14.2: Problems on Conditional Expectation, Regression
For the distributions in Exercises 1-3
- Determine the regression curve of \(Y\) on \(X\) and compare with the regression line of \(Y\) on \(X\).
- For the function \(Z = g(X, Y)\) indicated in each case, determine the regression curve of \(Z\) on \(X\).
Exercise \(\PageIndex{1}\)
(See 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 |
The regression line of \(Y\) on \(X\) is \(u = 0.5275 t + 0.6924\).
\(Z = X^2Y + |X + Y|\)
- Answer
-
The regression line of \(Y\) on \(X\) is \(u = 0.5275t + 0.6924\).
npr08_07 Data are in X, Y, P jcalc - - - - - - - - - - - EYx = sum(u.*P)./sum(P); disp([X;EYx]') -3.1000 -0.0290 -0.5000 -0.6860 1.2000 1.3270 2.4000 2.1960 3.7000 3.8130 4.9000 2.5700 G = t.^2.*u + abs(t+u); EZx = sum(G.*P)./sum(P); disp([X;EZx]') -3.1000 4.0383 -0.5000 3.5345 1.2000 6.0139 2.4000 17.5530 3.7000 59.7130 4.9000 69.1757
Exercise \(\PageIndex{2}\)
(See 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 |
The regression line of \(Y\) on \(X\) is \(u = -0.2584 t + 5.6110\).
\(Z = I_Q (X, Y) \sqrt{X} (Y - 4) + I_{Q^c} (X, Y) XY^2\) \(Q = \{(t, u) : u \le t \}\)
- Answer
-
The regression line of \(Y\) on \(X\) is \(u = -0.2584 t + 5.6110\).
npr08_08 Data are in X, Y, P jcalc - - - - - - - - - - - - EYx = sum(u.*P)./sum(P); disp([X;EYx]') 1.0000 5.5350 3.0000 5.9869 5.0000 3.6500 7.0000 2.3100 9.0000 2.0254 11.0000 2.9100 13.0000 3.1957 15.0000 0.9100 17.0000 1.5254 19.0000 0.9100 M = u<=t; G = (u-4).*sqrt(t).*M + t.*u.^2.*(1-M); EZx = sum(G.*P)./sum(P); disp([X;EZx]') 1.0000 58.3050 3.0000 166.7269 5.0000 175.9322 7.0000 185.7896 9.0000 119.7531 11.0000 105.4076 13.0000 -2.8999 15.0000 -11.9675 17.0000 -10.2031 19.0000 -13.4690
Exercise \(\PageIndex{3}\)
(See 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 |
The regression line of \(Y\) on \(X\) is \(u = -0.7793t + 4.3051\).
\(Z = (Y -2.8)/X\)
- Answer
-
The regression line of \(Y\) on \(X\) is \(u = -0.7793t + 4.3051\).
npr08_09 Data are in X, Y, P jcalc - - - - - - - - - - - - EYx = sum(u.*P)./sum(P); disp([X;EYx]') 1.0000 3.8333 1.5000 3.1250 2.0000 2.5175 2.5000 2.3933 3.0000 2.3900 G = (u - 2.8)./t; EZx = sum(G.*P)./sum(P); disp([X;EZx]') 1.0000 1.0333 1.5000 0.2167 2.0000 -0.1412 2.5000 -0.1627 3.0000 -0.1367
For the joint densities in Exercises 4-11 below
- Determine analytically the regression curve of \(Y\) on \(X\) and compare with the regression line of \(Y\) on \(X\).
- Check these with a discrete approximation.
Exercise \(\PageIndex{4}\)
(See Exercise 10 from " Problems On Random Vectors and Joint Distributions ", Exercise 20 from " Problems on Mathematical Expectation ", and Exercise 23 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} (t, u) = 1\) for \(0 \le t \le 1\). \(0 \le u \le 2(1 - t)\).
The regression line of \(Y\) on \(X\) is \(u = 1 - t\).
\(f_X (t) = 2(1 - t)\), \(0 \le t \le 1\)
- Answer
-
The regression line of \(Y\) on \(X\) is \(u = 1 - t\).
\(f_{Y|X} (u|t) = \dfrac{1}{2(1 - t)}\). \(0 \le t \le 1\), \(0 \le u \le 2(1 - t)\)
\(E[Y|X = t] = \dfrac{1}{2(1 - t)} \int_{0}^{2(1-t)} udu = 1 - t\), \(0 \le t \le 1\)
tuappr: [0 1] [0 2] 200 400 u<=2*(1-t) - - - - - - - - - - - - - EYx = sum(u.*P)./sum(P); plot(X,EYx) % Straight line thru (0,1), (1,0)
Exercise \(\PageIndex{5}\)
(See Exercise 13 from " Problems On Random Vectors and Joint Distributions ", Exercise 23 from " Problems on Mathematical Expectation ", and Exercise 24 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} (t, u) = \dfrac{1}{8} (t+u)\) for \(0 \le t \le 2\), \(0 \le u \le 2\).
The regression line of \(Y\) on \(X\) is \(u = -t/11 + 35/33\).
