12.3 Problems on variance, covariance, linear regression (Page 4/5)

Applied probability Page 4 / 5

(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 performa 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


npr08_09 Data are in X, Y, P
jcalc- - - - - - - - - - - -
EX = dot(X,PX);EY = dot(Y,PY);
VX = dot(X.^2,PX) - EX^2VX =   0.3319
CV = total(t.*u.*P) - EX*EYCV =  -0.2586
a  = CV/VXa  =  -0.77937/6;
b = EY - a*EXb  =   4.3051       % Regression line: u = at + b

Got questions? Get instant answers now!

For the joint densities in Exercises 23-30 below

Determine analytically $Var [X]$ , $Cov [X, Y]$ , and the regression line of Y on X .
Check these with a discrete approximation.

(See Exercise 10 from "Problems On Random Vectors and Joint Distributions", and Exercise 20 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = 1$ for $0 \leq t \leq 1$ , $0 \leq u \leq 2 (1 - t)$ .

E [X] = \frac{1}{3}, E [X^{2}] = \frac{1}{6}, E [Y] = \frac{2}{3}

E [X Y] = \int_{0}^{1} \int_{0}^{2 (1 - t)} t u d u d t = 1 / 6

Cov [X, Y] = \frac{1}{6} - \frac{1}{3} \cdot \frac{2}{3} = - 1 / 18 Var [X] = 1 / 6 - {(1 / 3)}^{2} = 1 / 18

a = Cov [X, Y] / Var [X] = - 1 b = E [Y] - a E [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^2VX =   0.0556
CV = total(t.*u.*P) - EX*EYCV =  -0.0556
a = CV/VXa =   -1.0000
b = EY - a*EXb =    1.0000

Got questions? Get instant answers now!

(See Exercise 13 from "Problems On Random Vectors and Joint Distributions", and Exercise 23 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = \frac{1}{8} (t + u)$ for $0 \leq t \leq 2$ , $0 \leq u \leq 2$ .

E [X] = E [Y] = \frac{7}{6}, E [X^{2}] = \frac{5}{3}

E [X Y] = \frac{1}{8} \int_{0}^{2} \int_{0}^{2} t u (t + u) d u d t = 4 / 3, Cov [X, Y] = - 1 / 36, Var [X] = 11 / 36

a = Cov [X, Y] / 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

Got questions? Get instant answers now!

(See Exercise 15 from "Problems On Random Vectors and Joint Distributions", and Exercise 25 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = \frac{3}{88} (2 t + 3 u^{2})$ for $0 \leq t \leq 2$ , $0 \leq u \leq 1 + t$ .

E [X] = \frac{313}{220}, E [Y] = \frac{1429}{880}, E [X^{2}] = \frac{49}{22}

E [X Y] = \frac{3}{88} \int_{0}^{2} \int_{0}^{1 + t} t u (2 t + 3 u^{2}) d u d t = \frac{2153}{880} Cov [X, Y] = \frac{26383}{1933600}, Var [X] = \frac{9831}{48400}

a = Cov [X, Y] / Var [X] = \frac{26383}{39324} b = E [Y] - a E [X] = \frac{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

Got questions? Get instant answers now!

(See Exercise 16 from "Problems On Random Vectors and Joint Distributions", and Exercise 26 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = 12 t^{2} u$ on the parallelogram with vertices

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

E [X] = \frac{2}{5}, E [Y] = \frac{11}{15}, E [X^{2}] = \frac{2}{5}

E [X Y] = 12 \int_{- 1}^{0} \int_{0}^{t + 1} t^{3} u^{2} d u d t + 12 \int_{0}^{1} \int_{t}^{1} t^{3} u^{2} d u d t = \frac{2}{5}

Cov [X, Y] = \frac{8}{75}, Var [X] = \frac{6}{25}

a = Cov [X, Y] / 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

Got questions? Get instant answers now!

(See Exercise 17 from "Problems On Random Vectors and Joint Distributions", and Exercise 27 from "Problems on Mathematical Expectation"). $f_{X Y} (t, u) = \frac{24}{11} t u$ for $0 \leq t \leq 2$ , $0 \leq u \leq min {1, 2 - t}$ .

E [X] = \frac{52}{55}, E [Y] = \frac{32}{55}, E [X^{2}] = \frac{627}{605}

E [X Y] = \frac{24}{11} \int_{0}^{1} \int_{0}^{1} t^{2} u^{2} d u d t + \frac{24}{11} \int_{1}^{2} \int_{0}^{2 - t} t^{2} u^{2} d u d t = \frac{28}{55}

Cov [X Y] = - \frac{124}{3025}, Var [X] = \frac{431}{3025}

a = Cov [X, Y] / Var [X] = - \frac{124}{431} b = E [Y] - a E [X] = \frac{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

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

©flickr: eLife	Mycobacterial Infections (Mandell 8th Ch 251-254) By Cath Yu Start Quiz
	5 AP 05 Integumentary System Essay By OpenStax Start Flashcards
	4 Microbiology Final 4 By Madison Christian Start Quiz
	How much do you love him? By Zarina Chocolate Start Quiz
	8 AP 08 Appendicular Skeleton Essay By OpenStax Start Flashcards
	3 Neuroanatomy 03 Ventricular System By Stephen Voron Start Quiz
	5 Lec:5 Case-control Studies By Janet Forrester Start Quiz
	19 Sociology 19 Health and Medicine MCQ By OpenStax Start Quiz
	Principles of economics By OpenStax Read Online Course
	4 Microeconomics 04 Labor Financial Markets By OpenStax Start Flashcards