Post on 01-May-2015
Disegno del modello di analisi dei dati sperimentali
Lezione 2
interpolare un modello ai dati e valutare
i relativi parametri
-4 -3 -2 -1 0 1 2 3 4
x
0
10
20
30
40
y
(-2,16)
(-1,7)
(0,4) (1,6)
(2,10)
2210 xxy
dove x1 = x ed x2 = x12
22110 xx
-4 -3 -2 -1 0 1 2 3 4
x
0
10
20
30
40
y
(-2,16)
(-1,7)
(0,4) (1,6)
(2,10)i
2210 xxy
dove x1 = x ed x2 = x12
22110 xx
εi è il residuo per la i-ma osservazione
Il modello migliore interpolante è un modello che minimizza la somma delle deviazioni quadrate fra il i valori osservati ed i valori predetti dal modello, i.e.
n
ii
1
2min
Come fare i calcoli
2210 xxy
dove x1 = x ed x2 = x12
22110 xx
(x,y) = (-2,16) => y = β0(1) + β1(-2) + β2(4) + ε = 16
(x,y) = (-1,7) => y = β0(1) + β1(-1) + β2(1) + ε = 7
(x,y) = (0,4) => y = β0(1) + β1(0) + β2(0) + ε = 4
(x,y) = (1,6) => y = β0(1) + β1(1) + β2(1) + ε = 6
(x,y) = (2,10) => y = β0(1) + β1(2) + β2(4) + ε = 10
x0 x1 x2 y
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
Matrice X Transposta
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
14
10
7
1
010
10
7
10
35
17
)'( 1XX
Matrice Inversa di X’X
100
010
001
34010
0100
1005
14
10
7
1
010
10
7
10
35
17
)'()'( 1 XXXX
(X’X)-1 è called il inverse matrix di X’X.
It è defined as
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
14
10
7
1
010
10
7
10
35
17
)'( 1XX
117
13
43
10
6
4
7
16
41014
21012
11111
'YX
421
111
001
111
421210 xxx
X
10
6
4
7
16
Y
41014
21012
11111
X'
34010
0100
1005
421
111
001
111
421
41014
21012
11111
XX'
14
10
7
1
010
10
7
10
35
17
)'( 1XX
117
13
43
10
6
4
7
16
41014
21012
11111
'YX
214.2
3.1
171.4
117
13
43
14
10
7
1
010
10
7
10
35
17
)'()'(ˆ
ˆ
ˆ
ˆ 1
2
1
0
YXXX
2214.23.1171.4ˆ xxy
Matrice di Varianza - Covarianza
stima della varianza residua (s2)Somma degli scarti quadratici
Y)(X''βYY' ˆˆ1
22
1
n
ii
n
iii yySSE
657.1343.455457
117
13
43
214.23.1171.4
10
6
4
7
16
1064716
829.035
657.12
pn
SSEs
gradi di libertà per s2
Varianza dei parametri stimati
Matrice di Varianza - Covarianza:
222120
121110
0201001
14
10
7
1
010
10
7
10
35
17
)'(
ccc
ccc
ccc
XX
402.0829.035
17)ˆ( 2
000 scV
083.0829.010
1)ˆ( 2
111 scV
059.0829.014
1)ˆ( 2
222 scV 242.0)ˆ()ˆ(
288.0)ˆ()ˆ(
634.0)ˆ()ˆ(
22
11
00
VSE
VSE
VSE
Covarianza dei parametri stimati
Matrice di Varianza - Covarianza
222120
121110
0201001
14
10
7
1
010
10
7
10
35
17
)'(
ccc
ccc
ccc
XX
0829.00)ˆ,ˆ( 210
20110 scscCov
118.0829.07
1)ˆ,ˆ( 2
202
0220 scscCov
0829.00)ˆ,ˆ( 221
21221 scscCov
limiti di confidenza per βi
)ˆ( iSE
1)ˆ(ˆ)ˆ(ˆ
,, ipniiipni VtVtP
303.42,05.0 t
95.0)260.3167.1(
95.0)061.0539.2(
95.0)901.6441.1(
2
1
0
P
P
P
Varianza della retta predetta
• Assumiamo che si voglia to predire y per un assegnato valore di x
