Ratio Estimation
(precise auxillary information)

Auxillary Information

  • stratification used coarse aux. information to improve estimator precision
    • groupings
  • Can we use continuous aux. information?
    • Ratio Estimator

Fundamental Information

  • We sample via probabilistic sampling and observe our primary variable of interest (\(y_i\))
  • We also record a secondary variable - that correlates with the primary (\(x_i\))
  • We also know the true population mean/total for the secondary variable (\(\mu_x\))

Is this useful to us??

Relevant Wildlife/Fish/Habitat studies???

Example

We take morphometric data on all Lowland Tree Kangaroos in a forest of Papua New Guinea (6 individuals).

However, what if you end up releasing 2 individuals before getting their weight, but all individuals height was measured.

What was the mean weight of all individuals in the forest?

Example (truth)

Indiv height weight
1 32 25
2 24 20
3 28 26
4 20 13
5 36 33
6 25 26

\(\mu_{weight} =\) 23.8333333

\(\mu_{height} =\) 27.5

Example (samples)

Weight

Sample.Number Indiv.1 Indiv.2 Indiv.3 Indiv.4 means
1 25 25 25 25 25.00
2 25 25 25 25 25.00
3 25 25 20 20 22.50
4 20 20 26 20 21.50
5 20 20 20 20 20.00
6 20 26 26 26 24.50
7 13 26 26 26 22.75
8 13 13 26 26 19.50
9 26 13 13 33 21.25
10 13 13 33 33 23.00
11 13 13 33 33 23.00
12 33 13 33 26 26.25
13 33 26 26 33 29.50
14 26 26 26 33 27.75
15 26 26 26 26 26.00

\(E[\hat{\mu}_{weight}] =\) 23.8333333 \(=\mu_{weight}\)

Example (samples)

Height

Sample.Number Indiv.1 Indiv.2 Indiv.3 Indiv.4 means
1 32 32 32 32 32.00
2 32 32 32 32 32.00
3 32 32 24 24 28.00
4 24 24 28 24 25.00
5 24 24 24 24 24.00
6 24 28 28 28 27.00
7 20 28 28 28 26.00
8 20 20 28 28 24.00
9 28 20 20 36 26.00
10 20 20 36 36 28.00
11 20 20 36 36 28.00
12 36 20 36 25 29.25
13 36 25 25 36 30.50
14 25 25 25 36 27.75
15 25 25 25 25 25.00

\(E[\hat{\mu}_{height}] =\) 27.5 \(=\mu_{height}\)

Ratio Estimator of population mean

\(\hat{\mu}_{r}\) = sample ratio \(\times\) population mean of aux

\(\hat{\mu}_{r}\) = sample primary mean / sample aux. mean \(\times\) population mean of aux . . .

\(\hat{\mu}_{r}\) = \(r \times \mu_{x}\)

\(\mu_{x} = \frac{\sum_{i=1}^N x_i}{N}\)

\(\hat{r} = \frac{\sum_{i=1}^n y_i}{\sum_{i=1}^n x_i} = \frac{\hat{\mu}_{primary}}{\hat{\mu}_{secondary}} =\frac{\hat{\mu}_{weight}}{\hat{\mu}_{height}}= \frac{\bar{y}}{\bar{x}}\)

Ratio Estimator of population mean

\[ \hat{\sigma}^2_{\hat{\mu_r}} = \left(\frac{N-n}{N}\right)\frac{\hat{\sigma}^2_r}{n} \]

\[ \hat{\sigma}^2_r = \frac{1}{n-1}\sum_{i=1}^n\left(y_i-rx_i\right)^2 \]

Example (ratio estimator)

Sample.Number Primary.Weight Aux.Height Ratio
1 32.00 25.00 21.48438
2 32.00 25.00 21.48438
3 28.00 22.50 22.09821
4 25.00 21.50 23.65000
5 24.00 20.00 22.91667
6 27.00 24.50 24.95370
7 26.00 22.75 24.06250
8 24.00 19.50 22.34375
9 26.00 21.25 22.47596
10 28.00 23.00 22.58929
11 28.00 23.00 22.58929
12 29.25 26.25 24.67949
13 30.50 29.50 26.59836
14 27.75 27.75 27.50000
15 25.00 26.00 28.60000

\(E[\hat{\mu}_{r}] =\) 23.8683977 \(\neq\) 23.83333

Comparisons

  • Sample average of primary (weight) is unbiased
    • min: 19.5
    • max: 29.5
    • Exp.Var: 3.3777778
  • Ratio estimator of primary (weight) is biased
    • min: 21.484375
    • max: 28.6
    • Exp.Var: 0.5236118

Considerations

Ratio Estimator

  • design based
  • estimators are not unbiased (almost at large sample size; Thompson, Ch. 7)
  • improves precision
  • depends on linear correlation b/w primary and auxiliary information
  • requires: when \(x_i = 0\) then \(y_i=0\); i.e., intercept is zero
  • rarely do we measure one thing when sampling

Classic Example

Pierre-Simon Laplace

  • Wanted a census of France in 1802
    • Primary: Population Size
    • Auxiliary: Births
  • Had all birth records from church records
  • Sampled 30 communities and calculated a total of 2,037,615 people with 71,866 births
    • 2,037,615/71,866 = 28.35 people per birth
  • Reasoned he could multiply the population number of births by 28.35 to obtain an estimate of population size.

Classic Wildlife Example

Estimating total population

  • We mark and release \(n_1\) individuals
  • We go and recapture the same population, catching \(n_2\) individuals
  • The number of marked individuals caught again is \(m_2\)
  • Assume equal catchability of individuals and by sample

\[ \frac{m_2}{n_2} = \frac{n_1}{N} \] . . .

Lincoln-Peterson Abundance Estimator

Classic Wildlife Example

\[ \hat{N} = \frac{n_1n_2}{m_2} \] . . .

\[ \hat{N} = \frac{n_1}{\hat{p}} \] \[ \hat{p} = \frac{m_2}{n_2} \]

Classic Wildlife Example

Model-based Ratio Estimators