Estimating parameters using a different method?

I have a probability distribution which has two parameters $a$ and $b$ I have re-parametrized the distribution such that the new distribution has two parameters $c$ and $d$ where:

$c=a$ but $d = \frac{1}{b} - 1$

I can easily estimate $a$ and $b$ but I need to make inferences on $c$ and $d$, specially $d$.

So my question is: If I estimate $b$ and use relationship $d= \frac{1}{b} -1$ to find out the estimated values of $d$, is this correct? Is there any information loss?

Also, how can I find the standard errors and confidence intervals of $d$ if those exist for $b$?

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How are you estimating these parameters? If you are using maximum likelihood, then you can just transform the original estimator and this is justified by the Invariance Principle. In a Bayesian framework this is also reasonable. –  user10525 Aug 16 '12 at 8:04
I am not only using ML estimation matheods, but I also use methods like quasi-likelihood, extended quasi likelihood and quadratic estimating equations etc.. Does the Invariance Principal works for these methods too? –  Arun Aug 16 '12 at 8:16
I do not know. Why don't you reparameterise the model before conducting inference? –  user10525 Aug 16 '12 at 8:19
Yes, I can.. But R codes are not available for re-parameterized models.. –  Arun Aug 16 '12 at 8:26
Then, if you need to know about the invariance of certain methods, you should specify this clearly in your question. Otherwise the question sounds too broad. Not all the estimation approaches are invariant. If your context allows it, you may consider using bootstrap for constructing confidence intervals. –  user10525 Aug 16 '12 at 8:28

Like @Procrastinator says, it all depends on the way you estimate the parameters, and what properties you expect from the esimtator. I will try to illustrate this with an example. If you have an exponential distribution parametrized like

$$f(x) = \frac{1}{\theta} e^{- x/\theta},$$

$\theta$ is the expected value, so you could use the sample mean to estimate it, and this would be an unbiased estimator. Indeed, the sum of $n$ such exponential variables is a $\Gamma(n, \theta)$, with expected value $n\theta$, so that the mean has expected value $\theta$.

If you reparametrize the distribution as

$$f(x) = \lambda e^{-\lambda x},$$

you might be tempted to estimate $\lambda$ with the inverse of the sample mean. We saw that the sample sum is distributed as $\Gamma(n, 1/\lambda)$, so the expected value of the inverse of the sample sum $y$ is

$$E(Y) = \frac{\lambda^n}{\Gamma(n)} \int \frac{1}{y} y^{n-1}e^{-\lambda y} dy = \frac{\Gamma(n-1)\lambda^n}{\Gamma(n)\lambda^{n-1}} = \frac{\lambda}{n-1},$$

so the expected value of the inverse of the sample mean is $\lambda \frac{n}{n-1}$, which is biased.

If it is important for you that the estimator is unbiased, then this approach is not right. But if instead, you care that the estimator is a maximum likelihood estimator (which is usually asympotically unbiased), then the approach is right because the inverse of the sample mean is the maximum likelihood estimator of $\lambda$.

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