Confidence limits for the inverse of an estimated value

[Stephen Tashi]

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.

 

[viraltux]

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

 

[Calvadosser]

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.

 

[DrDu]

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.

I cannot give you spontaneously a general definition, but, in the relevant context of estimation of CI's for a single parameter via Maximum Likelihood, the situation is easy to grasp:
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 $\lambda$.