[R] R numerical integration
savicky at cs.cas.cz
Sat Mar 24 09:46:58 CET 2012
On Fri, Mar 23, 2012 at 01:27:57PM -0700, casperyc wrote:
> Hi all,
> Is there any other packages to do numerical integration other than the
> default 'integrate'?
> Basically, I am integrating:
> integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value
> The integration is ok provided sigma is >0.
> However, when mu=-1.645074 and sigma=17535.26
> It stopped working. On the other hand, Maple gives me a value of
> It is an important line of the coding that I am doing and I am looking for
> some package that is able to do numerical integration efficiently (fast and
> accurate to a tol=1e-4). I have tried 'cubature', which does not give me
> anything even after 10 minutes.
Integrating with infinite limits is necessarily a heuristic. If you
want to bound the absolute error (not the relative error), then it is
sufficient to integrate over the interval
[mu - bound*sigma, mu + bound*sigma]
where "bound" is such that the integral of the tails of
dnorm(, mu=1, sigma=1) outside of the interval is less than the
required absolute error. This follows from the fact that the
multiplier 1/(1+exp(-x)) is at most 1. For example, with bound = 10,
the absolute error due to limiting the interval is at most
2*pnorm(-10) = 1.523971e-23. The total error will contain this
and the error of the numerical algorithm.
mu <- -1.645074
sigma <- 17535.26
integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)), mu - 10*sigma, mu + 10*sigma)
0.5 with absolute error < 3.7e-05
The exact result is at most 0.5. This may be seen by shifting the
integrated function so that the mean of the dnorm() becomes 0.
We get the following function f_1(x), which has the same integral
as the original function
f_1(x) = dnorm(x, 0, sigma)/(1+exp(-x-mu))
Consider the function f_1(-x). It has also the same integral as the
original function. So, the function f_2(x) = f_1(x) + f_1(-x) has
twice the integral of the original function. Since dnorm(-x, 0, sd) =
dnorm(x, 0, sd), the function f_2(x) has the form
f_2(x) = dnorm(x, 0, sigma)*(1/(1+exp(-x-mu)) + 1/(1+exp(x-mu)))
Since mu < 0, we have for all x
1/(1+exp(-x-mu)) + 1/(1+exp(x-mu)) <= 1/(1+exp(-x)) + 1/(1+exp(x)) = 1
Consequently, the integral of f_2(x) is at most the integral of
dnorm(x, 0, sigma), which is 1. The integral of the original
function is one half of this, so it is at most 0.5. The result
0.5005299403 Maple is more than 0.5 only due to a numeric error.
Hope this helps.
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