regression line: u = at + b
Exercise \(\PageIndex{22}\)
(See Exercise 9 from " Problems On Random Vectors and Joint Distributions ", and Exercise 19 from " Problems on Mathematical Expectation "). 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 (in file npr08_09.m ):
\(P(X = t, Y = u)\)
| 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 |
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npr08_09 Data are in X, Y, P jcalc - - - - - - - - - - - - EX = dot(X,PX); EY = dot(Y,PY); VX = dot(X.^2,PX) - EX^2 VX = 0.3319 CV = total(t.*u.*P) - EX*EY CV = -0.2586 a = CV/VX a = -0.77937/6; b = EY - a*EX b = 4.3051 % Regression line: u = at + b
For the joint densities in Exercises 23-30 below
- Determine analytically \(\text{Var} [X]\) \(\text{Cov} [X, Y]\), and the regression line of \(Y\) on \(X\).
- Check these with a discrete approximation.
Exercise \(\PageIndex{23}\)
(See Exercise 10 from " Problems On Random Vectors and Joint Distributions ", and Exercise 20 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = 1\) for \(0 \le t \le 1\), \(0 \le u \le 2(1 - t)\).
\(E[X] = \dfrac{1}{3}\), \(E[X^2] = \dfrac{1}{6}\), \(E[Y] = \dfrac{2}{3}\)
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\(E[XY] = \int_{0}^{1} \int_{0}^{2(1-t)} tu\ dudt = 1/6\)
\(\text{Cov} [X, Y] = \dfrac{1}{6} - \dfrac{1}{3} \cdot \dfrac{2}{3} = -1/18\) \(\text{Var} [X] = 1/6 - (1/3)^2 = 1/18\)
\(a = \text{Cov} [X, Y] /\text{Var} [X] = -1\) \(b = E[Y] - aE[X] = 1\)
tuappr: [0 1] [0 2] 200 400 u<=2*(1-t) EX = dot(X,PX); EY = dot(Y,PY); VX = dot(X.^2,PX) - EX^2 VX = 0.0556 CV = total(t.*u.*P) - EX*EY CV = -0.0556 a = CV/VX a = -1.0000 b = EY - a*EX b = 1.0000
Exercise \(\PageIndex{24}\)
(See Exercise 13 from " Problems On Random Vectors and Joint Distributions ", and Exercise 23 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = \dfrac{1}{8} (t + u)\) for \(0 \le t \le 2\), \(0 \le u \le 2\).
\(E[X] = E[Y] = \dfrac{7}{6}\), \(E[X^2] = \dfrac{5}{3}\)
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\(E[XY] = \dfrac{1}{8} \int_{0}^{2} \int_{0}^{2} tu (t + u)\ dudt = 4/3\), \(\text{Cov} [X, Y] = -1/36\), \(\text{Var} [X] = 11/36\)
\(a = \text{Cov} [X, Y]/\text{Var} [X] = -1/11\), \(b = E[Y] - a E[X] = 14/11\)
tuappr: [0 2] [0 2] 200 200 (1/8)*(t+u) VX = 0.3055 CV = -0.0278 a = -0.0909 b = 1.2727
Exercise \(\PageIndex{25}\)
(See Exercise 15 from " Problems On Random Vectors and Joint Distributions ", and Exercise 25 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1 + t\).
\(E[X] = \dfrac{313}{220}\), \(E[Y] = \dfrac{1429}{880}\), \(E[X^2] = \dfrac{49}{22}\)
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\(E[XY] = \dfrac{3}{88} \int_{0}^{2} \int_{0}^{1+t} tu (2t + 3u^2)\ dudt = 2153/880\), \(\text{Cov} [X, Y] = 26383/1933600\), \(\text{Var} [X] = 9831/48400\)
\(a = \text{Cov} [X, Y]/\text{Var} [X] = 26383/39324\), \(b = E[Y] - a E[X] = 26321/39324\)
tuappr: [0 2] [0 3] 200 300 (3/88)*(2*t + 3*u.^2).*(u<=1+t) VX = 0.2036 CV = 0.1364 a = 0.6700 b = 0.6736
Exercise \(\PageIndex{26}\)
(See Exercise 16 from " Problems On Random Vectors and Joint Distributions ", and Exercise 26 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = 12t^2 u\) on the parallelogram with vertices
(-1, 0), (0, 0), (1, 1), (0, 1)
\(E[X] = \dfrac{2}{5}\), \(E[Y] = \dfrac{11}{15}\), \(E[X^2] = \dfrac{2}{5}\)
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\(E[XY] = 12 \int_{-1}^{0} \int_{0}^{t + 1} t^3 u^2\ dudt + 12 \int_{0}^{1} \int_{t}^{1} t^3 u^2 \ dudt = \dfrac{2}{5}\)
\(\text{Cov} [X, Y] = \dfrac{8}{75}\), \(\text{Var} [X] = \dfrac{6}{25}\)
\(a = \text{Cov} [X, Y]/\text{Var} [X] = 4/9\), \(b = E[Y] - a E[X] = 5/9\)
tuappr: [-1 1] [0 1] 400 200 12*t.^2.*u.*(u>= max(0,t)).*(u<= min(1+t,1)) VX = 0.2383 CV = 0.1056 a = 0.4432 b = 0.5553
Exercise \(\PageIndex{27}\)
(See Exercise 17 from " Problems On Random Vectors and Joint Distributions ", and Exercise 27 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1, 2 - t\}\).