\(f_{X} (t) = \dfrac{1}{4} (t + 1)\), \(0 \le t \le 2\)
- Answer
-
The regression line of \(Y\) on \(X\) is \(u = -t/11 + 35/33\).
\(f_{Y|X} (u|t) = \dfrac{(t + u)}{2(t + 1)}\) \(0 \le t \le 2\), \(0 \le u \le 2\)
\(E[Y|X = t] = \dfrac{1}{2(t + 1)} \int_{0}^{2} (tu + u^2)\ du = 1 + \dfrac{1}{3t+3}\) \(0 \le t \le 2\)
tuappr: [0 2] [0 2] 200 200 (1/8)*(t+u) EYx = sum(u.*P)./sum(P); eyx = 1 + 1./(3*X+3); plot(X,EYx,X,eyx) % Plots nearly indistinguishable
Exercise \(\PageIndex{6}\)
(See Exercise 15 from " Problems On Random Vectors and Joint Distributions ", Exercise 25 from " Problems on Mathematical Expectation ", and Exercise 25 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1 + t\).
The regression line of \(Y\) on \(X\) is \(u = 0.0958t + 1.4876\).
\(f_X (t) = \dfrac{3}{88} (1 + t) (1 + 4t + t^2) = \dfrac{3}{88} (1 + 5t + 5t^2 + t^3)\), \(0 \le t \le 2\)
- Answer
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The regression line of \(Y\) on \(X\) is \(u = 0.0958t + 1.4876\).
\(f_{Y|X} (u|t) = \dfrac{2t + 3u^2}{(1 + t)(1 + 4t + t^2)}\) \(0 \le u \le 1 + t\)
\(E[Y|X = t] = \dfrac{1}{(1 + t) (1 + 4t + t^2)} \int_{0}^{1 + t} (2tu + 3u^3)\ du\)
\(= \dfrac{(t + 1)(t + 3) (3t+1)}{4(1 + 4t +t^2)}\), \(0 \le t \le 2\)
tuappr: [0 2] [0 3] 200 300 (3/88)*(2*t + 3*u.^2).*(u<=1+t) EYx = sum(u.*P)./sum(P); eyx = (X+1).*(X+3).*(3*X+1)./(4*(1 + 4*X + X.^2)); plot(X,EYx,X,eyx) % Plots nearly indistinguishable
Exercise \(\PageIndex{7}\)
(See Exercise 16 from " Problems On Random Vectors and Joint Distributions ", Exercise 26 from " Problems on Mathematical Expectation ", and Exercise 26 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} (t, u) = 12t^2u\) on the parallelogram with vertices
(-1, 0), (0, 0), (1, 1), (0, 1)
The regression line of \(Y\) on \(X\) is \(u = (4t + 5)/9\).
\(f_{X} (t) = I_{[-1, 0]} (t) 6t^2 (t + 1)^2 + I_{(0, 1]} (t) 6t^2 (1 - t^2)\)
- Answer
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The regression line of \(Y\) on \(X\) is \(u = (23t + 4)/18\).
\(f_{Y|X} (u|t) = I_{[-1, 0]} (t) \dfrac{2u}{(t + 1)^2} + I_{(0, 1]} (t) \dfrac{2u}{(1 - t^2)}\) on the parallelogram
\(E[Y|X = t] = I_{[-1, 0]} (t) \dfrac{1}{(t + 1)^2} \int_{0}^{t + 1} 2u\ du + I_{(0, 1]} (t) \dfrac{1}{(1 - t^2)} \int_{t}^{1} 2u \ du\)
\(= I_{[-1, 0]} (t) \dfrac{2}{3} (t + 1) + I_{(0, 1]} (t) \dfrac{2}{3} \dfrac{t^2 + t + 1}{t + 1}\)
tuappr: [-1 1] [0 1] 200 100 12*t.^2.*u.*((u<= min(t+1,1))&(u>=max(0,t))) EYx = sum(u.*P)./sum(P); M = X<=0; eyx = (2/3)*(X+1).*M + (2/3)*(1-M).*(X.^2 + X + 1)./(X + 1); plot(X,EYx,X,eyx) % Plots quite close
Exercise \(\PageIndex{8}\)
(See Exercise 17 from " Problems On Random Vectors and Joint Distributions ", Exercise 27 from " Problems on Mathematical Expectation ", and Exercise 27 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{1, 2 - t\}\).
