Horvitz-Thompson estimator example
For illustration, assume a population of N = 3 elements from which a sample of n = 2 is selected. The population is given in Table 1. The target variable is “production” and an ancillary information is available: the stand size. This example is from the lecture notes for the teaching modul "Forest Inventory" by Kleinn et. al[1]).
The parametric mean, total and variance of the variable Y (production) are, respectively,
- \[\mu = 62/3 = 20.67, \tau = 62, \sigma^2 = \sum_{i=1}^N (y_i - \mu)^2 = 150.22\]
Two options of unequal selection probabilities are listed and analyzed (see Table 1): option I is proportional to size (area) and option II is inversely proportional to size (which is included here for illustration purposes only).
Table 1. Example population of N=3
Stand Production (total) Area Selection probabilities Option I Option II 1 6 10 0.1 0.6 2 20 30 0.3 0.3 3 36 60 0.6 0.1
In Table 2 all possible samples of size n = 2 are listed in the first column, where also the order of the sample is considered. For simple random sampling with replacement, the probability for each combination of elements is (1/3)², as given constantly in column 4. In column 2 and 3, the estimated mean and total is given for each sample of two. The mean of all nine estimated means and totals (this is the population of all possible sample results for n = 2) are the expected values of the mean \(E(\bar y)\) and total \(E(\hat \tau)\). Here \(E(\bar y) = 20.667 = \mu\) and \(E(\hat \tau) = 62 = \tau\). That is, the estimator is unbiased. At the bottom of Table 2, the error variance is calculated, for sampling with replacement and also for sampling without replacement (for which, however, the selection probabilities are not specified in the table!). For sampling with replacement, the standard error of the total is \(\sqrt {V(\hat \tau)} = 26\),
which results either from the known formula \(\sqrt {V(\hat \tau)} = \sqrt {N^2 \frac {\sigma^2} {n}}\)
or from simply calculating the parametric variance of the population of all nine estimated totals \(\hat \tau\)
For the selection with unequal probabilities proportional to size (option I), only the estimation of the total is presented, applying the Hansen-Hurwitz estimator. It is yet immediately seen from column 5 that the 9 estimated totals vary much less than for simple random sampling. In column 6 the selection probabilities are given. It is seen that the different combinations of n = 2 have considerably different probabilities of being selected. The mean of the estimated totals must here be calculated as weighted mean and yields \(E(\hat \tau) = \sum p(s)\hat \tau = 62\), so that we empirically show that the estimator is unbiased. The standard error of the estimated total is
- \[\sqrt {V(\hat \tau)} = \sqrt {\sum p(s)(\hat \tau)^2} = \sqrt {E(\hat \tau^2) - \{E(\hat \tau)\}^2} = (3848.667 - 62^2)^\frac {1}{2} = 4.667^\frac {1}{2} = 2.16 \]
Table 2. Ilustration of sampling with equal and unequal probabilities.