\(E[X] = \dfrac{52}{55}\), \(E[Y] = \dfrac{32}{55}\), \(E[X^2] = \dfrac{627}{605}\)
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\(E[XY] = \dfrac{24}{11} \int_{0}^{1} \int_{0}^{1} t^2 u^2\ dudt + \dfrac{24}{11} \int_{1}^{2} \int_{0}^{2-t} t^2 u^2 \ dudt = \dfrac{28}{55}\)
\(\text{Cov} [X, Y] = -\dfrac{124}{3025}\), \(\text{Var} [X] = \dfrac{431}{3025}\)
\(a = \text{Cov} [X, Y]/\text{Var} [X] = -\dfrac{124}{431}\), \(b = E[Y] - a E[X] = \dfrac{368}{431}\)
tuappr: [0 2] [0 1] 400 200 (24/11)*t.*u.*(u<=min(1,2-t)) VX = 0.1425 CV =-0.0409 a = -0.2867 b = 0.8535
Exercise \(\PageIndex{28}\)
(See Exercise 18 from " Problems On Random Vectors and Joint Distributions ", and Exercise 28 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{max } \{2 - t, t\}\).
\(E[X] = \dfrac{53}{46}\), \(E[Y] = \dfrac{22}{23}\), \(E[X^2] = \dfrac{9131}{5290}\)
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\(E[XY] = \dfrac{3}{23} \int_{0}^{1} \int_{0}^{2-t} tu (t + 2u)\ dudt + \dfrac{3}{23} \int_{1}^{2} \int_{0}^{t} tu (t + 2u) \ dudt = \dfrac{251}{230}\)
\(\text{Cov} [X, Y] = -\dfrac{57}{5290}\), \(\text{Var} [X] = \dfrac{4217}{10580}\)
\(a = \text{Cov} [X, Y]/\text{Var} [X] = -\dfrac{114}{4217}\), \(b = E[Y] - a E[X] = \dfrac{4165}{4217}\)
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
Exercise \(\PageIndex{29}\)
(See Exercise 21 from " Problems On Random Vectors and Joint Distributions ", and Exercise 31 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2t, 3 - t\}\).
\(E[X] = \dfrac{16}{13}\), \(E[Y] = \dfrac{11}{12}\), \(E[X^2] = \dfrac{2847}{1690}\)
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\(E[XY] = \dfrac{2}{13} \int_{0}^{1} \int_{0}^{3-t} tu (t + 2u)\ dudt + \dfrac{2}{13} \int_{1}^{2} \int_{0}^{2t} tu (t + 2u) \ dudt = \dfrac{431}{390}\)
\(\text{Cov} [X, Y] = -\dfrac{3}{130}\), \(\text{Var} [X] = \dfrac{287}{1690}\)
\(a = \text{Cov} [X, Y]/\text{Var} [X] = -\dfrac{39}{297}\), \(b = E[Y] - a E[X] = \dfrac{3733}{3444}\)
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
Exercise \(\PageIndex{30}\)
(See Exercise 22 from " Problems On Random Vectors and Joint Distributions ", and Exercise 32 from " Problems on Mathematical Expectation "). \(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2u^2\), for \(0 \le u \le 1\).
\(E[X] = \dfrac{243}{224}\), \(E[Y] = \dfrac{11}{16}\), \(E[X^2] = \dfrac{107}{70}\)
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\(E[XY] = \dfrac{3}{8} \int_{0}^{1} \int_{0}^{1} tu (t^2 + 2u)\ dudt + \dfrac{9}{14} \int_{1}^{2} \int_{0}^{1} t^3u^3 \ dudt = \dfrac{347}{448}\)
\(\text{Cov} [X, Y] = -\dfrac{103}{3584}\), \(\text{Var} [X] = \dfrac{88243}{250880}\)
\(a = \text{Cov} [X, Y]/\text{Var} [X] = -\dfrac{7210}{88243}\), \(b = E[Y] - a E[X] = \dfrac{105691}{176486}\)
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
Exercise \(\PageIndex{31}\)
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 \(E[W]\) and \(\text{Var} [W]\). Do this using icalc, then repeat with icalc3 and compare results.
