Re: [R] about R squared value
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 the formulas, they are the sum of squared differences which are neccessarily non-negative), SSE/SST 0 is impossible. Bernd __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] p-values and significance
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|) (Intercept) 0.0007941 0.0002900 2.738 0.006278 ** x.1 -0.0446746 0.0303192 -1.473 0.140901 x.2 0.1014467 0.0285513 3.553 0.000396 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.009774 on 1134 degrees of freedom (64 observations deleted due to missingness) Multiple R-Squared: 0.01336,Adjusted R-squared: 0.01162 F-statistic: 7.676 on 2 and 1134 DF, p-value: 0.0004883 summary(lm(...)) computes t-values and the resulting p-values for each regressor. The intercept is significant at 0.6%, similarly, x.2 is significant at 0.04%. Only x.1 is not significant at a conventional level of 5%. Its p is 14%. Overall significance of the model is given by the F stats (=7.676 at p less than 0.05%). Hope that helped. Bernd [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] - Nonparametric variance test
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 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] garch and extra explanatory variable
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 this be implemented in R? I checked garch of tseries. An extended formula cannot be specified. fitGarch of fseries might be able to do that. Unfortunately, I am not quite sure how to specify in the fseries package. Or would the estimation have do be done manually? Comments and hints highly appreciated! Thanks! Bernd [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculating the Sharpe ratio
Hi Mark,
thanks for your email.
I used your formula for cumul. returns and plugged them into sharpe:
mysharpe - function(x){
+ return(sharpe(cret(x), r=0, scale=1))
+ }
whereby cret is my cumul. returns function as defined by:
cret
function(x){
cumprod(diff(log(x))+1)-1
}
For the index series Index I obtain a sharpe ratio (r=0 and scale=1) of:
mysharpe(Index)
[1] 0.8836429
Do you reckon this result and the method above are correct?
Many thanks in advance!
Bernd
Leeds, Mark (IED) schrieb:
If the doc says to use cumulated and you didn't, then I supsect the call
to shaprp in
Tseries is not correct. Also, to get PercetnREturns, I hope you did
diff(log(series))
Where series is an object containing prices. It's not so clear
Form your email.
If you want to send in cumulative returns ( which
You should do if the doc says to ) you just take the returns ( by doing
above )
and then , add 1 to each element, do a cumprod and then subtract 1 so
something like :
rtns-diff(log(priceseries)
oneplusrtns-1+rtns
cumprodrtns-cumprod(oneplusreturns)
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 PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Bernd Dittmann
Sent: Monday, February 19, 2007 8:39 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Calculating the Sharpe ratio
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 cumulated returns but rather the daily returns of the
FTSE stock index.
Is this way of calculating the Sharpe ratio correct?
Here are my own data:
yearIndexPercentReturns
19851170.091
1986129.90.11
1987149.90.154
1988184.80.233
1989223.10.208
1990223.20
1991220.5-0.012
1992208.1-0.056
1993202.1-0.029
1994203.10.005
1995199.6-0.017
1996208.60.045
1997221.70.063
1998233.70.054
1999250.50.072
2000275.10.098
2001298.60.085
2002350.60.174
2003429.10.224
2004507.60.183
2005536.60.057
2006581.30.083
I calculated the Sharpe ratio in two different ways:
(1) using natural logs as approximation of % returns, using sharpe of
tseries.
(2) using the % returns using a variation the sharpe function.
In both cases I used the risk free rate r=0 and scale=1 since I am using
annual data already.
My results:
METHOD 1: sharpe:
index - log(Index)
sharpe(index, scale=1)
[1] 0.9614212
METHOD 2: my own %-based formula:
mysharp
function(x, r=0, scale=sqrt(250))
{
if (NCOL(x) 1)
stop(x is not a vector or univariate time series) if (any(is.na(x)))
stop(NAs in x) if (NROW(x) ==1)
return(NA)
else{
return(scale * (mean(x) - r)/sd(x))
}
}
mysharp(PercentReturns, scale=1)
[1] 0.982531
Both Sharp ratios differ only slightly since logs approximate percentage
changes (returns).
Are both methods correct, esp. since I am NOT using cumulated returns as
the manual says?
If cumulated returns were supposed to be used, could I cumulate the
%-returns with cumsum(PercentReturns)?
Many thanks in advance!
