Download presentation
Presentation is loading. Please wait.
Published byBrett Kittridge Modified over 9 years 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 1.000000 0.636290 INDEX 0.636290 1.000000
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 = 0.022218 – – x = 0.014361 – x / y ) = (0.02543/0.043669) = –intercept = 0.0064 –slope = 1.094
20
20
21
21 Vs. 0.0064 Vs. 1.094
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 = 1.39 -0.0216*education - 0.0105*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
© 2025 SlidePlayer.com. Inc.
All rights reserved.