The regression line of \(Y\) on \(X\) is \(u = (-124t + 368)/431\)
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{11} t + I_{(1, 2]} (t) \dfrac{12}{11} t (2 - t)^2\)
- Answer
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The regression line of \(Y\) on \(X\) is \(u = (-124t + 368)/431\)
\(f_{Y|X} (u|t) = I_{[0, 1]} (t) 2u + I_{(1, 2]} (t) \dfrac{2u}{(2 - t)^2}\)
\(E[Y|X = t] = I_{[0, 1]} (t) \int_{0}^{1} 2u^2 \ du + I_{(1, 2]} (t) \dfrac{1}{(2 - t)^2} \int_{0}^{2 - t} 2u^2 \ du\)
\(= I_{[0, 1]} (t) \dfrac{2}{3} + I_{(1, 2]} (t) \dfrac{2}{3} (2 - t)\)
tuappr: [0 2] [0 1] 200 100 (24/11)*t.*u.*(u<=min(1,2-t)) EYx = sum(u.*P)./sum(P); M = X <= 1; eyx = (2/3)*M + (2/3).*(2 - X).*(1-M); plot(X,EYx,X,eyx) % Plots quite close
Exercise \(\PageIndex{9}\)
(See Exercise 18 from " Problems On Random Vectors and Joint Distributions ", Exercise 28 from " Problems on Mathematical Expectation ", and Exercise 28 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{max } \{2 - t, t\}\).
The regression line of \(Y\) on \(X\) is \(u = 1.0561 t - 0.2603\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{6}{23} (2 - t) + I_{(1, 2]} (t) \dfrac{6}{23} t^2\)
- Answer
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The regression line of \(Y\) on \(X\) is \(u = 1.0561 t - 0.2603\).
\(f_{Y|X} (u|t) = I_{[0, 1]} (t) \dfrac{t+2u}{2(2-t)} + I_{(1, 2]} (t) \dfrac{t + 2u}{2t^2}\) \(0 \le u \le \text{max } (2 - t, t)\)
\(E[Y|X = t] = I_{[0, 1]} (t) \dfrac{1}{2(2 - t)} \int_{0}^{2 - t} (tu + 2u^2) \ du + I_{(1, 2]} (t) \dfrac{1}{2t^2} \int_{0}^{t} (tu + 2u^2)\ du\)
\(= I_{[0, 1]} (t) \dfrac{1}{12} (t - 2) ( t - 8) + I_{(1, 2]} (t) \dfrac{7}{12} t\)
tuappr: [0 2] [0 2] 200 200 (3/23)*(t+2*u).*(u<=max(2-t,t)) EYx = sum(u.*P)./sum(P); M = X<=1; eyx = (1/12)*(X-2).*(X-8).*M + (7/12)*X.*(1-M); plot(X,EYx,X,eyx) % Plots quite close
Exercise \(\PageIndex{10}\)
(See Exercise 21 from " Problems On Random Vectors and Joint Distributions ", Exercise 31 from " Problems on Mathematical Expectation ", and Exercise 29 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2t, 3 - t\}\).
The regression line of \(Y\) on \(X\) is \(u = -0.1359 t + 1.0839\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{13} t^2 + I_{(1, 2]} (t) \dfrac{6}{13} (3 - t)\)
- Answer
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The regression line of \(Y\) on \(X\) is \(u = -0.1359 t + 1.0839\).
\(f_{Y|X} (t|u) = I_{[0, 1]} (t) \dfrac{t + 2u}{6t^2} + I_{(1,2]} (t) \dfrac{t + 2u}{3(3 - t)}\) \(0 \le u \le \text{max } (2t, 3 - t)\)
\(E[Y|X = t] = I_{[0, 1]} (t) \dfrac{1}{6t^2} \int_{0}^{t} (tu + 2u^2)\ du + I_{(1, 2]} (t) \dfrac{1}{3(3 - t)} \int_{0}^{3 - t} (tu + 2u^2)\ du\)
\(= I_{[0, 1]} (t) \dfrac{11}{9} t + I_{(1, 2]} (t) \dfrac{1}{18} (t^2 - 15t + 36)\)
tuappr: [0 2] [0 2] 200 200 (2/13)*(t+2*u).*(u<=min(2*t,3-t)) EYx = sum(u.*P)./sum(P); M = X<=1; eyx = (11/9)*X.*M + (1/18)*(X.^2 - 15*X + 36).*(1-M); plot(X,EYx,X,eyx) % Plots quite close
Exercise \(\PageIndex{11}\)
(See Exercise 22 from " Problems On Random Vectors and Joint Distributions ", Exercise 32 from " Problems on Mathematical Expectation ", and Exercise 30 from " Problems on Variance, Covariance, Linear Regression "). \(f_{XY} 9t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2u^2\). for \(0 \le u \le 1\).
The regression line of \(Y\) on \(X\) is \(u = 0.0817t + 0.5989\).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 1) + I_{(1, 2]} (t) \dfrac{3}{14} t^2\)
- Answer
-
The regression line of \(Y\) on \(X\) is \(u = 0.0817t + 0.5989\).
\(f_{Y|X} (t|u) = I_{[0, 1]} (t) \dfrac{t^2 + 2u}{t^2 + 1} + I_{(1, 2]} (t) 3u^2\) \(0 \le u \le 1\)
\(E[Y|X = t] = I_{[0, 1]} (t) \dfrac{1}{t^2 + 1} \int_{0}^{1} (t^2u + 2u^2)\ du + I_{(1, 2]} (t) \int_{0}^{1} 3u^3 \ du\)
\(= I_{[0, 1]} (t) \dfrac{3t^2 + 4}{6(t^2 + 1)} + I_{(1, 2]} (t) \dfrac{3}{4}\)
tuappr: [0 2] [0 1] 200 100 (3/8)*(t.^2 + 2*u).*(t<=1) + ... (9/14)*t.^2.*u.^2.*(t>1) EYx = sum(u.*P)./sum(P); M = X<=1; eyx = M.*(3*X.^2 + 4)./(6*(X.^2 + 1)) + (3/4)*(1 - M); plot(X,EYx,X,eyx) % Plots quite close
For the distributions in Exercises 12-16 below
- Determine analytically \(E[Z|X = t]\)
- Use a discrete approximation to calculate the same functions.
