[R] iteration scheme

[Clive Jenkins]

To make your formulae more readable, I substituted
 x  for Lam(m)
 x' for Lam(m+1)
 k  for Ybar
The Newton-Raphson iteration formula is now
 x' = x - [x - k*(1 - exp(-x))]/[1 - k*exp(-x)]
If we differentiate the numerator (top) of the correction term, 
 x - k*(1 - exp(-x))
we get
 1 - k*exp(-x)
which is the denominator (bottom). Therefore the function f(x) must be
 x - k*(1 - exp(-x))
Putting this back into your notation, we get
 Lam(m) - Ybar*(1 - exp(-Lam(m)))
instead of what you suggested:
 Lam(m)/(1-exp(-Lam(m))

I hope this makes sense in terms of the Poisson distribution and MASS3
p.95

Best regards,
Clive Jenkins

Troels Ring wrote:
> 
> Dear friends.
> On p 95 in 3. ed. MASS a zero-truncated Poisson distribution is analyzed. I
> understand the probability distribution and expected mean. The Newton
> iteration scheme is
> Lam(m+1)=Lam(m)-[Lam(m)-Ybar(1-exp(-Lam(m))]/[1-Ybar*exp(-Lam(m)], and I
> suppose the latter part should be f(Lam(m))/f ' (Lam(m)) and f(Lam(m)) is
> Lam(m)/(1-exp(-Lam(m)), right ? But then
> f ' (Lam(m)) is (exp(Lam(m)*(exp(lam(m)-Lam(m)-1)/(exp(lam(m)-1)^2 and the
> iterations scheme is not right ? So I got this wrong. If possible without
> too much trouble, please tell me how this is done.
> I did check that the formula works OK and leads to the right answer.
> 
> Best wishes
> 
> Troels

-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !)  To: r-help-request at stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._