# Confidence limits for the inverse of an estimated value

1. Jun 26, 2012

I am aware that, in statistics, things get difficult as soon as they get nonlinear. And taking the reciprocal of a quantity is a nonlinear operation.

I have some data that would form a nice looking straight line, except for random error scattering it around the line. I have a total of about fifty points. If I fit a regresssion line to the data, I can find an estimate of the slope of the line. In my particular case, the slope of the line (if I knew it precisely) would give me the coefficient λ in a 1st order linear differential equation dC(t)/dt = -λC(t). Thus the regression analysis gives me an estimate for λ.

There is a standard formula for calculating confidence limits on the estimate of the slope of a line computed via a regression analysis. This formula gives me the upper and lower confidence limits λ$_{lower}$ and λ$_{upper}$ on my estimate of λ.

The solution for the differential equation is C(t) = C(0) exp(-t/T), where the time constant, T = 1/λ. It is the time constant T that is the thing of real interest because this will tell me how long a system takes to settle following a disturbance.

Here is my question.

(A) What is the "best", in some appropriate sense, estimate for T? Is it simply 1/(my regression estimate for λ)?

(B) If so, what are the confidence limits for my estimate of T? Are they simply the inverses, 1/λ$_{lower}$ and 1/ λ$_{upper}$. of my confidence limits on λ?

Thank you for any help. I assume it is a simple and straightforward question but I have not succeeded in finding the answer nor in working it out myself.

2. Jun 26, 2012

### viraltux

To enter into a discussion on what means the "best" estimator would require a whole course in statistical inference, but for the sake of this problem let's say yes, $\frac{1}{\hat\lambda}$ is the "best".
That's another yes; in statistics we make data go through all kind of transformations so that we can fit it into statistical models of our choice; that's actually what you are doing with your equation so that you are able to fit your data into a linear model. Once you have fitted the model you need to apply the inverse transformation to the results, including confidence intervals, so yeah, your procedure is correct, you're fine.

Last edited: Jun 26, 2012
3. Jun 27, 2012

### chiro

To follow on with what viraltux said, it depends ultimately on how specific transformations preserve information about the probabilities and subsequent information.

With some things if you are given say x, and you need to find T = f(x), then you can apply the transformations to give new results which will conserve the probabilistic properties under that transformation.

But other times, they don't. One example is with a technique known as highest posterior density in Bayesian analysis which doesn't.

If you want to look into this problem in general, find out frameworks which deal with transformation of statistics, intervals, and other measures that conserve the probabilistic information under transformation.

4. Jun 27, 2012

### haruspex

I'm not sure about that. Take a simple example: X is uniform on [0,1].
The average is 0.5, so the 'correct' time constant is 2.
If we take 2 samples of X, X1 and X2, and calculate the time constant as the inverse of their mean, what is the expected value of the result?
$\int^{1}_{x_{1}=0}\int^{1}_{x_{2}=0}2/(x_1+x_2)dx_2.dx_{1} = 2\int^{1}_{x_{1}=0}[ln(x_{1}+x_2)]^{1}_{x_{2}=0}dx_{1}$
$= 2\int^{1}_{x_{1}=0}(ln(x_{1}+1)-ln(x_{1}))dx_1$
$= 2[(x_{1}+1)ln(x_{1}+1)-x_{1}-x_{1}ln(x_{1})+x_{1}]^{1}_{x_{1}=0}$
= 4 ln(2) =~ 2.77

5. Jun 27, 2012

### viraltux

There are a few things sure in this world; the Sun is hot, the water is wet, and haruspex will be there to keep you on track! hahaha

What you are doing is $\frac{\frac{2}{x_{1,1}+x_{2,1}} + \frac{2}{x_{1,2}+x_{2,2}}+ ... + \frac{2}{x_{1,n}+x_{2,n}}}{n}≈2.77$ but what you should do is $\frac{n}{\frac{x_{1,1}+x_{2,1}}{2} + \frac{x_{1,2}+x_{2,2}}{2}+ ... + \frac{x_{1,n}+x_{2,n}}{2}}≈2$

But anyway, it is true that inference is a whole world; you may get really seemingly crazy estimators once you squeeze a problem, but for the problem presented by the OP as such what he/she is doing is just OK.

