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Topic: Behrens-Fisher by Intra-Permutations
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Luis A. Afonso

Posts: 4,615
From: LIsbon (Portugal)
Registered: 2/16/05
Behrens-Fisher by Intra-Permutations
Posted: Nov 25, 2012 6:03 PM
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Generalizing the Fisher?s Exact Permutation Method (FEPM): Application to the Behrens-Fisher Problem



Introduction

We intend to devise a paradigm to solve the Behrens-Fisher problem (BFP) consisting in to find a confidence interval for the difference on means of two independent normal samples, X~N(mu1, sigma1):n, Y~N(mu2, sigma2):m by permuting freely the items of the samples. This feature is akin to the Fisher?s Permutation but this one is restricted to distributions with same dispersions. The present permutation method only allows switches among the items the sample they belong. An important feature is that, for each pseudo-sample, the individual dispersion is kept unchanged.


The Intra-Permutation Method.
The arrival coefficients W( )

Let be X= X1, ?, Xm and chose at random and exhaustively all m without replacement, affecting each one by the index Wx(j)= j/(m*(m+1)/2 where j is the order the item is chosen. The same for Y= Y1, . . .,Yn, Wy(j)= j/(n*(n+1)/2.
Noting that
_____mmX = E(Sum (Wx(j)*X(j))) = E(Xhat)
_____mmY = E(Sum (Wy(j)*Y(j))) = E(Yhat)
Where
_______Xhat= (X1 + ?+ Xn)/m
_______Yhat= (Y1 + ? +Ym)/n
we view mmX - mmY as an element that can sample the r.v. D = E(X) - E(Y), obtaining D* and the respective 5% CI, as shown below by 40000 repetitions.


Results

X~N(4, 10^2):20, Y=N(0, 1^2):40

__Xhat___Yhat____D_____D*_____5%CI*____centre*
__4.099__0.093__4.007__4.000 __[1.29, 6.77]__4.03
__6.630_-0.055__6.685__6.694__ [4.48, 8.86]__6.67
__3.907_-0.020__3.927__3.902__ [2.04, 5.79]__3.92

X~N(4, 10^2):40, Y~N(0, 1^2):20

__4.394__0.395__3.999__4.002__[2.35, 5.71]__4.03
__1.450_-0.409__1.859__1.859__[0.29, 4.83]__2.56**
__4.866_-0.255__5.121__5.129__[3.34, 6.96]__5.15

X~N(4, 10^2):100, Y~N(0, 1^2):50

__3.493_-0.113__3.606__3.605__[2.40, 4.83]__3.61
__2.963_-0.126__3.089__3.088__[2.04, 4.14]__3.09
__5.785__0.041_ 5.744__5.745__[4.58, 6.92]__5.75

X~N(5, 7^2): 30, Y~N(1, 1^2):20

__4.278__1.262__3.015__3.019__[1.57, 4.48]__3.02
__6.279__0.509__5.771__5.780__[3.99, 7.67]__5.83
__2.633__0.692__1.941__1.936__[0.69, 3.24]__1.96



Comments


The pair of source samples from which the permutations are get, do have difference of means noted by D. The procedure allows us to obtain 5% significance Confidence Intervals for the difference of means, which centres are well in accordance with D, being of course irrelevant given the evaluation procedure. One (out of 12) doesn?t follow this regularity, when 1/100 standard deviations is present.
The CI bounds are those the empirical distribution provides. Summing-up we feel that intra-permutations could be a far reaching method if properly scrutinized.

Luis A. Afonso



REM "BF-intra"
CLS
REM
INPUT " m1 , stdev1 , n1 "; m1, s1, n1
INPUT " m2 , stdev2 , n2 "; m2, s2, n2
REM
DIM X(n1), Y(n2), XX(n1), YY(n2)
REM
DIM W(8001), W1(n1), W2(n2)
REM
pi = 4 * ATN(1)
REM
INPUT " How many "; many
REM
REM
RANDOMIZE TIMER
sumX = 0: sumY = 0
FOR i = 1 TO n1
a = RND
aa = SQR(-2 * LOG(a))
X(i) = m1 + s1 * aa * COS(2 * pi * RND)
sumX = sumX + X(i)
NEXT i
FOR i = 1 TO n1
W1(i) = i / (.5 * n1 * (n1 + 1))
NEXT i
FOR i = 1 TO n2
W2(i) = i / (.5 * n2 * (n2 + 1))
NEXT i
REM
FOR i = 1 TO n2
a = RND
aa = SQR(-2 * LOG(a))
Y(i) = m2 + s2 * aa * COS(2 * pi * RND)
sumY = sumY + Y(i)
NEXT i
FOR i = 1 TO n2
W2(i) = i / (.5 * n2 * (n2 + 1))
NEXT i
U = sumX / n1: V = sumY / n2
LOCATE 9, 33
PRINT USING "##.### "; U; V; U - V
REM
REM
FOR j = 1 TO many
RANDOMIZE TIMER
LOCATE 10, 48: PRINT USING "#######"; many - j
FOR ii = 1 TO n1: XX(ii) = X(ii): NEXT ii
FOR ii = 1 TO n2: YY(ii) = Y(ii): NEXT ii
sumXX = 0: Ex = 0: sumYY = 0: Ey = 0
FOR k = 1 TO n1 - 1
1 g = INT(RND * n1) + 1
IF XX(g) = 7777777 THEN GOTO 1
sumXX = sumXX + XX(g)
Ex = Ex + W1(k) * XX(g)
XX(g) = 7777777
NEXT k
remain = sumX - sumXX
Ex = Ex + W1(n1) * remain
REM
FOR k = 1 TO n2 - 1
2 g = INT(RND * n2) + 1
IF YY(g) = 7777777 THEN GOTO 2
sumYY = sumYY + YY(g)
Ey = Ey + W2(k) * YY(g)
YY(g) = 7777777
NEXT k
remain = sumY - sumYY
Ey = Ey + W2(n2) * remain
E = Ex - Ey
IF E < 0 THEN GOTO 11
W = INT(100 * E) + 1
W(W) = W(W) + 1
md = md + E / many
11 NEXT j
REM
c(1) = .025 * many
c(2) = .975 * many
FOR cc = 1 TO 2
su = 0
FOR t = 0 TO 8000
su = su + W(t)
IF su > c(cc) THEN GOTO 22
NEXT t
22 tw(cc) = t / 100: rr(cc) = su / many
NEXT cc
LOCATE 10, 40: COLOR 14
PRINT USING " ##.### "; md
LOCATE 13, 20
PRINT USING " [ ##.## (#.###)"; tw(1); rr(1);
PRINT USING " ##.## (#.###) ]"; tw(2); rr(2)
COLOR 7
END



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