--- [email protected] wrote:
> 1. Yes, but more generally, the ihs fuctions is
> ln(theta*z + sqrt((theta^2 * z^2
> + 1))/theta; where theta is scale parameter that
> could be estimated.
Is IHS = log(z + sqrt(z^2 + 1)) then only an
approximation? If so, I assume that it is good when z
is large relative to 1/theta. But how can we know
ahead of time how good the approximation will be on
our data?
Regards,
Ricardo.
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