Berend Hasselman <bhh <at> xs4all.nl> writes:
>
> Your function miBeta returns a scalar when the argument mu is a vector.
> Use Vectorize to vectorize it. Like this
>
> VmiBeta <- Vectorize(miBeta,vectorize.args=c("mu"))
> VmiBeta(c(420,440))
>
> and draw the curve with this
>
> curve(VmiBeta,xlim=c(370,430), xlab="mu", ylab="L(mu)")
>
> Berend
>
Taking into account what you have pointed out, I reprogrammed my function
as follows, as an alternative solution to yours:
zetas <- function(alpha) {z <- qnorm(alpha/2); c(z,-z)}
# First transformation function
Tzx <- function(z, sigma_p, mu_p) sigma_p*z + mu_p
# Second transformation function
Txz <- function(x, sigma_p, mu_p) (x - mu_p)/sigma_p
BetaG <- function(mu, alpha, n, sigma, mu_0) {
lasZ <- zetas(alpha) # Zs corresponding to alpha
sigma_M <- sigma/sqrt(n) # sd of my distribution
lasX <- Tzx(lasZ, sigma_M, mu_0) # Transformed Zs into Xs
# Now I consider mu to be a vector composed of m's
NewZ <- lapply(mu, function(m) Txz(lasX, sigma_M, m))
# NewZ is a list, the same length as mu, with 2D vectors
# The result will be a vector, the same length as mu (and NewZ)
sapply(NewZ, function(zz) pnorm(zz[2]) - pnorm(zz[1]))
}
miBeta <- function(mu) BetaG(mu, 0.05, 36, 48, 400)
plot(miBeta,xlim=c(370,430), xlab="mu", ylab="L(mu)")
I hope this is useful to people following this discussion,
-Sergio.
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