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14.2 Problems on Conditional Expectation, Regression

Conditional expectation given X computed from earlier joint distributions, tracing the regression curve and comparing it with the best linear fit.

Applied Probability, 14.2 Problems on Conditional Expectation, Regression

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24 exercises

14.2 Exercise 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"). fXY(t,u)=1f_{X Y} \left(t , u\right) = 1 for 0≤t≤10 \leq t \leq 1, 0≤u≤2(1-t)0 \leq u \leq 2 \left(1 - t\right).

The regression line of Y on X is u=1-tu = 1 - t.

f X ( t ) = 2 ( 1 - t ) , 0 ≤ t ≤ 1 f_{X} \left(t\right) = 2 \left(1 - t\right) , 0 \leq t \leq 1
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14.2 Exercise 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"). fXY(t,u)=18(t+u)f_{X Y} \left(t , u\right) = \frac{1}{8} \left(t + u\right) for 0≤t≤20 \leq t \leq 2, 0≤u≤20 \leq u \leq 2.

The regression line of Y on X is u=-t/11+35/33u = - t / 11 + 35 / 33.

f X ( t ) = 1 4 ( t + 1 ) , 0 ≤ t ≤ 2 f_{X} \left(t\right) = \frac{1}{4} \left(t + 1\right) , 0 \leq t \leq 2
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14.2 Exercise 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"). fXY(t,u)=388(2t+3u2)f_{X Y} \left(t , u\right) = \frac{3}{88} \left(2 t + 3 u^{2}\right) for 0≤t≤20 \leq t \leq 2, 0≤u≤1+t0 \leq u \leq 1 + t.

The regression line of Y on X is u=0.0958t+1.4876u = 0 . 0958 t + 1 . 4876.

f X ( t ) = 3 88 ( 1 + t ) ( 1 + 4 t + t 2 ) = 3 88 ( 1 + 5 t + 5 t 2 + t 3 ) , 0 ≤ t ≤ 2 f_{X} \left(t\right) = \frac{3}{88} \left(1 + t\right) \left(1 + 4 t + t^{2}\right) = \frac{3}{88} \left(1 + 5 t + 5 t^{2} + t^{3}\right) , 0 \leq t \leq 2
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14.2 Exercise 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"). fXY(t,u)=12t2uf_{X Y} \left(t , u\right) = 12 t^{2} u on the parallelogram with vertices

( - 1 , 0 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 0 , 1 ) \left(- 1 , 0\right) , \left(0 , 0\right) , \left(1 , 1\right) , \left(0 , 1\right)

The regression line of Y on X is u=(4t+5)/9u = \left(4 t + 5\right) / 9.

f X ( t ) = I [ - 1 , 0 ] ( t ) 6 t 2 ( t + 1 ) 2 + I ( 0 , 1 ] ( t ) 6 t 2 ( 1 - t 2 ) f_{X} \left(t\right) = I_{\left[- 1 , 0\right]} \left(t\right) 6 t^{2} \left(t + 1\right)^{2} + I_{\left(0 , 1\right]} \left(t\right) 6 t^{2} \left(1 - t^{2}\right)
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14.2 Exercise 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"). fXY(t,u)=2411tuf_{X Y} \left(t , u\right) = \frac{24}{11} t u for 0≤t≤20 \leq t \leq 2, 0≤u≤min{1,2-t}0 \leq u \leq \min \left\{1 , 2 - t\right\}.

The regression line of Y on X is u=(-124t+368)/431u = \left(- 124 t + 368\right) / 431

f X ( t ) = I [ 0 , 1 ] ( t ) 12 11 t + I ( 1 , 2 ] ( t ) 12 11 t ( 2 - t ) 2 f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{12}{11} t + I_{\left(1 , 2\right]} \left(t\right) \frac{12}{11} t \left(2 - t\right)^{2}
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14.2 Exercise 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"). fXY(t,u)=323(t+2u)f_{X Y} \left(t , u\right) = \frac{3}{23} \left(t + 2 u\right) for 0≤t≤20 \leq t \leq 2, 0≤u≤max{2-t,t}0 \leq u \leq \max \left\{2 - t , t\right\}.

