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 $b$?
 A: 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 $\lambda$.
