Heteroskedasticity
As I was suffering from food poisoning in Florida last week, I had many hours in between vomiting bouts to while away woozily. So I thought about heteroskedasticity. Suppose you are estimating the temperature of lake water using various thermometers, and the observations are 50, 52, and 64. We will assume the thermometers are all unbiased, so if x_i is observation i and T is the true temperature, E(x_i) = T. The obvious estimator is the sample mean, (50+52+64)/3 = 55.3 (about). This is also the best estimator if the thermometers are all equally good. But what if they are not? That is, what if the disturbances are heteroskedastic?
If we know the variances of the disturbances of each thermometer, it is well known how to proceed, though I forget exactly how at the moment. What you do is to weight each observation by its accuracy. Thus, the right way might be that if the variances of the thermometers are 60, 30, and 10, and the disturbances are normally distributed, you should use the estimator (50/60 + 52/30 + 64/10) /(1/60+ 1/30+ 1/10) = 59.8 (about). But usually you won't know the variances.
So you need to estimate the variances, which you *can* do, if you have at least 3 observations. Consider observation 1, which is 50. What is a good estimate of its variance? First, we need to estimate the population mean. We can do that from observations 2 and 3, which have a sample mean of (52+64)/2 = 58. Then we use our one data point left for estimating the variance, 50. to estimate the variance to be (58-50)^2/1 = 64. We can do the same for observation 2, which uses the sample mean of (50+64)/2 = 57 so the variance is (57-52)^2/1 = 25. Finally we do it for observation 3, which uses the sample mean of (50+52)/2 = 51 so the variance is (64 - 51)^2/1 = 169. Now we estimates of the variance, so we can do a weighted average, as before.
There is a problem here, though. The reason that with homoskedasticity the sample mean is the best estimator is that it makes the fullest use of the available information. It weights all three observations equally, making use of all of them. Another unbiased estimator is to use just the first two observations, and always throw out the third one, whatever it may be. That wastes information. Similarly, weighting is wasting, if the data is homoskedastic. So there is a cost to weighting.
Weighting also has a benefit, because it lets us use our best information most heavily. If we know exactly how good the information is, there is no downside to weighting. But we don't. We're estimating it. So when we weight, we are hoping our estimation error is small enough that it's worth messing with equal weighting. Is it small enough? That needs proving.
The proof shouldn't be too difficult. But it is too hard to do in my head. I need to look at the proof of the standard weighting scheme when you know the variances. That, I think, comes from doing a calculus minimization of mean squared error for an unbiased estimator.
There is a problem, here, though. How good my estimator is depends on the true state of the world. Suppose the data really *is* homoskedastic. Then my estimator will just make things worse. The same is true if the data is just slightly heteroskedastic but has high variances, because in a given sample, it may well happen that one observation is an outlier and I will weight it less heavily than the other two. So the quality of the estimator depends on the amount of heteroskedasticity, which we don't know in advance.
The only way to resolve that problem, I think, is to go Bayesian. If we have a distribution of possible variances (and, more important, of variances of variances), we can find out whether the estimator does well in expectation against that distribution. For example, we might assume that the variance of each observation is drawn randomly and independently from 0 to 100. Better, probably, would be to see how well the estimator does for a variety of true states of the world and then to eyeball it and decide whether it's worth using.