Exercise \(\PageIndex{12}\)
\(f_{XY} (t, u) = \dfrac{3}{88} (2t + 3u^2)\) for \(0 \le t \le 2\), \(0 \le u \le 1 + t\), (see Exercise 37 from " Problems on Mathematical Expectation ", and Exercise 14.2.6 ).
\(f_{X} (t) = \dfrac{3}{88} (1 + t) (1 + 4t + t^2) = \dfrac{3}{88} (1 + 5t + 5t^2 + t^3)\), \(0 \le t \le 2\)
\(Z = I_{[0, 1]} (X) 4X + I_{(1, 2]} (X) (X + Y)\)
- Answer
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\(Z = I_M (X) 4X + I_N (X) (X + Y)\). Use of linearity, (CE8) , and (CE10) gives
\(E[Z|X = t] = I_M (t) 4t + I_N(t) (t + E[Y|X = t])\)
\(= I_M (t) 4t + I_N (t) (t + \dfrac{(t + 1)(t + 3) (3t + 1)}{4(1 + 4t + t^2)})\)
% Continuation of Exercise 14.2.6 G = 4*t.*(t<=1) + (t + u).*(t>1); EZx = sum(G.*P)./sum(P); M = X<=1; ezx = 4*X.*M + (X + (X+1).*(X+3).*(3*X+1)./(4*(1 + 4*X + X.^2))).*(1-M); plot(X,EZx,X,ezx) % Plots nearly indistinguishable
Exercise \(\PageIndex{13}\)
\(f_{XY} (t, u) = \dfrac{24}{11} tu\) for \(0 \le t \le 2\), \(0 \le u \text{min } \{1, 2 - t\}\) (see Exercise 38 from " Problems on Mathematical Expectaton ", Exercise 14.2.8 ).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{11} t + I_{(1, 2]} (t) \dfrac{12}{11} t (2 - t)^2\)
\(Z = I_{M} (X, Y) \dfrac{1}{2} X + I_M (X, Y) Y^2\), \(M = \{(t ,u): u > t\}\)
- Answer
-
\(Z = I_{M} (X, Y) \dfrac{1}{2} X + I_M (X, Y) Y^2\), \(M = \{(t ,u): u > t\}\)
\(I_M(t, u) = I_{[0, 1]} (t) I_{[t, 1]} (u)\) \(I_{M^c} (t, u) = I_{[0, 1]} (t) I_{[0, t]}(u) + I_{(1, 2]} (t) I_{[0, 2 - t]} (u)\)
\(E[Z|X = t] = I_{[0, 1]} (t) [\dfrac{t}{2} \int_{t}^{1} 2u\ du + \int_{0}^{t} u^2 \cdot 2u\ du] + I_{(1, 2]} (t) \int_{0}^{2 - t} u^2 \cdot \dfrac{2u}{(2 - t)^2}\ du\)
\(= I_{[0, 1]} (t) \dfrac{1}{2} t (1 - t^2 + t^3) + I_{(1, 2]} (t) \dfrac{1}{2} (2- t)^2\)
% Continuation of Exercise 14.2.8 Q = u>t; G = (1/2)*t.*Q + u.^2.*(1-Q); EZx = sum(G.*P)./sum(P); M = X <= 1; ezx = (1/2)*X.*(1-X.^2+X.^3).*M + (1/2)*(2-X).^2.*(1-M); plot(X,EZx,X,ezx) % Plots nearly indistinguishable
Exercise \(\PageIndex{14}\)
\(f_{XY} (t, u) = \dfrac{3}{23} (t + 2u)\) for \(0 \le t \le 2\), \(0 \le u \le \text{max } \{2 - t, t\}\) (see Exercise 39 from " Problems on Mathematical Expectaton ", and Exercise 14.2.9 ).
