Hi Nitish,
R^2 cannot take values of greater than 1.
Per definition (see
http://en.wikipedia.org/wiki/Coefficient_of_determination)
R^2 := 1- SSE/SST
whereby
SSE = sum of squared errors
SST = total sum of squares
For R^2 > 1 would require SSE/SST <0.
Since SSE and SST are non-negative (check
Hi Paul,
here's a lm model to illustrate this:
> summary(lm(y~x.1+x.2))
Call:
lm(formula = y ~ x.1 + x.2)
Residuals:
Min 1Q Median 3QMax
-0.0561359 -0.0054020 0.0004553 0.0056516 0.0515817
Coefficients:
Estimate Std. Error t value Pr(>|t|)
Hi useRs,
can a variance test for 2 non-normal samples be tested in R? Also, thus
far I have not been able to find the Friedman two way analysis of variance.
For normal r.v., the var.test is available, but are there any tests
available for non-normal samples?
Thanks!
Bernd
__
Hi useRs,
a daily garch(1,1) model can be extended whereby the variance equation
incorporates say higher frequency volatility measure.
The variance equation would look something like:
s(t)2 = garch(1,1) + a*v(t-1)
whereby v(t-1) would be the intraday vola of yesterday ("a" the coef.).
How can
plusreturns)
>
> cumrtns<-cumprodrtns-1.
>
> Then, the elements in cumrtns represent the cumulative reeturn upto that
> point.
>
> But, test it out with an easy example to make sure because I didn't.
>
>
>
>
> -Original Message-
> From: [EMAIL PRO
Hi useRs,
I am trying to calculate the Sharpe ratio with "sharpe" of the library
"tseries".
The documentation requires the univariate time series to be a
portfolio's cumulated returns. In this case, the example given
data(EuStockMarkets)
dax <- log(EuStockMarkets[,"FTSE"])
is however not the
mp;w plot for each group.
How could I possibly do that? Also, what would be the most convenient
approach.
Looking forward to your suggestions.
Many thanks in advance!
Sincerely,
Bernd Dittmann
__
R-help@stat.math.ethz.ch mailing list
https://stat.et
e result correctly.
Is it indeed implied that, if the condition of dx[1:2746] > 15 is
fulfilled, then dx[2:2747] changes by 3.49333 the next period?
Alternatively, if this period's charge is =< 15, then there is no
significant change (-0.04129, t=-0.360) the next period.
0.5612
Description:
Sat Apr 08 19:11:40 2006
I checked with the help pages of the adfTest and fMultivar, but can
simply not figure out why I am receiving these error messages above.
How could I fix this?
Many thanks!
Sincerely,
Bernd Dittmann
Gabor Grothendieck schrieb:
> Try this:
>
>
0,nrow=100)
> bar <- apply(foo,2,adf.test)
> sapply(bar, "[[", "statistic")
> sapply(bar, "[[", "p.value")
>
>
> HTH, Andy
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Bernd
each day's calculations.
How can such a test be done in R? More specifically, how can it be
programmed to iteratively perform the test and also how to extract the
t-values on a daily basis?
Thank you.
Sincerely,
Bernd Dittmann
__
R
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