Recognitions:

## Confidence limits for the inverse of an estimated value

 Quote by DrDu Even when one uses MLE estimates the transformed confidence intervals are not identical to the confidence intervals for 1/ parameter, especially if the CI's are the shortest possible for one parameter they transformed CI's are not the shortest for 1/parameter.
I don't know if this thread is helping the original poster Calvadosser, but it is very informative!

In that spirit, I'll ask about the technical definition of a "shortest possible" confidence interval. Suppose we have some algorithm that takes the sample data and computes an interval and the algorithm has the property that there is a 95% chance that the population parameter we are estimating is in that interval. There is no requirement that all the intervals the algorithm produces have the same length. Only If we take particular sample data do we get an interval of a particular length and (from the frequentist point of view) we don't know that there is a 95% chance that the population parameter is in that particular interval. If I am comparing two such algorithms, I can compare the expected lengths of the intervals they produce. As I interpret your statement, it refers to expected lengths of confidence intervals. Is that correct?

Recognitions:
 Quote by mfb I would use the approximation dC/dt = 1/2 (C(t+1)-C(t-1)), but chances are good that it does not change much (something you can add as cross-check).
And there are quite a variety of additional mathematical refinements that could be introduced. Real world problems can be modeled different sets of mathematical assumptions and the only way that people get confident about their answers is to work them many different ways using a variety of different assumptions and get roughly the same answer everytime. (I've observed that this is particularly true in bureacracies. Most managers don't work by trying to pick the best answers. They force analysts to solve problems many different ways and they pick the answer that comes up most often.) Calvadosser must let us know if he's interested in fancier math.

How to treat the C(t) data is an interesting question. I looked at the data on the web and the higher resolution data is (of course) wavy. So what will C(n) be? I suppose you could set it to be the moving average taken over a year's time prior to year n.

 $$\lambda = \frac{\frac{dC}{dt}-Fa}{C(t)}$$ This gives you a direct estimate for λ and if you take the inverse value you get a direct estimate for T in each year. If your values do not have individual (and different) errors, the arithmetric average should be the best way to estimate your value.
From
$$- \lambda = \frac{\frac{dC}{dt}-Fa}{C(t)}$$

We can reduce the data
$$T_i= \frac{C(i)}{Fa_i - \frac{dC(i)}{dt}}$$

A statistical concern is that the $T_i$ and $T_{i+1}$ are not independent random samples since computing the values of $\frac{dC(i+1)}{dt}$ and $\frac{dC(i)}{dt}$ both involve using the datum $C(i)$.

I wonder if Calvadosser's method of using regression cleverly takes care that concern.

 Hi All, Would you mind to answer a question? Being OK: The OP's procedure is absolutely fine. Not OK: The OP's procedure is either wrong or misses so further analysis. IDK: I don't know / I am not sure. Where would you situate yourself? Also, could you tell about your background? I think this would help to situate the thread's audience. For instance, in my case it would be: OK | viraltux | Statistician / CS

 Quote by mfb I would use the approximation dC/dt = 1/2 (C(t+1)-C(t-1)), but chances are good that it does not change much (something you can add as cross-check). $$\lambda = \frac{\frac{dC}{dt}-Fa}{C(t)}$$ This gives you a direct estimate for λ, and if you take the inverse value you get a direct estimate for T in each year. If your values do not have individual (and different) errors, the arithmetric average should be the best way to estimate your value.
Thank you for the suggestion. I will give it a try as soon as I can (but not for a week or so because of things that have to take priority).

I was using C(t)-C(t-1) because that then covered the same time interval as Fa(t). Maybe I can use 1/2(Fa(t+1)+Fa(t-1)) to match up with 1/2 (C(t+1)-C(t-1)). I'll think it through.

Recognitions:
If $\lambda$ is the true and unknown parameter value and $\hat{\lambda}$ it's maximum likelihood estimate, furthermore $\hat{\sigma}$ it's estimated variance.
Then the two sided CI's are $\lambda_{u/l}=\hat{\lambda}\pm z \hat{\sigma}$ were z is the quantile of the normal distribution corresponding to a given strength $\alpha$. On the mean this interval will be of length $2z\sigma$ which is the shortest possible interval which covers $1-\alpha$ of the normal distribution.
Now if you are interested not in $\lambda$ but in $1/\lambda$, then the transformed CI's $1/\lambda_{u/l}$ do not span the shortest interval.
This can be seen as follows. By error propagation (or delta method, as statisticians tend to call it), $\hat{\sigma}_{1/\lambda}=1/\hat{\lambda}^2 \hat{\sigma}_\lambda$.
Hence the shortest CI's (on the mean) for $1/\lambda$ are
$1/\hat{\lambda}\pm z \hat{\sigma}_{1/\lambda}$ which does not coincide with the transformed CI's of $\lambda$.