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x = [-5 -1 3 4 7]; px = 0.01*[15 20 30 25 10]; EX = dot(x,px) % Use of properties EX = 1.6500 VX = dot(x.^2,px) - EX^2 VX = 12.8275 EW = (3 - 4+ 2)*EX EW = 1.6500 VW = (3^2 + 4^2 + 2^2)*VX VW = 371.9975 icalc % Iterated use of 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; [W,PW] = csort(H,P); EW = dot(W,PW) EW = 1.6500 VW = dot(W.^2,PW) - EW^2 VW = 371.9975 icalc3 % Use of 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 S = 3*t - 4*u + 2*v; [w,pw] = csort(S,P); Ew = dot(w,pw) Ew = 1.6500 Vw = dot(w.^2,pw) - Ew^2 Vw = 371.9975
Exercise \(\PageIndex{32}\)
\(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1 + t\) (see Exercise 25 and Exercise 37 from " Problems on Mathematical Expectation ").
\(Z = I_{[0, 1]} (X) 4X + I_{(1, 2]} (X) (X+ Y)\)
\(E[X] = \dfrac{313}{220}\), \(E[Z] = \dfrac{5649}{1760}\), \(E[Z^2] = \dfrac{4881}{440}\)
Determine \(\text{Var} [Z]\) and \(\text{Cov} [X, Z]\). Check with discrete approximation.
- Answer
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\(E[XZ] = \dfrac{3}{88} \int_0^1 \int_{0}^{1+t} 4t^2 (2t + 2u^2)\ dudt + \dfrac{3}{88} \int_{1}^{2} \int_{0}^{1 + t} t (t + u) (2t + 3u^2)\ dudt = \dfrac{16931}{3520}\)
\(\text{Var} [Z] = E[Z^2] - E^2[Z] = \dfrac{2451039}{3097600}\) \(\text{Cov} [X,Z] = E[XZ] - E[X] E[Z] = \dfrac{94273}{387200}\)
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*EZ CV = 0.2445 % Theoretical 0.2435 VZ = total(G.^2.*P) - EZ^2 VZ = 0.7934 % Theoretical 0.7913
Exercise \(\PageIndex{33}\)
\(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1, 2 - t\}\) (see Exercise 27 and Exercise 38 from " Problems on Mathematical Expectation ").
\(Z = I_M (X,Y) (X + Y) + I_{M^c} (X, Y) 2Y\), \(M = \{(t, u): \text{max } (t, u) \le 1\}\)
\(E[X] = \dfrac{52}{55}\), \(E[Z] = \dfrac{16}{55}\), \(E[Z^2] = \dfrac{39}{308}\)
Determine \(\text{Var} [Z]\) and \(\text{Cov} [X, Z]\). Check with discrete approximation.
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\(E[XZ] = \dfrac{24}{11} \int_0^1 \int_{t}^{1} t (t/2) tu \ dudt + \dfrac{24}{11} \int_{0}^{1} \int_{0}^{t} tu^2tu \ dudt \dfrac{24}{11} \int_1^2 \int_{0}^{2 - t} t tu^2 tu\ dudt= \dfrac{211}{770}\)
\(\text{Var} [Z] = E[Z^2] - E^2[Z] = \dfrac{3557}{84700}\) \(\text{Cov} [Z,X] = E[XZ] - E[X] E[Z] = -\dfrac{43}{42350}\)
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^2 VZ = 0.0425 CV = total(t.*G.*P) - EZ*dot(X,PX) CV = -9.2940e-04
Exercise \(\PageIndex{34}\)
\(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{max } \{2 - t, t\}\) (see Exercise 28 and Exercise 39 from " Problems on Mathematical Expectation ").
\(Z = I_M (X, Y) (X+Y) + I_{M^c} (X, Y) 2Y\), \(M = \{(t, u): \text{max } (t, u) \le 1\}\)
\(E[X] = \dfrac{53}{46}\), \(E[Z] = \dfrac{175}{92}\), \(E[Z^2] = \dfrac{2063}{460}\)
Determine \(\text{Var} [Z]\) and \(\text{Cov} [Z]\). Check with discrete approximation.
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\(E[ZX] = \dfrac{3}{23} \int_{0}^{1} \int_{0}^{1} t (t + u) (t + 2u) \ dudt + \dfrac{3}{23} \int_{0}^{1} \int_{1}^{2 - t} 2tu(t + 2u) \ dudt +\)
\(\dfrac{3}{23} \int_{1}^{2} \int_{1}^{t} 2tu(t + 2u)\ dudt = \dfrac{1009}{460}\)
\(\text{Var} [Z] = E[Z^2] - E^2[Z] = \dfrac{36671}{42320}\) \(\text{Cov} [Z, X] = E[ZX] - E[Z] E[X] = \dfrac{39}{21160}\)
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
Exercise \(\PageIndex{35}\)
\(f_{XY} (t, u) = \dfrac{12}{179} (3t^2 + u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2, 3 - t\}\) (see Exercise 29 and Exercise 40 from " Problems on Mathematical Expectation ").