The regression line of Y on X is u=1.0561t-0.2603u = 1 . 0561 t - 0 . 2603.

f X ( t ) = I [ 0 , 1 ] ( t ) 6 23 ( 2 - t ) + I ( 1 , 2 ] ( t ) 6 23 t 2 f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{6}{23} \left(2 - t\right) + I_{\left(1 , 2\right]} \left(t\right) \frac{6}{23} t^{2}
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14.2 Exercise 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"). fXY(t,u)=213(t+2u)f_{X Y} \left(t , u\right) = \frac{2}{13} \left(t + 2 u\right), for 0≤t≤20 \leq t \leq 2, 0≤u≤min{2t,3-t}0 \leq u \leq \min \left\{2 t , 3 - t\right\}.

The regression line of Y on X is u=-0.1359t+1.0839u = - 0 . 1359 t + 1 . 0839.

f X ( t ) = I [ 0 , 1 ] ( t ) 12 13 t 2 + I ( 1 , 2 ] ( t ) 6 13 ( 3 - t ) f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{12}{13} t^{2} + I_{\left(1 , 2\right]} \left(t\right) \frac{6}{13} \left(3 - t\right)
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14.2 Exercise 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"). fXY(t,u)=I[0,1](t)38(t2+2u)+I(1,2](t)914t2u2f_{X Y} \left(t , u\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{3}{8} \left(t^{2} + 2 u\right) + I_{\left(1 , 2\right]} \left(t\right) \frac{9}{14} t^{2} u^{2},

for 0≤u≤10 \leq u \leq 1.

The regression line of Y on X is u=0.0817t+0.5989u = 0 . 0817 t + 0 . 5989.

f X ( t ) = I [ 0 , 1 ] ( t ) 3 8 ( t 2 + 1 ) + I ( 1 , 2 ] ( t ) 3 14 t 2 f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{3}{8} \left(t^{2} + 1\right) + I_{\left(1 , 2\right]} \left(t\right) \frac{3}{14} t^{2}
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14.2 Exercise 12

fXY(t,u)=388(2t+3u2)f_{X Y} \left(t , u\right) = \frac{3}{88} \left(2 t + 3 u^{2}\right) for 0≤t≤20 \leq t \leq 2, 0≤u≤1+t0 \leq u \leq 1 + t (see Exercise 37 from "Problems on Mathematical Expectation", and the exercise).

f X ( t ) = 3 88 ( 1 + t ) ( 1 + 4 t + t 2 ) = 3 88 ( 1 + 5 t + 5 t 2 + t 3 ) , 0 ≤ t ≤ 2 f_{X} \left(t\right) = \frac{3}{88} \left(1 + t\right) \left(1 + 4 t + t^{2}\right) = \frac{3}{88} \left(1 + 5 t + 5 t^{2} + t^{3}\right) , 0 \leq t \leq 2
Z = I [ 0 , 1 ] ( X ) 4 X + I ( 1 , 2 ] ( X ) ( X + Y ) Z = I_{\left[0 , 1\right]} \left(X\right) 4 X + I_{\left(1 , 2\right]} \left(X\right) \left(X + Y\right)
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14.2 Exercise 13

fXY(t,u)=2411tuf_{X Y} \left(t , u\right) = \frac{24}{11} t u for 0≤t≤20 \leq t \leq 2, 0≤u≤min{1,2-t}0 \leq u \leq \min \left\{1 , 2 - t\right\} (see Exercise 38 from "Problems on Mathematical Expectaton", the exercise).

f X ( t ) = I [ 0 , 1 ] ( t ) 12 11 t + I ( 1 , 2 ] ( t ) 12 11 t ( 2 - t ) 2 f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{12}{11} t + I_{\left(1 , 2\right]} \left(t\right) \frac{12}{11} t \left(2 - t\right)^{2}
Z = I M ( X , Y ) 1 2 X + I M c ( X , Y ) Y 2 , M = { ( t , u ) : u > t } Z = I_{M} \left(X , Y\right) \frac{1}{2} X + I_{M^{c}} \left(X , Y\right) Y^{2} , M = \left\{\left(t , u\right) : u > t\right\}
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14.2 Exercise 14

fXY(t,u)=323(t+2u)f_{X Y} \left(t , u\right) = \frac{3}{23} \left(t + 2 u\right) for 0≤t≤20 \leq t \leq 2, 0≤u≤max{2-t,t}0 \leq u \leq \max \left\{2 - t , t\right\} (see Exercise 39, and the exercise).