PS: If I ever go into space I want you to check the rocket engines; I know you won't let anything pass :tongue:

Last edited: Jun 27, 2012
6. Jun 27, 2012

### haruspex

viraltux, I believe I have correctly modelled the consequence of using OP's procedure. Try it in a spreadsheet. Generate 100 pairs of samples from U(0,1); for each pair take the average, then the inverse. You will almost always get a result > 2, often > 3.
Of course, this is fairly extreme. More samples in each set would give a smaller error.
Calvadosser, can you take a look at the distribution of the lambdas? If we know more about that we might be able either to suggest a better procedure or to put bounds on the error.

7. Jun 28, 2012

### viraltux

Hi haruspex,

The OP has a straight line set of measurements in its model and he estimates via linear regression its slope having the value $\hat{λ}$ and a standard error for the estimate based on a Gaussian distribution.

He does not have 100 slopes to work with which seems to be the way you are approaching the problem based on the example you post, so even if he wanted he could not do the inverse for every slope and see how the distribution of inverse slopes behaves (or the distribution of λ slopes for that matter).

But even so, let's change the problem and imaging that he can actually measure a set of, let's say, n lambdas ($λ_{1..n}$), and he wants to estimate $T=1/λ$

What you suggest is $\hat{T}= \frac{1/λ_1 + 1/λ_2 + ... +1/λ_n}{n}$ but this is wrong, this approach allows the function that calculates T to bias its own estimation.

In this situations we apply what is called in inference the invariance principle which basically states that if you have a function $f$ and you want to estimate $f(θ)$ then you do $\widehat{f(θ)} = f(\hat{θ})$, which in the OP case would be $\hat{T}=1/\hat{λ}$, and, by the way, this principle would hold for whatever distribution θ, λ might have.

Last edited: Jun 28, 2012
8. Jun 28, 2012

### haruspex

No, you misunderstand what I did.
To find an unbiased estimator for a statistic s, the following is standard procedure:
- construct some candidate function fs({xi}) of n observations
- compute E(fs) as a function of a presumed value s of the statistic
- see how E(fs) compares with s
E.g. with s being the mean and fmean({xi}) = (Ʃxi)/n gives E(fmean) = mean, but with s being the variance, we get fvar = (Ʃ(xi-Ʃxi)/n)2)/(n-1).
If L is a linear function and fs is an unbiased estimator for s then L(fs) is an unbiased estimator for L(s). But it does not work for nonlinear functions. E.g. for the standard deviation, √fvar is not an unbiased estimator for √var. (There are corrections that have been developed, but none are perfect.)
In my model for the OP procedure, I took just two observations, computed the mean, and inverted. There are two ways we can try to assess this as an estimator for 1/mean. In my first post I assessed it analytically. Since you thought the analysis flawed, I then assessed it numerically. To find E(f) I had to generate lots of pairs and take the average result.

Now, as I said, my model was rather extreme. It took a uniform distribution 'close' to the origin (i.e. the std dev is a large fraction of the mean) and only used one pair of observations to compute the mean. Changing either of those would reduce the error. To get a bound on the error in the OP procedure we need to know more about the distribution and the number of samples.

9. Jun 28, 2012

### viraltux

First, $\hat{λ}$ being a biased or unbiased estimator is absolutely irrelevant to the problem, the invariance principle holds for any estimator; biased, unbiased or otherwise.

Second, the fact that an estimator is biased does not make it bad, for instance, the maximum likelihood estimator for the variance of a Gaussian distribution is biased and yet is the one preferred in many areas of multivariate analysis.

OK, I think we are reaching the point of confusion here, seems to me that for some reason you think that a biased estimator is something bad that has to be fixed, right? that's why these examples showing how applying the invariance principle might return biased estimators, is this correct?

Well, so it seems that in order to fix this, which we don't have to, you are asking the OP about the distribution of λ when actually there is no such thing; we only have one unique value for λ which is estimated via MLE in a linear regression procedure.

What we get is the estimation of λ using this method (or whatever method the OP used for the linear regression) and the error associated with it, that is all we have; no distributions, no number of points... no nothing, that's it, and in this scenario you apply the invariance principle to estimate T which is what the OP is already doing.

And when we calculate the confidence interval for T we will see how the estimation of T is not centered in such interval, accounting this way for the bias that seems to be the issue in this discussion.