• Il valore scelto di x è chiamato a
• Possiamo ora scrivere l'equazione come:
2210
2210
ˆˆˆ
ˆˆˆˆ
aa
xxy
221100ˆˆˆ aaa 221100
ˆˆˆ aaa
βa' ˆ
ˆ
ˆ
ˆ
ˆ
2
1
0
210
aaay
21)ˆ( s)yV aX(X'a'
Ex. a = -4
1641)4(41 2210 aaaa'
091.13829.0
16
4
1
14
10
7
1
010
10
7
10
35
17
1641)ˆ( 21
s)yV aX(X'a'
618.3)ˆ()ˆ( yVySE
795.44
214.2
4.1
171.4
1641ˆ
ˆ
ˆ
ˆ
ˆ
2
1
0
210
βa'
aaay
nota! dovrebbe essere -1.3
V(x+y) = V(x) + V(y) + 2Cov(x,y)V(x-y) = V(x) + V(y) – 2Cov(x,y)
)ˆ,ˆ(2)ˆ,ˆ(2)ˆ,ˆ(2
)ˆ()ˆ()ˆ(
)ˆˆˆ()ˆ(
221122001100
221100
221100
aaCovaaCovaaCov
aVaVaV
aaaVyV
V(ax) = a2V(x)Cov(ax,by) = abCov(x,y)
)ˆ,ˆ(2)ˆ,ˆ(2)ˆ,ˆ(2
)ˆ()ˆ()ˆ()ˆ(
212120201010
2221
210
20
CovaaCovaaCovaa
VaVaVayV
091.13118.032059.0256083.016402.0
016)4(2)118.0(16120)4(12059.016083.0)4(402.01)ˆ( 22
yV
Una via alternativa del calcolo
La varianza di una nuova osservazione di y
212 )1()ˆ()( syVsyV aX)(X'a'
a = -4
V(y) = (1+15.80)0.829 = 13.92SE(y) = 3.73
Varianza della rettaVarianza di nuova oss
limiti di confidenza
1))ˆ(ˆ)ˆ(ˆ( ,, ySEtyyySEtyP
95% limiti di confidenza limiti di confidenza per la retta:
a = -4
95.0)37.60)(23.29(
)62.3303.48.44)(62.3303.48.44(
yEP
yEP
95% limiti di confidenza per singole osservazioni:
95.0)85.60)(75.28(
)73.3303.48.44)(73.3303.48.44(
yEP
yEP
limiti di confidenza al 95%
-4 -3 -2 -1 0 1 2 3 4
x
-10
0
10
20
30
40
y
PredettiOss.Limiti per la curvaLimiti per singole oss.
come fare questo in SAS?
DATA eks21;
INPUT x y;
CARDS;
-2 16
-1 7
0 4
1 6
2 10
;
PROC GLM;
MODEL y = x x*x/solution ;
OUTPUT out= new p= yhat L95M= low_mean U95M = up_mean L95 = low U95 = upper;
RUN;
PROC PRINT;
RUN;
Number di observations in data set = 5 General Linear Models ProcedureDependent Variable: Y Source DF Sum di Squares Mean Square F Value Pr > F Model 2 85.54285714 42.77142857 51.62 0.0190Error 2 1.65714286 0.82857143 Corrected Total 4 87.20000000 R-Square C.V. Root MSE Y Mean 0.980996 10.58441 0.91025899 8.60000000 Source DF Type I SS Mean Square F Value Pr > F X 1 16.90000000 16.90000000 20.40 0.0457X*X 1 68.64285714 68.64285714 82.84 0.0119 Source DF Type III SS Mean Square F Value Pr > F X 1 16.90000000 16.90000000 20.40 0.0457X*X 1 68.64285714 68.64285714 82.84 0.0119 T per H0: Pr > |T| Std Error ofParameter Estimate Parameter=0 Estimate INTERCEPT 4.171428571 6.58 0.0224 0.63438867X -1.300000000 -4.52 0.0457 0.28784917X*X 2.214285714 9.10 0.0119 0.24327695 OBS X Y YHAT LOW_MEAN UP_MEAN LOW UPPER 1 -2 16 15.6286 11.9426 19.3145 10.2503 21.0068 2 -1 7 7.6857 5.2988 10.0726 3.0991 12.2723 3 0 4 4.1714 1.4419 6.9010 -0.6024 8.9453 4 1 6 5.0857 2.6988 7.4726 0.4991 9.6723 5 2 10 10.4286 6.7426 14.1145 5.0503 15.8068
s2
s
DATA eks21;
INPUT x y;
CARDS;
-4 .