\(f_X(t) = I_{[0, 1]} (t) \dfrac{6}{23} (2 - t) + I_{(1, 2]} (t) \dfrac{6}{23} t^2\)
\(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y) 2Y\), \(M = \{(t, u): \text{max } (t, u) \le 1\}\)
- Answer
-
\(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y) 2Y\), \(M = \{(t, u): \text{max } (t, u) \le 1\}\)
\(I_M (t, u) = I_{[0, 1]} (t) I_{[0, 1]} (u)\) \(I_{M^c} (t, u) = I_{[0, 1]} (t) I_{[1, 2 -t]} (u) + I_{(1,2]} (t) I_{[0, 1]} (u)\)
\(E[Z|X = t] = I_{[0, 1]} (t) \dfrac{1}{2(2 - t)} \int_{0}^{1} (t + u) (t + 2u)\ du + \dfrac{1}{2 - t} \int_{1}^{2 - t} u (t + 2u)\ du] + I_{(1, 2]} (t) 2E [Y|X = t]\)
\(= I_{[0, 1]} (t) \dfrac{1}{12} \cdot \dfrac{2t^3 - 30t^2 + 69t - 60}{t - 2} + I_{(1, 2]} (t) \dfrac{7}{6} 2t\)
% Continuation of Exercise 14.2.9 M = X <= 1; Q = (t<=1)&(u<=1); G = (t+u).*Q + 2*u.*(1-Q); EZx = sum(G.*P)./sum(P); ezx = (1/12)*M.*(2*X.^3 - 30*X.^2 + 69*X -60)./(X-2) + (7/6)*X.*(1-M); plot(X,EZx,X,ezx)
Exercise \(\PageIndex{15}\)
\(f_{XY} (t, u) = \dfrac{2}{13} (t + 2u)\), for \(0 \le t \le 2\), \(0 \le u \le \text{min } \{2t, 3 - t\}\). (see Exercise 31 from " Problems on Mathematical Expectaton ", and Exercise 14.2.10 ).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{12}{13} t^2 + I_{(1, 2]} (t) \dfrac{6}{13} (3 - t)\)
\(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y) 2Y^2\), \(M = \{(t, u): t \le 1, u \ge 1\}\)
- Answer
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\(Z = I_M (X, Y) (X + Y) + I_{M^c} (X, Y) 2Y^2\), \(M = \{(t, u): t \le 1, u \ge 1\}\)
\(I_M(t, u) = I_{[0, 1]} (t0 I_{[1, 2]} (u)\) \(I_{M^c} (t, u) = I_{[0, 1]} (t) I_{[0, 1)} (u) + I_{(1, 2]} (t) I_{[0, 3 - t]} (u)\)
\(E[Z|X = t] = I_{[0, 1/2]} (t) \dfrac{1}{6t^2} \int_{0}^{2t} 2u^2 (t + 2u) \ du +\)
\(I_{(1/2, 1]} (t) [\dfrac{1}{6t^2} \int_{0}^{1} 2u^2 (t + 2u)\ du + \dfrac{1}{6t^2} \int_{1}^{2t} (t + u) (t + 2u)\ du] + I_{(1, 2]} (t) \dfrac{1}{3 (3 - t)} \int_{0}^{3 - t} 2u^2 (t + 2u)\ du\)
\(= I_{[0, 1/2]} (t) \dfrac{32}{9} t^2 + I_{(1/2, 1]} (t) \dfrac{1}{36} \cdot \dfrac{80t^3 - 6t^2 - 5t + 2}{t^2} + I_{(1, 2]} (t) \dfrac{1}{9} (- t^3 + 15t^2 - 63t + 81)\)
tuappr: [0 2] [0 2] 200 200 (2/13)*(t + 2*u).*(u<=min(2*t,3-t)) M = (t<=1)&(u>=1); Q = (t+u).*M + 2*(1-M).*u.^2; EZx = sum(Q.*P)./sum(P); N1 = X <= 1/2; N2 = (X > 1/2)&(X<=1); N3 = X > 1; ezx = (32/9)*N1.*X.^2 + (1/36)*N2.*(80*X.^3 - 6*X.^2 - 5*X + 2)./X.^2 ... + (1/9)*N3.*(-X.^3 + 15*X.^2 - 63.*X + 81); plot(X,EZx,X,ezx)
Exercise \(\PageIndex{16}\)
\(f_{XY} (t, u) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 2u) + I_{(1, 2]} (t) \dfrac{9}{14} t^2 u^2\), for \(0 \le u \le 1\). (see Exercise 32 from " Problems on Mathematical Expectaton ", and Exercise 14.2.11 ).
\(f_X (t) = I_{[0, 1]} (t) \dfrac{3}{8} (t^2 + 1) + I_{(1, 2]} (t) \dfrac{3}{14} t^2\)
\(Z = I_M (X, Y) X + I_{M^c} (X, Y) XY\), \(M = \{(t, u): u \le \text{min } (1 , 2 - t)\}\)
- Answer
-
\(Z = I_M (X, Y) X + I_{M^c} (X, Y) XY\), \(M = \{(t, u): u \le \text{min } (1 , 2 - t)\}\)
\(E[|X = t] = I_{[0, 1]} (t) \int_{0}^{1} \dfrac{t^3+ 2tu}{t^2 + 1} \ du + I_{(1, 2]} (t) [\int_{0}^{2 - t} 3tu^2\ du + \int_{2 - t}^{1} 3tu^3\ du]\)
\(= I_{[0, 1]} (t) t + I_{(1, 2]} (t) (-\dfrac{13}{4} t+ 12t^2 - 12t^3 + 5t^4 - \dfrac{3}{4} t^5)\)
tuappr: [0 2] [0 1] 200 100 (t<=1).*(t.^2 + 2*u)./(t.^2 + 1) +3*u.^2.*(t>1) M = u<=min(1,2-t); G = M.*t + (1-M).*t.*u; EZx = sum(G.*P)./sum(P); N = X<=1; ezx = X.*N + (1-N).*(-(13/4)*X + 12*X.^2 - 12*X.^3 + 5*X.^4 - (3/4)*X.^5); plot(X,EZx,X,ezx)
Exercise \(\PageIndex{17}\)
Suppose \(X\) ~ uniform on 0 through \(n\) and \(Y\) ~ conditionally uniform on 0 through \(i\), given \(X = i\).