10. Jun 28, 2012

### haruspex

I thought the invariance principle was a pragmatic one rather than mathematically proven. Do I have that wrong?
No, I don't go that far. I'm saying it's a reason to stop and think about what you're going to do with the answer. The search for unbiased estimators is grounded, in my view, on the typical case that an error of +ε costs about the same as an error of -ε. In the present case, if taking the parameter to be 1/(λ+ε) (instead of 1/λ) has about the same cost as taking it to be 1/(λ-ε) then the OP procedure is fine. But if it's (1/λ)+ε and (1/λ)-ε are about equal in cost then it's not.
That was not my intention, whatever the wording. I was after the scatter in the data.

11. Jun 28, 2012

### chiro

I too am interested with regards to the invariance principle, because although I know it exists for many probabilistic and statistical situations involving transformations, I've never actually never checked it out which is contributing to my ignorance.

Could you point out a reference or two for this viraltux (maybe even a book that covers it?)

12. Jun 29, 2012

### viraltux

Yes you have it wrong haruspex, it is mathematically sound, if given the trillion of times that I've used it, and the quadrillions of times that I've seen it in publications, if now it turned out to be mathematically flawed my heart would skip a beat. :tongue:

Hi chiro, I didn't find the paper I wanted but anyway, this one proves that applying functions to MLE estimators return the MLE estimator of the function (which would be the OP case) and has an example too:

http://www.stats.ox.ac.uk/~dlunn/b8_02/b8pdf_6.pdf

This "cost about the same" is too ambiguous but I can see where the problem is, you have 1/(λ+ε) and you reason something like "well, if ε>0 I'm fine but, hey, if ε<0 the estimation for 1/(λ+ε) might go way to high! so... how about if I just adjust the estimator a little bit, something like 1/(λ+ε+τ) with τ>0 to make the estimator unbiased so when the bad news ε<0 come I am in the safe side." Something in these lines is what you are thinking right?

OK, I am going to briefly (and dramatically) describe what happened back in the days when the "bias vs unbiased" lesson came up in my faculty.

- professor: "what is best, a biased or an unbiased estimator?"
- students: "Unbiased", "of course" "I agree", "What kind of question is that?"
- professor: OK, why?
- students: :uhh: :uhh: :uhh: :uhh: :uhh:
- professor: soooooooooooooo...
- students: Well, if you got it biased, and you know it is biased, and you can even calculate how biased it is, well, then you can take the bias away! why would anyone want to use a calculation that is known to be consistently higher or lower?
- professor: That's right, why would anyone do that?
- students: :uhh: :shy: :grumpy: :zzz:

I am sure this situation is a classic in every inference course.

Anyway, also a classic is to take the estimation of the Gaussian variance as an example, let's consider the following three estimators for the variance

$S_{unbiased} = \frac{1}{n-1} \sum_{i=1}^n\left(x_i - \overline{x} \right)^ 2$
$S_{MLE} = \frac{1}{n} \sum_{i=1}^n\left(x_i - \overline{x} \right)^ 2$
$S_{LSE} = \frac{1}{n+1} \sum_{i=1}^n\left(x_i - \overline{x} \right)^ 2$

Well, turns out that among these three the unbiased estimator is the one with the highest error! (in terms of least square error). OK, this is hard to believe and very counter intuitive because, geeee, we know it is biased!! fix it!!! right? Well, you do that, and you introduce error.

You know, I had the mathematical proof in front of my eyes and I still had to run a simulation to believe it, but it is true!!

Then one student (in this case it was me) asked "But everyone in school, engineering, physicist... they all use the unbiased version to estimate the variance, why on Earth we don't all use the LSE version with the lowest error!?"

professor: "Not everything that shines is gold" and he went on with the class.

Oh well, I had to do all kind of guesses about my professor statement on why people don't widely use the LSE version, but I will not go on with this here, for now, suffice to say that what the OP is doing is OK, and that unbiased doesn't mean the "best".

Last edited: Jun 29, 2012
13. Jun 29, 2012

### chiro

Thank you viraltux for that result, I'll have to remember that because it's going to be very useful. I'm sure it's probably in my introductory stats book, but it's definitely good to know that it holds.

14. Jun 29, 2012

### viraltux

Don't be surprised if it is not because I think this is one of these things most of the time courses take for granted and they simply state it, at least I don't remember myself the professors proved it in class... Glad to know the prove is there though! :tongue:

15. Jun 29, 2012

My thanks for the replies - and for the discussion, some of which (but not all) is over my head. I had originally supposed that more or less the same question is asked very frequently.