-3.5 .
-3 .
-2.5 .
-2 16
-1.5 .
-1 7
-0.5 .
0 4
0.5 .
1 6
1.5 .
2 10
2.5 .
3 .
3.5 .
4 .
;
PROC GLM;
MODEL y = x x*x/solution ;
OUTPUT out= new p= yhat L95M= low_mean U95M = up_mean L95 = low U95 = upper;
RUN;
PROC PRINT;
RUN;
OBS X Y YHAT LOW_MEAN UP_MEAN LOW UPPER 1 -4.0 . 44.8000 29.2321 60.3679 28.7470 60.8530 2 -3.5 . 35.8464 24.1430 47.5499 23.5050 48.1878 3 -3.0 . 28.0000 19.6000 36.4000 18.7318 37.2682 4 -2.5 . 21.2607 15.5647 26.9568 14.3481 28.1733 5 -2.0 16 15.6286 11.9426 19.3145 10.2503 21.0068 6 -1.5 . 11.1036 8.5369 13.6702 6.4210 15.7862 7 -1.0 7 7.6857 5.2988 10.0726 3.0991 12.2723 8 -0.5 . 5.3750 2.7660 7.9840 0.6691 10.0809 9 0.0 4 4.1714 1.4419 6.9010 -0.6024 8.9453 10 0.5 . 4.0750 1.4660 6.6840 -0.6309 8.7809 11 1.0 6 5.0857 2.6988 7.4726 0.4991 9.6723 12 1.5 . 7.2036 4.6369 9.7702 2.5210 11.8862 13 2.0 10 10.4286 6.7426 14.1145 5.0503 15.8068 14 2.5 . 14.7607 9.0647 20.4568 7.8481 21.6733 15 3.0 . 20.2000 11.8000 28.6000 10.9318 29.4682 16 3.5 . 26.7464 15.0430 38.4499 14.4050 39.0878 17 4.0 . 34.4000 18.8321 49.9679 18.3470 50.4530
Un problema più complesso
-50
0
50
100
150
0 20 40 60 80 100 120
x
y
Interpola con un modello questi dati
DATA polynom;
INPUT x y;
CARDS;
0 8.62
10 -3.99
20 6.80
30 -7.70
40 3.44
50 12.01
60 23.37
70 9.25
80 34.93
90 70.05
100 126.70
;
DATA add;
SET polynom;
x2 = x**2;
x3 = x**3;
x4 = x**4;
PROC REG;
MODEL y = x x2 x3 x4;
RUN;
il SAS System 08:22 Tuesday, October 29, 2002 1 il REG Procedure Model: MODEL1 Dependent Variable: y Analysis di Varianza Sum di Mean Source DF Squares Square F Value Pr > F Model 4 15449 3862.13306 56.59 <.0001 Error 6 409.47543 68.24591 Corrected Total 10 15858 Root MSE 8.26111 R-Square 0.9742 Dependent Mean 25.77091 Adj R-Sq 0.9570 Coeff Var 32.05594 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 8.92923 7.90689 1.13 0.3019 x 1 -1.90184 1.21774 -1.56 0.1694 x2 1 0.09562 0.05335 1.79 0.1232 x3 1 -0.00165 0.00082091 -2.01 0.0917 x4 1 0.00000999 0.00000407 2.45 0.0495
polinomio di quarto ordine
il SAS System 08:22 Tuesday, October 29, 2002 2 Procedure REG Model: MODEL1 Dependent Variable: y Analysis di Varianza Sum di Mean Source DF Squares Square F Value Pr > F Model 3 15037 5012.44667 42.75 <.0001 Error 7 820.66769 117.23824 Corrected Total 10 15858 Root MSE 10.82766 R-Square 0.9482 Dependent Mean 25.77091 Adj R-Sq 0.9261 Coeff Var 42.01505 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 1.73490 9.62511 0.18 0.8621 x 1 0.59619 0.87649 0.68 0.5182 x2 1 -0.02928 0.02099 -1.39 0.2057
x3 1 0.00035168 0.00013776 2.55 0.0379
polinomio di terzo ordine
The SAS System 08:22 Tuesday, October 29, 2002 3 il REG Procedure Model: MODEL1 Dependent Variable: y Analysis di Varianza Sum di Mean Source DF Squares Square F Value Pr > F Model 2 14273 7136.65872 36.03 <.0001 Error 8 1584.69025 198.08628 Corrected Total 10 15858 Root MSE 14.07431 R-Square 0.9001 Dependent Mean 25.77091 Adj R-Sq 0.8751 Coeff Var 54.61318 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 14.39524 10.72255 1.34 0.2163 x 1 -1.41540 0.49888 -2.84 0.0219 x2 1 0.02347 0.00480 4.88 0.0012