\(Z = I_M (X, Y) (X+Y) + I_{M^c} (X, Y) 2Y^2\), \(M = \{(t, u): t \le 1, u \ge 1\}\)
\(E[X] = \dfrac{2313}{1790}\), \(E[Z] = \dfrac{1422}{895}\), \(E[Z^2] = \dfrac{28296}{6265}\)
Determine \(\text{Var} [Z]\) and \(\text{Cov} [X, Z]\). Check with discrete approximation.
- Answer
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\(E[ZX] = \dfrac{12}{179} \int_{0}^{1} \int_{1}^{2} t (t + u) (3t^2 + u) \ dudt + \dfrac{12}{179} \int_{0}^{1} \int_{0}^{1} 2tu^2 (3t^2 + u) \ dudt +\)
\(\dfrac{12}{179} \int_{1}^{2} \int_{0}^{3 - t} 2tu^2(3t^2 + u)\ dudt = \dfrac{24029}{12530}\)
\(\text{Var} [Z] = E[Z^2] - E^2[Z] = \dfrac{11170332}{5607175}\) \(\text{Cov} [Z, X] = E[ZX] - E[Z] E[X] = -\dfrac{1517647}{11214350}\)
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
Exercise \(\PageIndex{36}\)
\(f_{XY} (t, u) = \dfrac{12}{227} (3t + 2tu)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1 + t, 2\}\) (see Exercise 30 and Exercise 41 from " Problems on Mathematical Expectation ").
\(Z = I_M (X, Y) X + I_{M^c} (X, Y) XY\), \(M = \{(t, u): u \le \text{min } (1, 2 - t)\}\)
\(E[X] = \dfrac{1567}{1135}\), \(E[Z] = \dfrac{5774}{3405}\), \(E[Z^2] = \dfrac{56673}{15890}\)
Determine \(\text{Var} [Z]\) and \(\text{Cov} [X, Z]\). Check with discrete approximation.
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\(E[ZX] = \dfrac{12}{227} \int_{0}^{1} \int_{0}^{1} t^2 (3t + 2tu) \ dudt + \dfrac{12}{227} \int_{1}^{2} \int_{0}^{2-t} t^2(3t + 2tu) \ dudt +\)
\(\dfrac{12}{227} \int_{0}^{1} \int_{1}^{1 + t} t^2 u(3t + 2tu)\ dudt + \dfrac{12}{227} \int_{1}^{2} \int_{2 - t}^{2} t^2 u(3t + 2tu)\ dudt = \dfrac{20338}{7945}\)
\(\text{Var} [Z] = E[Z^2] - E^2[Z] = \dfrac{112167631}{162316350}\) \(\text{Cov} [Z, X] = E[ZX] - E[Z] E[X] = \dfrac{5915884}{27052725}\)
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*EZ CV = 0.2188 VZ = total(G.^2.*P) - EZ^2 VZ = 0.6907
Exercise \(\PageIndex{37}\)
(See Exercise 12.4.20, and Exercises 9 and 10 from " Problems on Functions of Random Variables "). For the pair \(\{X, Y\}\) in Exercise 12.4.20, let
\(Z = g(X, Y) = 3X^2 + 2XY - Y^2\)
\(W = h(X, Y) = \begin{cases} X & \text{for } X + Y \le 4 \\ 2Y & \text{for } X + Y > 4 \end{cases} = I_M (X, Y) X + I_{M^c} (X, Y) 2Y\)
Determine the joint distribution for the pair \(\{Z, W\}\) and determine the regression line of \(W\) on \(Z\).
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npr08_07 Data are in X, Y, P jointzw Enter joint prob for (X,Y) P Enter values for X X Enter values for Y Y Enter expression for g(t,u) 3*t.^2 + 2*t.*u - u.^2 Enter expression for h(t,u) t.*(t+u<=4) + 2*u.*(t+u>4) Use array operations on Z, W, PZ, PW, v, w, PZW EZ = dot(Z,PZ) EZ = 5.2975 EW = dot(W,PW) EW = 4.7379 VZ = dot(Z.^2,PZ) - EZ^2 VZ = 1.0588e+03 CZW = total(v.*w.*PZW) - EZ*EW CZW = -12.1697 a = CZW/VZ a = -0.0115 b = EW - a*EZ b = 4.7988 % Regression line: w = av + b