You have very kindly:

- Reassured me that what I propose is, at very least, not a stupid thing to do.

- Shown me that the question is deeper than I had imagined.

I'll add it to me ever-growing list of interesting things to look into - when other things don't take up all the available time.

16. Jun 29, 2012

### Stephen Tashi

How are the data points measured? Do you have seperate instruments to measure $\frac{dC(t)}{dt}$ and $C(t)$? Or are you computing $\frac{dC(t)}{dt}$ from differences in the $C(t)$ data?

(I'm wondering why you chose to use regression instead of fitting a curve to the C(t) data.)

17. Jun 29, 2012

### haruspex

Ok, I see. Invariance works for MLE (and for median-unbiased estimators, right?), but not for estimators in general.
Yes, but let me put it more precisely. In many cases, the cost of an error in one direction is about the same as the cost of the same magnitude error in the other direction. If so, an unbiased estimator is likely to be good (particularly if the cost is as the square of the error). In the present case, the question is whether this is true of the error in estimating λ (in which case the OP procedure is fine) or of the error in estimating 1/λ (in which case the OP procedure could be awful, depending on details).
I consider this to be the answer to your prof's challenge. Choice of estimator should depend on the cost function.
Fascinating - I wasn't aware of this, thanks.
However, in the absence of beliefs regarding cost function, I gather that the gold standard is not the estimator with the least MSE, but the unbiased estimator with least MSE (http://en.wikipedia.org/wiki/Mean_squared_error#Interpretation). OTOH, a cost function which grows faster than the square of the error might well indicate use of one of the biased estimators. Haven't tried to figure that out.

18. Jun 30, 2012

### viraltux

Considering the vast majority of estimators in general linear models, time series, etc use MLE estimators, or some sort LE estimators equivalent to MLE is not like we should worry much about it.

But let's say that the OP is using a Bayesian method to estimate λ, could we say that 1/λ is a Bayesian estimator? Maybe not but, could I adopt now a "in your face Bayesian boy" attitude just because the Bayesian estimator happens not to be invariant in a transformation? Well... tempting , but not really, we could call this transform estimator something like Bayesian induced estimator or something like that, and it would be as legitimate as the MLE one; it just would have other properties and that would be all.

It'd be nice to have a table of invariant properties for families of estimators though.

No, if the cost is an squared of the error then the SLSE is the estimator we would choose to estimate the variance, yet, it turns out that SLSE is the one among the three I presented with the biggest bias! In this cost scenario the unbiased estimator would be the worst choice.

These estimators are called MVUE and they gather much attention in the inference world because they have very nice mathematical properties. When working with biased estimator the bias make the mathematics much harder, but this does not make them better than biased estimators in a broad sense.

So the answer to what estimator is the best is always problem dependent since every bit of information you have about the problem might make you jump from one to another, or even to force you to develop one Taylor-made estimator for the problem in hand.

19. Jun 30, 2012

### haruspex

No, here I'm referring to estimates of the mean, not the variance. It seemed to me that the average of the data should minimise the expected value of the square of the error in the estimate of the true mean. But I admit that was just a guess. As far as I can make out now, it is the case amongst unbiased estimates of the mean, but I can find nothing regarding biased estimators that might produce a lower MSE for the mean. All the discussions on minimising MSE appear to centre on estimating the variance.
I tried applying a constant factor to the 1/$\hat{λ}$ estimator for 1/λ in my simple model. As I should have realised, the integral is unbounded, so the MSE estimator amongst that subset is 0. I changed the uniform distributions to be (1,2) and the factor came out as about 1/1.03.
So I think we've arrived on the same page: the least cost estimator depends on a number of details of the specific problem at hand.

20. Jul 1, 2012

### viraltux

Oh, but it does not matter about mean or variance you see, we are talking about estimators; what that estimator estimates is irrelevant. When we estimate something we always want to estimate the expected value of that something; so if we want to estimate the E(X) we may use the arithmetic mean.... mmmm

OK, put it this way, imaging the random variable $Y = (X-E(X))^2$ and now I ask you to estimate the mean of Y, would it change anything that the mean of Y happens to be the variance of X? No. So if you want the LSE estimator for the mean of Y, that is E(Y), you end up with a biased estimator.

Last edited: Jul 1, 2012