Thanks Michael,
Op 11/14/2011 3:30 PM, R. Michael Weylandt schreef:
You need to explicitly pass th to your function with the ... argument
of integrate.
That was a point I was missing!
Thanks again,
This solved my problems for this time.
Gerrit.
E<- function(th){
integrate(function(x,th) x*g(x, th), 0, Inf, th)$value
}
Also, it's value, not Value, which might be producing errors of another sort.
Michael
On Mon, Nov 14, 2011 at 9:16 AM, Gerrit Draisma<gdrai...@xs4all.nl> wrote:
Thanks Michael,
I see now how to include integrate function in the EV function.
And apologies:
I now realize that my code was sloppy.
I intended to write
E<- function(th) {
+ integrate( f = function(x,th){x*g(x,th)},
+ 0,Inf)$Value}
E(1/10)
But that does not work either,
Gerrit.
Op 11/14/2011 2:50 PM, R. Michael Weylandt schreef:
Try this:
EV<- function(lamb){
fnc<- function(x) x * dexp(x, lamb)
integrate(fnc, 0, Inf)$value
}
Your problem is that there's nothing to translate th to lambda in your
code for E.
Michael
On Mon, Nov 14, 2011 at 5:32 AM, Gerrit Draisma<gdrai...@xs4all.nl>
wrote:
Hallo,
I am trying to define expectation as an integral
and use uniroot to find the distribution parameter
for a given expectation.
However I fail to understand how to define properly
the functions involved and pass the parameters correctly.
Can anyone help me out?
Thanks,
Gerrit Draisma.
This what I tried:
=======
# exponential density
g<- function(x,lambda){ lambda *exp(-lambda*x) }
# expectation with lambda=1/10
integrate(f = function(x,lambda=1/10) {x*g(x,lambda)}, 0,Inf)
10 with absolute error< 6.7e-05
# *how to write this as a function?*
E<- function(lambda) {
+ integrate( f = function(x,th){x*g(x,lambda)},
+ 0,Inf)$Value}
E(1/10)
NULL
# *how to include this function in uniroot to find lambda*
# *for a given expectation?*
mu<- 10
uniroot(f<-function(th){E(th)-mu},lower=1,upper=100)
Error in if (is.na(f.lower)) stop("f.lower = f(lower) is NA") :
argument is of length zero
========
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