Re: Welch formula doesn´t converge
- From: "licas_@xxxxxxxxxxx" <licas_@xxxxxxxxxxx>
- Date: Sat, 16 Feb 2008 07:14:43 -0800 (PST)
On 15 Fev, 21:41, Jack Tomsky <jtom...@xxxxxxxxxxxxx> wrote:
Welch formula doesn´t converge...
Given that the estimator of the variance of the
difference of means for equal variance normal
Populations,
___var = [(ssX+ssY)/(nX+nY-2)]*(1/nX + 1/nY)
when dividing the difference on observed means: Xhat
- Yhat, follows exactly a T Student Distribution
(with nX+nY-2 df) it was expected that the Welch´s
formula:
___df = (vX/ nX + vY/ nY) ^2 / u
___ u =
= (1/(nX-1))* (vX/ nX) ^2 + (1/(nY-1))*(vY/ nY)^2
___________________(A)
Data:
Experiment 1 (10´000 pairs of samples)
______X ~ N(0, sd=1): 10____Y ~ N(0, sd=5): 10
__9_0.293___10_0.565__11_0.105__12_0.025 ...
______X ~ N(0, 1):10____Y~N(0, 1): 10
__. . . __13_0.035___14_0.063___15_0.100___16_0.147
__17_0.246___18_0.383___
It's clearly seen that if (A) converges relative to
the Population sd, then the frequency for df =
10+10-2 would be 1.000, 0.000 otherwise. BUT IN FACT
the frequencies vary from df=9 to df=18 when 10´000
pairs of simulated samples are used.
___100´000 sample pairs
______X ~ N(0, 1):10____Y~N(0, 1): 10
__. . . __13_0.037___14_0.062___15_0.095___16_0.149
__17_0.251___18_0.380___
Luis Amaral Afonso
REM "DID"
CLS
DEFDBL A-Z
PRINT " WELCH FORMULA : degrees of freedom
freedom "
INPUT " sX , nX "; sX, nX
INPUT " sY , nY "; sY, nY
DIM x(nX), y(nY), df(nX + nY)
all = 100000
pi = 4 * ATN(1)
FOR rpt = 1 TO all: RANDOMIZE TIMER
swX = 0: sswX = 0: swY = 0: sswY = 0
FOR i = 1 TO nX
aa = SQR(-2 * LOG(RND))
x(i) = sX * aa * COS(2 * pi * RND)
x = x(i)
swX = swX + x: sswX = sswX + x * x
NEXT i
FOR i = 1 TO nY
aa = SQR(-2 * LOG(RND))
y(i) = sY * aa * COS(2 * pi * RND)
y = y(i)
swY = swY + y: sswY = sswY + y * y
NEXT i
vX = (sswX - swX * swX / nX) / (nX - 1)
vY = (sswY - swY * swY / nY) / (nY - 1)
v1 = (1 / (nX - 1)) * ((vX / nX) ^ 2)
vv = (1 / (nY - 1)) * ((vY / nY) ^ 2)
a = (vX / nX + vY / nY) ^ 2
df = a / (v1 + vv)
u = INT(df + .5)
REM PRINT USING "## "; df;
df(u) = df(u) + 1
IF df > (nX + nY - 3) THEN g = g + 1
NEXT rpt
LOCATE 10, 1
FOR t = 0 TO nX + nY
IF df(t) = 0 THEN GOTO 40
PRINT USING "## #.### "; t; df(t) / all;
40 NEXT t: END
The Welsh df is not supposed to converge to a constant. It's a random variable which depends on the sample standard deviations and the sample sizes. It takes all values between min(N1, N2) - 1 and N1+ N2 - 2.
Jack (moderator)- Ocultar texto citado -
- Mostrar texto citado -
NO
When the underlying Normal Distributions tends to equal variances the
degrees of freedom MUST tend to a T Distribution with
nX + nY -2 degrees of freedom.
BUT ON CONTRARY THIS NEVER HAPPENS, CONCLUSION:
The Welch solution is not credible
Luis Amaral Afonso
.
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