R help - Jan 2008 - Looking for simpler solution to probabilistic question

Hi

I have two processes which take with a certain probability (p1 and p2) x 
number of years to complete (age1 and age2). As soon as thge first 
process is completed, the second one begins. I want to calculate the 
time it takes for the both processes to be completed.

I have the following script which gives me the answer, butI think there 
must be a more elegant way of doing the calculations between the #####

Any ideas welcom,

Rainer


age1 <- 1:20
p1 <- dnorm(age1,10,2)

age2 <- 10:30
p2 <- dnorm(age2,15,1)

######
age <- matrix(age1, ncol=length(age1), nrow=length(age2), byrow=TRUE) + age2

p <- matrix(p1, ncol=length(p1), nrow=length(p2), byrow=TRUE) + p2
p <- as.numeric(tapply(p, age, sum))
p <- p / sum(p)
#####

age <- sort(unique(as.numeric(age)))

plot(age, p)

lines(age1, p1)
lines(age2, p2)

matrix(age1, ncol=length(age1), nrow=length(age2), byrow=TRUE) + age2

can be replaced by

outer(age2, age1, "+")

and the same for 'p'.


Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: "Rainer M Krug" <R.M.Krug at gmail.com>
To: "r-help" <r-help at stat.math.ethz.ch>
Sent: Tuesday, January 15, 2008 1:24 PM
Subject: [R] Looking for simpler solution to probabilistic question

> Hi
>
> I have two processes which take with a certain probability (p1 and 
> p2) x
> number of years to complete (age1 and age2). As soon as thge first
> process is completed, the second one begins. I want to calculate the
> time it takes for the both processes to be completed.
>
> I have the following script which gives me the answer, butI think 
> there
> must be a more elegant way of doing the calculations between the 
> #####
>
> Any ideas welcom,
>
> Rainer
>
>
> age1 <- 1:20
> p1 <- dnorm(age1,10,2)
>
> age2 <- 10:30
> p2 <- dnorm(age2,15,1)
>
> ######
> age <- matrix(age1, ncol=length(age1), nrow=length(age2), 
> byrow=TRUE) + age2
>
> p <- matrix(p1, ncol=length(p1), nrow=length(p2), byrow=TRUE) + p2
> p <- as.numeric(tapply(p, age, sum))
> p <- p / sum(p)
> #####
>
> age <- sort(unique(as.numeric(age)))
>
> plot(age, p)
>
> lines(age1, p1)
> lines(age2, p2)
>
> ______________________________________________
> R-help at r-project.org 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.
> 

Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm

G'day Rainer,

On Tue, 15 Jan 2008 14:24:08 +0200
Rainer M Krug <R.M.Krug at gmail.com> wrote:
> I have two processes which take with a certain probability (p1 and
> p2) x number of years to complete (age1 and age2). As soon as thge
> first process is completed, the second one begins. I want to
> calculate the time it takes for the both processes to be completed.
> 
> I have the following script which gives me the answer, butI think
> there must be a more elegant way of doing the calculations between
> the #####
The code between the #### together with the first line after the second
#### could be just shortened to:
> ager <- range(age1) + range(age2)
> ager <- ager[1]:ager[2]
> pp1 <- c(cumsum(p1), rev(cumsum(rev(p1))))
> pp2 <- c(cumsum(p2[-21]), rev(cumsum(rev(p2)))[-1])
> pr <- pp1+pp2
> pr <- pr/sum(pr)
> all.equal(p, pr)
[1] TRUE
> all.equal(age, ager)
[1] TRUE


If this is more elegant is probably in the eye of the beholder, but it
should definitely use less memory. :)

BTW, I am intrigued, in which situation does this problem arise?  The
time it takes the second process to finish seems to depend in a curious
way on the time it took the first process to complete.....

Cheers,

	Berwin

=========================== Full address ============================Berwin A
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