1. I assume you are looking at the cumulative DISTRIBUTION function (cdf), F(x), which is the probability that a random variable X is less than or equal to x; the cdf is the integral of the density function.

2. The maximum likelihood nonparametric estimate of the cdf is the "empirical cdf" (ecdf) in library(stepfun). However, this is NOT a function appropriate for "integrate". The following looks like it gives me the correct answer:


pecdf <- function(x, y=1:3, lower.tail=TRUE){ nx <- length(x) F. <- rep(NA, nx) for(i in 1:nx){ F.[i] <- sum(y<=x[i]) } F. <- F./length(y) if(lower.tail) F. else (1-F.) }

pecdf(0:4)
pecdf(0:4, lower.tail=FALSE)

Omega <-
function(r){
 numerator <- integrate(pecdf, 1, r)
 denominator <- integrate(pecdf, r, 3, lower.tail=FALSE)
 numerator$value/denominator$value
}

Omega(2)

3. For "data" y = 1:3, I get the following expressions for the numerator and denominator:

numerator(r) = ((ifelse(r<=2, (5-2*r), (3-r))/3)

denominator(r) = ((ifelse(r<=2, (r-1), (2*r-3))/3)

One could probably develop a more general form of this for arbitrary "y".

hope this helps. spencer graves

s viswanath wrote:

Hi,

I am interested in looking at cumulative density functions. If F(x) is a cumulative density of monthly fund returns over the interval of a to b, and I am interested in returns above and below a specified point r, then I would like to find the number that is made up of

1.(integral from r to b)(1-F(x))dx 2. (integral from a to r)(F(x)dx)

3. the ratio of #1/#2 above


In financial literature this ratio has been called the Omega function.


My first guess in obtaining this equation using R
is to use the integrate function but I am have two problems:


I. can I use a nonparametric density in the integrate function(how?), II. how can i get the ratio of #3 above as the integrate function gives the number plus the absolute error


integrate(dnorm,-4,1.96)


0.9749704 with absolute error < 2.1e-07
so using a ratio  in #3 above i get the following error:
: non-numeric argument to binary operator



Thank you in advance.
Sri Viswanath

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