Horvitz-Thompson estimator example

Revision as of 21:19, 11 January 2011

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.

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).

Figure 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)
	\(\bar y\)	\(\hat \tau\)	Prob(wr)	\( \hat \tau\)	Prob.	\(\hat \tau\)	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\[\sqrt {V(\bar y)} = 6.1283 = \sqrt {\frac {\sigma^2}{n}} * \sqrt {\frac {N-n}{N-1}}\]				\( E(\hat \tau) = \sum p_i \hat \tau = 62 \) \(\sqrt {v(\hat \tau)} = \sqrt {\sum p_i (\hat \tau - \tau)^2}\) \(= \sqrt {E(\hat \tau^2)-\{E(\hat \tau)\}^2}\) \(=(3848.667 - 62^2)^\frac {1}{2}\) \(= 4.667^\ {1}{2} = 2.16\)		\( E(\hat \tau)= 62 \) \(E(\hat \tau^2) = 9098.7\) \(V(\hat \tau) = 5254.6\) \(\sqrt {V(\hat \tau)} = 72.49 \)		\( E(\hat \tau) = 62 \) \(V(\hat \tau) = 356.1\) \(\sqrt {V(\hat \tau)} = 18.9\)

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