In the sample Simple random sampling Unequal prob. Option (I) Unequal prob. Option (I) Horwitz-Thompson Option (I) \(\boldsymbol {\bar y}\) \(\boldsymbol {\hat \tau}\) Prob(wr) \(\boldsymbol {\hat \tau}\) Prob. \(\boldsymbol {\hat \tau}\) Prob. 1-2 13 39 \( (1/3)^2\) 63.333 0.03 38.333 0.18 70.79 2-1 13 39 \( (1/3)^2\) 63.333 0.03 38.333 0.18 70.79 1-3 21 63 \( (1/3)^2\) 60 0.06 185 0.06 74.44 3-1 21 63 \( (1/3)^2\) 60 0.06 185 0.06 74.44 2-3 28 84 \( (1/3)^2\) 63.333 0.18 213.33 0.03 82.07 3-2 28 84 \( (1/3)^2\) 63.333 0.18 213.33 0.03 82.07 1-1 6 18 \( (1/3)^2\) 60 0.01 10 0.36 31.58 2-2 20 60 \( (1/3)^2\) 66.667 0.09 66.667 0.09 39.22 3-3 36 108 \( (1/3)^2\) 60 0.36 360 0.01 42.86 \( E(\bar y) = 20.667 \quad E(\hat \tau) = 62\) \(\sqrt {V(\bar y)} = 8.667 = \sqrt {\frac {\sigma^2}{n}}\) \(\sqrt {V(\hat \tau)} = 26\)
Without replacement\[\sqrt {V(\bar y)} = 6.1283 = \sqrt {\frac {\sigma^2}{n}} * \sqrt {\frac {N-n}{N-1}}\]
\( E(\hat \tau) = \sum p_i \hat \tau = 62 \) \(\sqrt {v(\hat \tau)} = \sqrt {\sum p_i (\hat \tau - \tau)^2}\)
\(= \sqrt {E(\hat \tau^2)-\{E(\hat \tau)\}^2}\)
\(=(3848.667 - 62^2)^\frac {1}{2}\)
\(= 4.667^\ {1}{2} = 2.16\)
\( E(\hat \tau)= 62 \) \(E(\hat \tau^2) = 9098.7\)
\(V(\hat \tau) = 5254.6\)
\(\sqrt {V(\hat \tau)} = 72.49 \)
\( E(\hat \tau) = 62 \) \(V(\hat \tau) = 356.1\)
\(\sqrt {V(\hat \tau)} = 18.9\)
That means, with the same sampling effort n = 2, by assigning intelligently the selection probabilities, the standard error for the estimation of the total is, in comparison to simple random sampling, reduced from \(\sqrt {V(\hat \tau)} = 26\) to \(\sqrt {V(\hat \tau)} = 2.16\)!! Not in all cases can such a tremendous gain in precision be achieved. The performance of unequal probability sampling is the better the more we are able to assign selection probabilities as proportional as possible to the target variable. That is exactly the reason why Bitterlich sampling is so superior for the estimation of basal area (and lesser so for the estimation, for example, of number of stems).
For illustration purposes, let’s have a look at the case where the selection probabilities were (not very intelligently) assigned inversely proportional to size. That is, the elements that contain less information about the target variable have the highest selection probability. In that case, the standard error of the estimated total is with \(\sqrt {V(\hat \tau)} = 72.49\) much higher than for the other approaches and one sees that there is much variability among the estimated totals. However, also this – imprecise – estimation is unbiased, as is seen \(E(\hat \tau) = \sum p(s)\hat \tau = 62\).
Table 3. Calculation of inclusion and joint inclusion probabilities for application of the Horvitz-Thompson estimator as given in Table 2
Stand \(y_i\) \(p_i\) \(\pi_i\) Elements i and j \( \pi_{ij}\) 1 6 0.1 0.19 1-2 0.06 2 20 0.3 0.51 1-3 0.12 3 36 0.6 0.84 2-3 0.36 with \(\pi_i = 1 - (1-p_i)^2\) with \(\pi_{ij} = \pi_i +\pi_j - \{1-(1-p_i-p_j)^2\}\)
While also the Horvitz-Thompson estimator is unbiased for the total, we see that it produces a much higher standard error of the estimated mean (\(\sqrt {V(\hat \tau)} = 18.9\)).
This is an example for a situation in which we have the choice to use the one or the other estimator. Of course, we will use the estimator that produces the smallest standard error, in our case the Hansen-Hurwitz estimator; by that we can be sure that our estimation is, on the average, closer to the unknown true value than with the other estimator.
Eventually, the last column, the Horvitz-Thompson estimator is used to evaluate the samples along selection probabilities proportional to size (option I). The corresponding inclusion and joint inclusion probabilities as required for application of the Horvitz-Thompson estimator are given in Table 3.
References
- ↑ Kleinn, C. 2007. Lecture Notes for the Teaching Module Forest Inventory. Department of Forest Inventory and Remote Sensing. Faculty of Forest Science and Forest Ecology, Georg-August-Universität Göttingen. 179 S.
