Download presentation

Presentation is loading. Please wait.

Published byBrett Kittridge Modified about 1 year ago

1
1 The Vision Thing Power Thirteen Bivariate Normal Distribution

2
2 Outline Circles around the origin Circles translated from the origin Horizontal ellipses around the (translated) origin Vertical ellipses around the (translated) origin Sloping ellipses

3
3 x y x = 0, x 2 =1 y = 0, y 2 =1 x, y = 0

4
4 x y x = a, x 2 =1 y = b, y 2 =1 x, y = 0 a b

5
5 x y x = 0, x 2 > y 2 y = 0 x, y = 0

6
6 x y x = 0, x 2 < y 2 y = 0 x, y = 0

7
7 x y x = a, x 2 > y 2 y = b x, y > 0 a b

8
8 x y x = a, x 2 > y 2 y = b x, y < 0 a b

9
9 Why? The Bivariate Normal Density and Circles f(x, y) = {1/[2 x y ]}*exp{(-1/[2(1- )]* ([(x- x )/ x ] 2 -2 ([(x- x )/ x ] ([(y- y )/ y ] + ([(y- y )/ y ] 2 } If means are zero and the variances are one and no correlation, then f(x, y) = {1/2 }exp{(-1/2 )*(x 2 + y 2 ), where f(x,y) = constant, k, for an isodensity ln2 k =(-1/2)*(x 2 + y 2 ), and (x 2 + y 2 )= -2ln2 k=r 2

10
10 Ellipses If x 2 > y 2, f(x,y) = {1/[2 x y ]}*exp{(-1/2)* ([(x- x )/ x ] 2 + ([(y- y )/ y ] 2 }, and x* = (x- x ) etc. f(x,y) = {1/[2 x y ]}exp{(-1/2)* ([x*/ x ] 2 + [y*/ y ] 2 ), where f(x,y) =constant, k, and ln{k [2 x y ]} = (-1/2) ([x*/ x ] 2 + [y*/ y ] 2 ) and x 2 /c 2 + y 2 /d 2 = 1 is an ellipse

11
11 x y x = 0, x 2 < y 2 y = 0 x, y < 0 Correlation and Rotation of the Axes Y’ X’

12
12 Bivariate Normal: marginal & conditional If x and y are independent, then f(x,y) = f(x) f(y), i.e. the product of the marginal distributions, f(x) and f(y) The conditional density function, the density of y conditional on x, f(y/x) is the joint density function divided by the marginal density function of x: f(y/x) = f(x, y)/f(x)

13
Conditional Distribution f(y/x)= 1/[ y ]exp{[-1/2(1- y 2 ]* [y- y - x- x )( y / x )]} the mean of the conditional distribution is: y + (x - x ) )( y / x ), i.e this is the expected value of y for a given value of x, x=x*: E(y/x=x*) = y + (x* - x ) )( y / x ) The variance of the conditional distribution is: VAR(y/x=x*) = x 2 (1- ) 2

14
14 x y x = a, x 2 > y 2 y = b x, y > 0 xx yy Regression line intercept: y - x ( y / x ) slope: ( y / x )

15
15 Bivariate Regression: Another Perspective Regression line is the E(y/x) line if y and x are bivariate normal –intercept: y - x x / y ) –slope: x / y )

16
16 Example: Lab Six

17
17 Example: Lab Six

18
18 Correlation Matrix GEINDEX GE INDEX

19
19 Bivariate Regression: Another Perspective Regression line is the E(y/x) line if y and x are bivariate normal –intercept: y - x x / y ) –slope: x / y ) – y = – – x = – x / y ) = ( / ) = –intercept = –slope = 1.094

20
20

21
21 Vs Vs

22
22 Bivariate Normal Distribution and the Linear probability Model

23
23 income education x = a, x 2 > y 2 y = b x, y > 0 mean income players Mean educ. Players Mean Educ Non-Players Mean income non Non-Players Players

24
24 income education x = a, x 2 > y 2 y = b x, y > 0 mean income players Mean educ. Players Mean Educ Non-Players Mean income Non-Players Non-Players Players

25
25 income education x = a, x 2 > y 2 y = b x, y > 0 mean income players Mean educ. Players Mean Educ Non-Players Mean income Non-Players Non-Players Players Discriminating line

26
26 Discriminant Function, Linear Probability Function, and Decision Theory, Lab 6 Expected Costs of Misclassification –E(C) = C(P/N)P(P/N)P(N)+C(N/P)P(N/P)P(P) Assume C(P/N) = C(N/P) Relative Frequencies P(N)=23/100~1/4, P(P)=77/100~3/4 Equalize two costs of misclassification by setting fitted value of P(P/N), i.e.Bern to 3/4 –E(C) = C(P/N)(3/4)(1/4)+C(N/P)(1/4)(3/4)

27
27 income education x = a, x 2 > y 2 y = b x, y > 0 mean income players Mean educ. players Mean Educ Non-Players Mean income Non-Players Non-Players Players Discriminating line Note: P(P/N) is area of the non-players distribution below (southwest) of the line

28
28 Set Bern = 3/4 = *education *income, solve for education as it depends on income and plot

29
29 7 non-players misclassified, as well as 14players misclassified

30
30

31
31 Decision Theory Moving the discriminant line, I.e. changing the cutoff value from 0.75 to 0.5, changes the numbers of those misclassified, favoring one population at the expense of another you need an implicit or explicit notion of the costs of misclassification, such as C(P/N) and C(N/P) to make the necessary judgement of where to draw the line

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google