f X ( t ) = I [ 0 , 1 ] ( t ) 6 23 ( 2 - t ) + I ( 1 , 2 ] ( t ) 6 23 t 2 f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{6}{23} \left(2 - t\right) + I_{\left(1 , 2\right]} \left(t\right) \frac{6}{23} t^{2}
Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y , M = { ( t , u ) : max ( t , u ) ≤ 1 } Z = I_{M} \left(X , Y\right) \left(X + Y\right) + I_{M^{c}} \left(X , Y\right) 2 Y , M = \left\{\left(t , u\right) : \max \left(t , u\right) \leq 1\right\}
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14.2 Exercise 15

fXY(t,u)=213(t+2u)f_{X Y} \left(t , u\right) = \frac{2}{13} \left(t + 2 u\right), for 0≤t≤20 \leq t \leq 2, 0≤u≤min{2t,3-t}0 \leq u \leq \min \left\{2 t , 3 - t\right\}. (see Exercise 31 from "Problems on Mathematical Expectation", and the exercise).

f X ( t ) = I [ 0 , 1 ] ( t ) 12 13 t 2 + I ( 1 , 2 ] ( t ) 6 13 ( 3 - t ) f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{12}{13} t^{2} + I_{\left(1 , 2\right]} \left(t\right) \frac{6}{13} \left(3 - t\right)
Z = I M ( X , Y ) ( X + Y ) + I M c ( X , Y ) 2 Y 2 , M = { ( t , u ) : t ≤ 1 , u ≥ 1 } Z = I_{M} \left(X , Y\right) \left(X + Y\right) + I_{M^{c}} \left(X , Y\right) 2 Y^{2} , M = \left\{\left(t , u\right) : t \leq 1 , u \geq 1\right\}
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14.2 Exercise 16

fXY(t,u)=I[0,1](t)38(t2+2u)+I(1,2](t)914t2u2f_{X Y} \left(t , u\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{3}{8} \left(t^{2} + 2 u\right) + I_{\left(1 , 2\right]} \left(t\right) \frac{9}{14} t^{2} u^{2}, for 0≤u≤10 \leq u \leq 1. (see Exercise 32 from "Problems on Mathematical Expectation", and the exercise).

f X ( t ) = I [ 0 , 1 ] ( t ) 3 8 ( t 2 + 1 ) + I ( 1 , 2 ] ( t ) 3 14 t 2 f_{X} \left(t\right) = I_{\left[0 , 1\right]} \left(t\right) \frac{3}{8} \left(t^{2} + 1\right) + I_{\left(1 , 2\right]} \left(t\right) \frac{3}{14} t^{2}
Z = I M ( X , Y ) X + I M c ( X , Y ) X Y , M = { ( t , u ) : u ≤ min ( 1 , 2 - t ) } Z = I_{M} \left(X , Y\right) X + I_{M^{c}} \left(X , Y\right) X Y , M = \left\{\left(t , u\right) : u \leq \min \left(1 , 2 - t\right)\right\}
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14.2 Exercise 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}\left\{X , Y\right\} and then determine E[Y]E \left[Y\right].

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14.2 Exercise 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]E \left[Y\right].

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14.2 Exercise 22

E[Y|X=t]=10tE \left[\right. Y \left|\right. X = t \left]\right. = 10 t and X has density function fX(t)=4-2tf_{X} \left(t\right) = 4 - 2 t for 1≤t≤21 \leq t \leq 2. Determine E[Y]E \left[Y\right].

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14.2 Exercise 23

E[Y|X=t]=23(1-t)E \left[\right. Y \left|\right. X = t \left]\right. = \frac{2}{3} \left(1 - t\right) for 0≤t<10 \leq t < 1 and X has density function fX(t)=30t2(1-t)2f_{X} \left(t\right) = 30 t^{2} \left(1 - t\right)^{2} for 0≤t≤10 \leq t \leq 1. Determine E[Y]E \left[Y\right].

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14.2 Exercise 24

E[Y|X=t]=23(2-t)E \left[\right. Y \left|\right. X = t \left]\right. = \frac{2}{3} \left(2 - t\right) and X has density function fX(t)=1516t2(2-t)20≤t<2f_{X} \left(t\right) = \frac{15}{16} t^{2} \left(2 - t\right)^{2} 0 \leq t < 2. Determine E[Y]E \left[Y\right].