a. Determine \(E[Y]\) from \(E[Y|X = i]\).
b. Determine the joint distribution for \(\{X, Y\}\) for \(n = 50\) (see
Example 7
from "
Conditional Expectation, Regression
" for a possible approach). Use jcalc to determine \(E[Y]\); compare with the theoretical value.
- Answer
-
a. \(E[Y|X = i] = i/2\), so
\(E[Y] = \sum_{i = 0}^{n} E[Y|X = i] P(X = i) = \dfrac{1}{n + 1} \sum_{i = 1}^{n} i/2 = n/4\)
b. \(P(X = i) = 1/(n + 1)\), \(0 \le i \le n\), \(P(Y = k|X = i) = 1/(i + 1)\). \(0 \le k \le i\); hence
\(P(X = i, Y = k) = 1/(n + 1)(i + 1)\), \(0 \le i \le n\), \(0 \le k \le i\).n = 50; X = 0:n; Y = 0:n; P0 = zeros(n+1,n+1); for i = 0:n P0(i+1,1:i+1) = (1/((n+1)*(i+1)))*ones(1,i+1); end P = rot90(P0); jcalc: X Y P - - - - - - - - - - - EY = dot(Y,PY) EY = 12.5000 % Comparison with part (a): 50/4 = 12.5
Exercise \(\PageIndex{18}\)
Suppose \(X\) ~ uniform on 1 through \(n\) and \(Y\) ~ conditionally uniform on 1 through \(i\), given \(X = i\).
a. Determine \(E[Y]\) from \(E[Y|X = i]\).
b. Determine the joint distribution for \(\{X, Y\}\) for \(n = 50\) (see
Example 7
from "
Conditional Expectation, Regression
" for a possible approach). Use jcalc to determine \(E[Y]\); compare with the theoretical value.
- Answer
-
a. \(E[Y|X = i] = (i+1)/2\), so
\(E[Y] = \sum_{i = 0}^{n} E[Y|X = i] P(X = i) = \dfrac{1}{n + 1} \sum_{i = 1}^{n} \dfrac{i + 1}{2} = \dfrac{n +3}{4}\)
b. \(P(X = i) = 1/n\), \(1 \le i \le n\), \(P(Y = k|X = i) = 1/i\). \(1 \le k \le i\); hence
\(P(X = i, Y = k) = 1/ni\), \(1 \le i \le n\), \(1 \le k \le i\).n = 50; X = 1:n; Y = 1:n; P0 = zeros(n,n); for i = 1:n P0(i,1:i) = (1/(n*i))*ones(1,i); end P = rot90(P0); jcalc: P X Y - - - - - - - - - - - - EY = dot(Y,PY) EY = 13.2500 % Comparison with part (a): 53/4 = 13.25
Exercise \(\PageIndex{19}\)
Suppose \(X\) ~ uniform on 1 through \(n\) and \(Y\) ~ conditionally binomial \((i, p)\), given \(X = i\).
a. Determine \(E[Y]\) from \(E[Y|X = k]\).
b. Determine the joint distribution for \(\{X, Y\}\) for \(n = 50\) and \(p = 0.3\). Use jcalc to determine \(E[Y]\); compare with the theoretical value.
- Answer
-
a. \(E[Y|X = i] = ip\), so
\(E[Y] = \sum_{i = 1}^{n} E[Y|X = i] P(X = i) = \dfrac{p}{n} \sum_{i = 1}^{n} i = \dfrac{p(n + 1)}{2}\)
b. \(P(X = i) = 1/n\), \(1 \le i \le n\), \(P(Y = k|X = i)\) = ibinom\((i, p, 0:i)\), \(0 \le k \le i\).
n = 50; p = 0.3; X = 1:n; Y = 0:n; P0 = zeros(n,n+1); % Could use randbern for i = 1:n P0(i,1:i+1) = (1/n)*ibinom(i,p,0:i); end P = rot90(P0); jcalc: X Y P - - - - - - - - - - - EY = dot(Y,PY) EY = 7.6500 % Comparison with part (a): 0.3*51/2 = 0.765
Exercise \(\PageIndex{20}\)
A number \(X\) is selected randomly from the integers 1 through 100. A pair of dice is thrown \(X\) times. Let \(Y\) be the number of sevens thrown on the \(X\) tosses. Determine the joint distribution for \(\{X, Y\}\) and then determine \(E[Y]\).
