From: Spencer Graves <spencer.graves_at_pdf.com>

Date: Sun 29 Jan 2006 - 11:18:44 EST

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 Received on Sun Jan 29 11:25:59 2006

Date: Sun 29 Jan 2006 - 11:18:44 EST

- Have you looked at "cumsum"?
- What do you think you are computing when adding 100 to cumsum(log.returns)? To compute cumulative returns in percent from log.returns [or cumusum(log.returns)], compute exp(log.returns) or expm1(log.returns) = (exp(log.returns)-1). Similarly, to compute log.returns from simple.returns, compute log1p(simple.returns) = log(1+simple.returns) [making the obvious conversions between percentages and proportions]. Or am I missing something?

hope this helps, spencer graves

Gottfried Gruber wrote:

*> hi,
**>
*

> i calculate the log-returns in return1 and i want to get the performance for

*> the security. with only one security i have the following code
**>
**> # create matrix to keep performance
**> return100=matrix(rep(100,length(return1)+1))
**> # matrix for the sum
**> z1=matrix(rep(0,length(return1)+1))
**> # suming up the returns from current index to start
**> for (i in 1:length(return1)) {z1[i+1]=sum(return1[c(1:i)]) }
**> #adding both matrices
**> return100=return100+z1*100
**>
**> this works fine for a 1 x n matrix, but if i want the same for a n x m matrix
**> i assume the above code will get time-consuming. is there a trick to
**> linearize the for-loop or any other solution?
**>
**> thanks for any solution & effort,
**> tia gg
*

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 Received on Sun Jan 29 11:25:59 2006

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