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14.2 Exercise 25

Suppose the pair {X,Y}\left\{X , Y\right\} is independent, with X∼X \sim Poisson (μ)\left(\mu\right) and Y∼Y \sim Poisson (λ)\left(\lambda\right). Show that X is conditionally binomial (n,μ/(μ+λ))\left(n , \mu / \left(\mu + \lambda\right)\right), given X+Y=nX + Y = n. That is, show that

P ( X = k | X + Y = n ) = C ( n , k ) p k ( 1 - p ) n - k , 0 ≤ k ≤ n , for p = μ / ( μ + λ ) P \left(\right. X = k \left|\right. X + Y = n \left.\right) = C \left(n , k\right) p^{k} \left(1 - p\right)^{n - k} , 0 \leq k \leq n , \text{for} p = \mu / \left(\mu + \lambda\right)
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14.2 Exercise 26

Use the fact that g(X,Y)=g*(X,Y,Z)g \left(X , Y\right) = g^{\star} \left(X , Y , Z\right), where g*(t,u,v)g^{\star} \left(t , u , v\right) 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 X Z ] E \left[\right. g \left(X , Y\right) \left|\right. X = t , Z = v \left]\right. = E \left[\right. g \left(t , Y\right) \left|\right. X = t , Z = v \left]\right. a . s . \left[P_{X Z}\right]
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14.2 Exercise 27

Use the result of the exercise and properties (CE9a) and (CE10) to show that

E [ g ( X , Y ) | Z = v ] = ∫ E [ g ( t , Y ) | X = t , Z = v ] F X | Z ( d t | v ) a . s . [ P Z ] E \left[\right. g \left(X , Y\right) \left|\right. Z = v \left]\right. = \int E \left[\right. g \left(t , Y\right) \left|\right. X = t , Z = v \left]\right. F_{X \left|\right. Z} \left(\right. d t \left|\right. v \left.\right) a . s . \left[P_{Z}\right]
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14.2 Exercise 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∼T \sim uniform [0,1]\left[0 , 1\right]. Service time Y for a unit received in that period is conditionally exponential β(2-u)\beta \left(2 - u\right), given T=uT = u. Determine the distribution function for Y.

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14.2 Exercise 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∼H \sim uniform on[0.005, 0.01], and X is conditionally exponential (u)\left(u\right), given H=uH = u. Determine P(X>150)P \left(X > 150\right). Determine E[X|H=u]E \left[\right. X \left|\right. H = u \left]\right. and use this to determine E[X]E \left[X\right].

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14.2 Exercise 30

A system has n components. Time to failure of the ith component is Xi and the class

{Xi:1≤i≤n}\left\{X_{i} : 1 \leq i \leq n\right\} is iid exponential (λ)\left(\lambda\right). 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 ith component?

Suggestion. Note that W=XiW = X_{i} iff Xj>XiX_{j} > X_{i} for all j≠ij \neq i. Thus

{ W = X i } = { ( X 1 , X 2 , ⋯ , X n ) ∈ Q } , Q = { ( t 1 , t 2 , ⋯ t n ) : t k > t i , ∀ k ≠ i } \left\{W = X_{i}\right\} = \left\{\left(X_{1} , X_{2} , \hdots , X_{n}\right) \in Q\right\} , Q = \left\{\left(t_{1} , t_{2} , \hdots t_{n}\right) : t_{k} > t_{i} , \forall k \neq i\right\}
P ( W = X i ) = E [ I Q ( X 1 , X 2 , ⋯ , X n ) ] = E { E [ I Q ( X 1 , X 2 , ⋯ , X n ) | X i ] } P \left(W = X_{i}\right) = E \left[I_{Q} \left(X_{1} , X_{2} , \hdots , X_{n}\right)\right] = E \left\{E \left[\right. I_{Q} \left(X_{1} , X_{2} , \hdots , X_{n}\right) \left|\right. X_{i} \left]\right.\right\}
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Adapted from Applied Probability © 2009 Paul E. Pfeiffer. Licensed under CC BY 3.0. Source: Applied Probability. Imported at revision d7a0420.