Bernd
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
This is not an offer (or solicitation of an offer) to buy/sell the
securities/instruments mentioned or an official confirmation. Morgan Stanley
may deal as principal in or own or act as market maker for
securities/instruments mentioned or may advise the issuers. This is not
research and is not from MS Research but it may refer to a research
analyst/research report. Unless indicated, these views are the author's and
may differ from those of Morgan Stanley research or others in the Firm. We
do not represent this is accurate or complete and we may not update this.
Past performance is not indicative of future returns. For additional
information, research reports and important disclosures, contact me or see
https://secure.ms.com/servlet/cls. You should not use e-mail to request,
authorize or effect the purchase or sale of any security or instrument, to
send transfer instructions, or to effect any other transactions. We cannot
guarantee that any such requests received vi!
a e-mail will be processed in a timely manner. This communication is solely
[R] Calculating the Sharpe ratio
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 cumulated returns but rather the daily returns of the
FTSE stock index.
Is this way of calculating the Sharpe ratio correct?
Here are my own data:
yearIndexPercentReturns
19851170.091
1986129.90.11
1987149.90.154
1988184.80.233
1989223.10.208
1990223.20
1991220.5-0.012
1992208.1-0.056
1993202.1-0.029
1994203.10.005
1995199.6-0.017
1996208.60.045
1997221.70.063
1998233.70.054
1999250.50.072
2000275.10.098
2001298.60.085
2002350.60.174
2003429.10.224
2004507.60.183
2005536.60.057
2006581.30.083
I calculated the Sharpe ratio in two different ways:
(1) using natural logs as approximation of % returns, using sharpe of
tseries.
(2) using the % returns using a variation the sharpe function.
In both cases I used the risk free rate r=0 and scale=1 since I am using
annual data already.
My results:
METHOD 1: sharpe:
index - log(Index)
sharpe(index, scale=1)
[1] 0.9614212
METHOD 2: my own %-based formula:
mysharp
function(x, r=0, scale=sqrt(250))
{
if (NCOL(x) 1)
stop(x is not a vector or univariate time series)
if (any(is.na(x)))
stop(NAs in x)
if (NROW(x) ==1)
return(NA)
else{
return(scale * (mean(x) - r)/sd(x))
}
}
mysharp(PercentReturns, scale=1)
[1] 0.982531
Both Sharp ratios differ only slightly since logs approximate percentage
changes (returns).
Are both methods correct, esp. since I am NOT using cumulated returns as
the manual says?
If cumulated returns were supposed to be used, could I cumulate the
%-returns with cumsum(PercentReturns)?
Many thanks in advance!
Bernd
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] manual construction of boxwhisker plot
Dear useRs, how can I construct a boxwhisker plot based on the vector fivenum? The challenge I face is as follow: I have a table such as x | fivenum --- ... | (.) ... | (.) and so forth For each observation x I have generated a vector containing the fivenum estimates. The first challenge is to group my fivenum vectors into groups based on a selection criterion of x, say for 0 x 6, all fivenum vectors would be collected in that group. Once all my fivenum vectors are in their respective groups, I wish to generate a bw 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.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] question reg. conditional regression
Hi useRs, I have been running a regression of the following kind: summary(lm(dx[2:2747] ~ 0 + (dx[1:2746]15))) Call: lm(formula = dx[2:2747] ~ 0 + (dx[1:2746] 15)) Residuals: Min1QMedian3Q Max -46.35871 -3.15871 0.04129 3.04129 30.04129 Coefficients: Estimate Std. Error t value Pr(|t|) dx[1:2746] 15FALSE -0.041290.11467 -0.3600.719 dx[1:2746] 15TRUE 3.493330.88309 3.956 7.82e-05 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 5.924 on 2712 degrees of freedom Multiple R-Squared: 0.005784, Adjusted R-squared: 0.005051 F-statistic: 7.889 on 2 and 2712 DF, p-value: 0.0003835 In this model, I have lagged the differences series dx (whereby I define dx:= diff(x, difference = 1) )and regressed next period's change to this period's change on the condition that this period's change is greater than 15. As shown in the summary above, for dx[1:2746] 15 true, my coefficient is significant (t = 3.956). My question however is whether I interpret the 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. Thank you! Sincerely, Bernd Dittmann __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to implement an iterative unit root test
Thank you for your suggestion, Andy.
Luckily, the fMultivar package has already implemented such a rolling
function: rollFun.
Thus I tried the following:
myfunction - function(x, n = 5)
{
rollFun(x = x, n = n, FUN = adfTest)
}
This however does not return the tau values (or alternatively, the
p.values) I am looking for.