- Answer
-
a. \(P(X = i) = 1/n\), \(E[Y|X = i] = i/6\), so
\(E[Y] = \dfrac{1}{6} \sum_{i = 0}^{n} i/n = \dfrac{(n + 1)}{12}\)
b.
n = 100; p = 1/6; X = 1:n; Y = 0:n; PX = (1/n)*ones(1,n); P0 = zeros(n,n+1); % Could use randbern for i = 1:n P0(i,1:i+1) = (1/n)*ibinom(i,p,0:i); end P = rot90(P0); jcalc EY = dot(Y,PY) EY = 8.4167 % Comparison with part (a): 101/12 = 8.4167
Exercise \(\PageIndex{21}\)
A number \(X\) is selected randomly from the integers 1 through 100. Each of two people draw \(X\) times, independently and randomly, a number from 1 to 10. Let \(Y\) be the number of matches (i.e., both draw ones, both draw twos, etc.). Determine the joint distribution and then determine \(E[Y]\).
- Answer
-
Same as Exercise 14.2.20 , except \(p = 1/10\). \(E[Y] = (n + 1)/20\)
n = 100; p = 0.1; X = 1:n; Y = 0:n; PX = (1/n)*ones(1,n); P0 = zeros(n,n+1); % Could use randbern for i = 1:n P0(i,1:i+1) = (1/n)*ibinom(i,p,0:i); end P = rot90(P0); jcalc - - - - - - - - - - EY = dot(Y,PY) EY = 5.0500 % Comparison with part (a): EY = 101/20 = 5.05
Exercise \(\PageIndex{22}\)
\(E[Y|X = t] = 10t\) and \(X\) has density function \(f_X (t) = 4 - 2t\) for \(1 \le t \le 2\). Determine \(E[Y]\).
- Answer
-
\(E[Y] = \int E[Y|X = t] f_X (t)\ dt = \int_{1}^{2} 10t(4 - 2t) \ dt = 40/3\)
Exercise \(\PageIndex{23}\)
\(E[Y|X = t] = \dfrac{2}{3} (1 - t)\) for \(0 \le t < 1\) and \(X\) has density function \(f_X (t) = 30 t^2 ( 1 - t)^2\) for \(0 \le t \le 1\). Determine \(E[Y]\).
- Answer
-
\(E[Y] = \int E[Y|X =t] f_X (t)\ dt = \int_{0}^{1} 20t^2 (1 - t)^3\ dt = 1/3\)
Exercise \(\PageIndex{24}\)
\(E[Y|X = t] = \dfrac{2}{3} (2 - t)\) and \(X\) has density function \(f_X(t) = \dfrac{15}{16} t^2 (2 - t)^2\) \(0 \le t < 2\). Determine \(E[Y]\).
- Answer
-
\(E[Y] = \int E[Y|X =t] f_X(t)\ dt = \dfrac{5}{8} \int_{0}^{2} t^2 (2 - t)^3\ dt = 2/3\)
Exercise \(\PageIndex{25}\)
Suppose the pair \(\{X, Y\}\) is independent, with \(X\) ~ Poisson (\(\mu\)) and \(Y\) ~ Poisson \((\lambda)\). Show that \(X\) is conditionally binomial \((n, \mu/(\mu + \lambda))\), given \(X + Y = n\). That is, show that
\(P(X = k|X + Y = n) = C(n, k) p^k (1 - p)^{n - k}\), \(0 \le k \le n\), for \(p = \mu/(\mu + \lambda)\)
- Answer
-
\(X\) ~ Poisson (\(\mu\)), \(Y\) ~ Poisson \((\lambda)\). Use of property (T1) and generating functions shows that \(X + Y\) ~ Poisson \((\mu + \lambda)\)
\(P(X = k|X + Y = n) = \dfrac{P(X = k, X + Y = n)}{P(X+Y = n)} = \dfrac{P(X = k, Y = n - k)}{P(X + Y) = n}\)
\(= \dfrac{e^{-\mu} \dfrac{\mu^k}{k!} e^{-\lambda} \dfrac{\lambda^{n -k}}{(n - k)!}}{e^{-(\mu + \lambda)} \dfrac{(\mu + \lambda)^n}{n!}} = \dfrac{n!}{k! (n - k)!} \dfrac{\mu^k \lambda^{n - k}}{(\mu + \lambda)^n}\)
Put \(p = \mu/(\mu + \lambda)\) and \(q = 1 - p = \lambda/(\mu + \lambda)\) to get the desired result.
Exercise \(\PageIndex{26}\)
Use the fact that \(g(X, Y) = g^* (X, Y, Z)\), where \(g^* (t, u, v)\) does not vary with \(v\). Extend property (CE10) to show
\(E[g(X, Y)|X = t, Z = v] = E[g(t, Y)|X = t, Z = v]\) a.s. \([P_{XZ}]\)
- Answer
-
\(E[g(X,Y)|X = t, Z = v] = E[g^* (X, Z, Y)| (X, Z) = (t, v)] = E[g^* (t, v, Y)|(X, Z) = (t, v)]\)
\(= E[g(t, Y)|X = t, Z = v]\) a.s. \([P_{XZ}]\) by (CE10)
Exercise \(\PageIndex{27}\)
Use the result of Exercise 14.2.26 and properties (CE9a) and (CE10) to show that
\(E[g(X, Y)|Z = v] = \int E[g(t, Y)|X = t, Z =v] F_{X|Z} (dt|v)\) a.s. \([P_Z]\)
- Answer
-
By (CE9) , \(E[g(X, Y)|Z] = E\{E|g(X, Y)|X, Z]|Z\} = E[e(X, Z)|Z]\) a.s.
