14.1 Problems on conditional expectation, regression (Page 2/5)

Applied probability Page 2 / 5

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.

(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"). $f_{X Y} (t, u) = 1$ for $0 \leq t \leq 1$ , $0 \leq u \leq 2 (1 - t)$ .

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

f_{X} (t) = 2 (1 - t), 0 \leq t \leq 1

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

f_{Y | X} (u | t) = \frac{1}{2 (1 - t)}, 0 \leq t \leq 1, 0 \leq u \leq 2 (1 - t)

E [Y | X = t] = \frac{1}{2 (1 - t)} \int_{0}^{2 (1 - t)} u d u = 1 - t, 0 \leq t \leq 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)

Got questions? Get instant answers now!

(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"). $f_{X Y} (t, u) = \frac{1}{8} (t + u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq 2$ .

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

f_{X} (t) = \frac{1}{4} (t + 1), 0 \leq t \leq 2

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

f_{Y | X} (u | t) = \frac{(t + u)}{2 (t + 1)} 0 \leq t \leq 2, 0 \leq u \leq 2

E [Y | X = t] = \frac{1}{2 (t + 1)} \int_{0}^{2} (t u + u^{2}) d u = 1 + \frac{1}{3 t + 3} 0 \leq t \leq 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

Got questions? Get instant answers now!

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

The regression line of Y on X is $u = 0.0958 t + 1.4876$ .

f_{X} (t) = \frac{3}{88} (1 + t) (1 + 4 t + t^{2}) = \frac{3}{88} (1 + 5 t + 5 t^{2} + t^{3}), 0 \leq t \leq 2

The regression line of Y on X is $u = 0.0958 t + 1.4876$ .

f_{Y | X} (u | t) = \frac{2 t + 3 u^{2}}{(1 + t) (1 + 4 t + t^{2})} 0 \leq u \leq 1 + t

E [Y | X = t] = \frac{1}{(1 + t) (1 + 4 t + t^{2})} \int_{0}^{1 + t} (2 t u + 3 u^{3}) d u

= \frac{(t + 1) (t + 3) (3 t + 1)}{4 (1 + 4 t + t^{2})}, 0 \leq t \leq 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

Got questions? Get instant answers now!

(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"). $f_{X Y} (t, u) = 12 t^{2} u$ on the parallelogram with vertices

(- 1, 0), (0, 0), (1, 1), (0, 1)

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

f_{X} (t) = I_{[- 1, 0]} (t) 6 t^{2} {(t + 1)}^{2} + I_{(0, 1]} (t) 6 t^{2} (1 - t^{2})

The regression line of Y on X is $u = (23 t + 4) / 18$ .

f_{Y | X} (u | t) = I_{[- 1, 0]} (t) \frac{2 u}{{(t + 1)}^{2}} + I_{(0, 1]} (t) \frac{2 u}{(1 - t^{2})} on the parallelogram

E [Y | X = t] = I_{[- 1, 0]} (t) \frac{1}{{(t + 1)}^{2}} \int_{0}^{t + 1} 2 u d u + I_{(0, 1]} (t) \frac{1}{(1 - t^{2})} \int_{t}^{1} 2 u d u

= I_{[- 1, 0]} (t) \frac{2}{3} (t + 1) + I_{(0, 1]} (t) \frac{2}{3} \frac{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

Got questions? Get instant answers now!

(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"). $f_{X Y} (t, u) = \frac{24}{11} t u$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {1, 2 - t}$ .

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

f_{X} (t) = I_{[0, 1]} (t) \frac{12}{11} t + I_{(1, 2]} (t) \frac{12}{11} t {(2 - t)}^{2}

The regression line of Y on X is $u = (- 124 t + 368) / 431$ .

f_{Y | X} (u | t) = I_{[0, 1]} (t) 2 u + I_{(1, 2]} (t) \frac{2 u}{{(2 - t)}^{2}}

E [Y | X = t] = I_{[0, 1]} (t) \int_{0}^{1} 2 u^{2} d u + I_{(1, 2]} (t) \frac{1}{{(2 - t)}^{2}} \int_{0}^{2 - t} 2 u^{2} d u

= I_{[0, 1]} (t) \frac{2}{3} + I_{(1, 2]} (t) \frac{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

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Applied probability' conversation and receive update notifications?

Ask

	Nutrition Exam By Hannah Sheth Start Quiz
	College physics for ap® courses By OpenStax Read Online Course
	9 Physiotherapy Modalities By Rhodes Start Exam
	8 Arts Society: Theater 8 By Jonathan Long Start Quiz
	14 Dr Landholt Large Animal Medicine-GI quiz By Brooke Delaney Start Exam
	28 Biology 28 Invertebrates MCQ By OpenStax Start Quiz
	26 Toxicology Gustafson- Biotoxins and Venoms By Brooke Delaney Start Exam
	18 Sociology 18 Work and the Economy MCQ By OpenStax Start Quiz
	17 Biology 17 Biotechnology and Genomics MCQ By OpenStax Start Quiz
©flickr: anjelkam	Art By Caitlyn Gobble Start Exam