How do I need to define the function FUN to obtain these values?
Many thanks!
Sincerely,
Bernd
Andy Bunn schrieb:
Does this get you started?
library(tseries)
?adf.test
foo - matrix(rnorm(1000),ncol=10,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 Dittmann
Sent: Wednesday, April 05, 2006 8:58 PM
To: r-help@stat.math.ethz.ch
Subject: [R] How to implement an iterative unit root test
Hello,
How can an interative unit root test be implemented in R?
More specifically, given a time series, I wish to perform the Dickey
Fuller Test on a daily basis for say the last 100 observations. It would
be interative in the sense that this test would be repeated each day for
the last 100 observations.
Given the daily Dickey Fuller estimates of delta for the autoregressive
process
d(Y(t)) = delta * Y(t-1) + u(t)
, the significance of delta would be computed. If possible, I would like
to extract that value and record it in a table, that is a table
containing the tau-values of a 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-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to implement an iterative unit root test
Thank you for the suggestion.
I ran it and oddly enough I am getting contradicting results:
rollFun(x[1:100], 10, FUN = function(x) adfTest(x)$statistic)
NULL
Warning messages:
1: p-value smaller than printed p-value in: adfTest(x)
...
...
...
These error messages appear for each single calculation.
However, performing the unit root test for that very interval (although
not repetitive) the results are:
adfTest(x[1:100])
Title:
Augmented Dickey-Fuller Test
Test Results:
PARAMETER:
Lag Order: 1
STATISTIC:
Dickey-Fuller: -0.1627
P VALUE:
0.5612
Description:
Sat Apr 08 19:11:40 2006
I checked with the help pages of the adfTest and fMultivar, but can
simply not figure out why I am receiving these error messages above.
How could I fix this?
Many thanks!
Sincerely,
Bernd Dittmann
Gabor Grothendieck schrieb:
Try this:
library(fMultivar)
set.seed(1)
x - rnorm(25)
rollFun(x, 15, FUN = function(x) adf.test(x)$p.value)
[1] 0.1207730 0.3995849 0.3261577 0.4733004 0.5776586 0.6400228 0.6758550
[8] 0.6897812 0.3792858 0.6587171 0.5675147
rollFun(x, 15, FUN = function(x) adf.test(x)$statistic)
[1] -3.185471 -2.453590 -2.646336 -2.260086 -1.986146 -1.822440 -1.728381
[8] -1.691824 -2.506875 -1.773368 -2.012774
Also, rapply in the zoo package and running in the gtools package are two
other
rolling routines.
On 4/8/06, Bernd Dittmann [EMAIL PROTECTED] wrote:
Thank you for your suggestion, Andy.
Luckily, the fMultivar package has already implemented such a rolling
function: rollFun.
Thus I tried the following:
myfunction - function(x, n = 5)
{
rollFun(x = x, n = n, FUN = adfTest)
}
This however does not return the tau values (or alternatively, the
p.values) I am looking for.
How do I need to define the function FUN to obtain these values?
Many thanks!
Sincerely,
Bernd
Andy Bunn schrieb:
Does this get you started?
library(tseries)
?adf.test
foo - matrix(rnorm(1000),ncol=10,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 Dittmann
Sent: Wednesday, April 05, 2006 8:58 PM
To: r-help@stat.math.ethz.ch
Subject: [R] How to implement an iterative unit root test
Hello,
How can an interative unit root test be implemented in R?
More specifically, given a time series, I wish to perform the Dickey
Fuller Test on a daily basis for say the last 100 observations. It would
be interative in the sense that this test would be repeated each day for
the last 100 observations.
Given the daily Dickey Fuller estimates of delta for the autoregressive
process
d(Y(t)) = delta * Y(t-1) + u(t)
, the significance of delta would be computed. If possible, I would like
to extract that value and record it in a table, that is a table
containing the tau-values of a 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-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[[alternative HTML version deleted]]
___
Does your mail provider give you FREE antivirus protection?
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to implement an iterative unit root test
Hello, How can an interative unit root test be implemented in R? More specifically, given a time series, I wish to perform the Dickey Fuller Test on a daily basis for say the last 100 observations. It would be interative in the sense that this test would be repeated each day for the last 100 observations. Given the daily Dickey Fuller estimates of delta for the autoregressive process d(Y(t)) = delta * Y(t-1) + u(t) , the significance of delta would be computed. If possible, I would like to extract that value and record it in a table, that is a table containing the tau-values of a 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-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