By (CE10) ,
\(E[e(X, Z)|Z = v] = E[e(X, v)|Z = v] = \)
\(\int e(t, v) F_{X|Z} (dt|v)\) a.s.
By Exercise 14.2.26 ,
\(\int E[g(X, Y)|X = t, Z = v] F_{X|Z} (dt|v) = \)
\(\int E[g(t, Y)|X = t, Z = v] F_{X|Z} (dt|v)\) a.s. \([P_Z]\)
Exercise \(\PageIndex{28}\)
A shop which works past closing time to complete jobs on hand tends to speed up service on any job received during the last hour before closing. Suppose the arrival time of a job in hours before closing time is a random variable \(T\) ~ uniform [0, 1]. Service time \(Y\) for a unit received in that period is conditionally exponential \(\beta (2 - u)\), given \(T = u\). Determine the distribution unction for \(Y\).
- Answer
-
\(F_Y (v) = \int F_{Y|T} (v|u) f_T (u)\ du = \int_{0}^{1} (1 - e^{-\beta (2 - u)v})\ du = \)
\(1 - e^{-2\beta v} \dfrac{e^{\beta v} - 1}{\beta v} = 1 - e^{\beta v} [\dfrac{1 - e^{-\beta v}}{\beta v}]\), \(0 < v\)
Exercise \(\PageIndex{29}\)
Time to failure \(X\) of a manufactured unit has an exponential distribution. The parameter is dependent upon the manufacturing process. Suppose the parameter is the value of random variable \(H\) ~ uniform on[0.005, 0.01], and \(X\) is conditionally exponential \((u)\), given \(H = u\). Determine \(P(X > 150)\). Determine \(E[X|H = u]\) and use this to determine \(E[X]\).
- Answer
-
\(F_{X|H} (t|u) = 1 - e^{ut}\) \(f_{H} (u) = \dfrac{1}{0.05} = 200\), \(0.005 \le u \le 0.01\)
\(F_X (t) = 1 - 200 \int_{0.005}^{0.01} e^{-ut}\ du = 1 - \dfrac{200}{t} [e^{-0.005t} - e^{-0.01t}]\)
\(P(X > 150) = \dfrac{200}{150}[e^{-0.75} - e^{-1.5}] \approx 0.3323\)
\(E[X|H = u] = 1/u\) \(E[X] = 200 \int_{0.005}^{0.01} \dfrac{du}{u} = 200 \text{ln } 2\)
Exercise \(\PageIndex{30}\)
A system has \(n\) components. Time to failure of the \(i\)th component is \(X_i\) and the class
\(\{X_i: 1 \le i \le n\}\) is iid exponential (\(\lambda\)). The system fails if any one or more of the components fails. Let \(W\) be the time to system failure. What is the probability the failure is due to the \(i\)th component?
Suggestion . Note that \(W = X_i\) iff \(X_j > X_i\), for all \(j \ne i\). Thus
\(\{W = X_i\} = \{(X_1, X_2, \cdot\cdot\cdot, X_n) \in Q\}\), \(Q = \{(t_1, t_2, \cdot\cdot\cdot t_n): t_k > t_i, \forall k \ne i\}\)
\(P(W = X_i) = E[I_Q (X_1, X_2, \cdot\cdot\cdot, X_n)] = E\{E[I_Q (X_1, X_2, \cdot\cdot\cdot, X_n)|X_i]\}\)
- Answer
-
Let \(Q = \{(t_1, t_2, \cdot\cdot\cdot, t_n): t_k > t_i, k \ne i\}\). Then
\(P(W = X_i) = E[I_Q (X_1, X_2, \cdot\cdot\cdot, X_n)] = E\{E[I_Q (X_1, X_2, \cdot\cdot\cdot, X_n)|X_i]\}\)
\(= \int E[I_Q(X_1, X_2, \cdot\cdot\cdot, t_i, \cdot\cdot\cdot X_n)] F_X (dt)\)
\(E[I_Q (X_1, X_2, \cdot\cdot\cdot, t_i, \cdot\cdot\cdot, X_n)] = \prod_{k \ne i} P(X_k > t) = [1 - F_X (t)]^{n - 1}\)
If \(F_X\) is continuous, strictly increasing, zero for \(t < 0\), put \(u = F_X (t)\), \(du = f_X (t)\ dt\), \(t = 0\) ~ \(u = 0, t = \infty\) ~ \(u = 1\). Then
\(P(W = X_i) = \int_{0}^{1} (1 - u)^{n - 1}\ du = \int_{0}^{1} u^{n - 1}\ du = 1/n\)