polinomio di secondo ordine
The SAS System 08:22 Tuesday, October 29, 2002 4 il REG Procedure Model: MODEL1 Dependent Variable: y Analysis di Varianza Sum di Mean Source DF Squares Square F Value Pr > F Model 1 9547.03680 9547.03680 13.61 0.0050 Error 9 6310.97089 701.21899 Corrected Total 10 15858 Root MSE 26.48054 R-Square 0.6020 Dependent Mean 25.77091 Adj R-Sq 0.5578 Coeff Var 102.75361 Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 -20.81000 14.93704 -1.39 0.1970 x 1 0.93162 0.25248 3.69 0.0050
polinomio di primo ordine (una retta)
-50
0
50
100
150
0 20 40 60 80 100 120
x
y
True relationship:y = 5 + 0.1x – 0.02x2 + 0.0003x3 + εε è normally distributed with 0 mean ed σ = 10
-50
0
50
100
150
0 20 40 60 80 100 120
x
y
Estimated relationship:y = 14.395 – 1.415x + 0.0235x2
s = 14.07
-50
0
50
100
150
0 20 40 60 80 100 120
x
y
Estimated relationship:y = -20.81 + 0.932xs = 26.48
This è a better fit
than this
Notazioni Matriciali
Of particular interest to us è il fact that not even in regression analysis was much use made di matrix algebra. In fact one di us, as a statistics graduate student at Cambridge University in il early 1950s, had lectures on multiple regression that were couched in scalar notation!
This absence di matrices ed vectors è surely surprising when one thinks di A.C. Aitken. His two books, Matrices ed Determinants ed Statistical Mathematics were both first published in 1939, had fourth ed fifth editions, respectively, in 1947 ed 1948, ed are still in print. Yet, very surprisingly, il latter makes no use di matrices ed vectors which are so thoroughly dealt with in il former.
There were exceptions, di course, as have already been noted, such as Kempthorne (1952) ed his co-workers, e.g. Wilk ed Kempthorne (1955, 1956) – ed others, too. Even with matrix expressions available, arithmetic was a real problem. A regression analysis in il New Zealand Department di Agriculture in il mid-1950s involved 40 regressors. Using electromechanical calculators, two calculators (people) using row echelon methods needed six weeks to invert il 40 x 40 matrix. One person could do a row, then il other checked it (to a maximum capacity di 8 to 10 digits, hoping per 4- or 5-digit accuracy in il final result). That person did il next row ed passed it to il first person per checking; ed so on. This was il impasse: matrix algebra was appropriate ed not really difficult. But il arithmetic stemming therefrom could be a nightmare.
(From Linear Models 1945-1995 by Shayle R. Searle ed Charles E. McCulloch in Advances in Biometry (eds. Peter Armitage ed Herbert A. David), John Wiley & Sons, 1996)