6
Calibration with an Auxiliary Variable
6.1 Calibration with a qualitative variable
We assume that the sizes Nh , where h = 1, ..., H, of H types of a qualitative
variable are known in the population. The qualitative variable specifies H
parts Uh , where h = 1, ..., H, called post-strata in the population. If the sample
S is selected in accordance with a simple design without replacement, then
the size of the sample intersecting post-strata h, being nh = #(Uh ∩ S) has a
hypergeometric distribution. If we denote Yh as the true total of a variable y
over Uh , we can construct the post-stratified estimator of the total
H
Ypost = Nh Y h ,
h=1
nh >0
where Y h = Yh /N
h . With a simple design without replacement,
Y h =
1
yk .
nh
k∈Uh ∩S
With a simple design without replacement, the post-stratified estimator is
unbiased as soon as we keep to the conditions of nh non-null for all h, and it
is all the more precise since the auxiliary variable is ‘linked’ to the variable of
interest. If n is ‘large enough’, the variance of Ypost is approximately, for the
simple design without replacement:
var(Ypost )
H
n1 n 1
H
Nh 2 Nh
≈ N2 1− S + 1− 1− 2
Syh ,
N n N yh N n2 N
h=1 h=1
,210 6 Calibration with an Auxiliary Variable
and is estimated by
Ypost )
var(
H
n1 n 1
H
Nh 2 Nh
=N 2
1− s + 1− 1− s2yh ,
N n N yh N n2 N
h=1 h=1
where
1
2
Syh = (yk − Y h )2 ,
Nh − 1
k∈Uh
and
(yk − Y h )2 .
1
s2yh =
nh − 1
k∈Uh ∩S
6.2 Calibration with a quantitative variable
If the total X of a quantitative variable x is known, we can use this information
to construct a more precise estimator. If X π and Yπ designate respectively the
Horvitz-Thompson estimators of the totals of variables x and y, then we can
construct
• the difference estimator:
YD = Yπ + X − X
π ,
• the ratio estimator:
X
YR = Yπ ,
π
X
• the regression estimator:
Yreg = Yπ + (X − X
π )b̂,
where b̂ is an estimator of the affine regression coefficient of y over x:
Sxy
b= ,
Sx2
and
1
Sxy = (xk − X)(yk − Y ).
N −1
k∈U
We can choose, to estimate b:
1 Xπ Yπ
xk − yk −
πk Nπ Nπ
k∈S
b̂ = 2 .
1 π
X
xk −
πk π
N
k∈S
, Exercise 6.1 211
All of these estimators satisfy a fundamental property of calibration, as they
estimate with null variance the total X (we are speaking about estimators
calibrated on x):
XD = X R = X
reg = X.
We can show that:
yk − xk
• var(YD ) = var ,
πk
⎛k∈S
⎞
Y
yk − xk
⎜ X ⎟
• var(YR ) ≈ var ⎝ ⎠ (n ‘large enough’),
πk
k∈S
(yk − Y ) − b(xk − X)
• var(Yreg ) ≈ var (n ‘large enough’),
πk
k∈S
which comes back to using the general expressions of Chapter 3 with new
individual variables. Thus, with simple random sampling, we estimate these
variances with:
n1 1
N2 1 − (yk − α − βxk )2 ,
N nn−1
k∈S
by holding:
• α = 0, β = 1 with YD ;
Y
• α = 0, β = with YR ;
X
(xk − X)(y k −Y)
α = Y − b X,
β = b = with Yreg .
k∈S
• 2
(xk − X)
k∈S
EXERCISES
Exercise 6.1 Ratio
In a population of 10 000 businesses, we want to estimate the average sales
Y . For that, we sample n = 100 businesses using simple random sampling.
Furthermore, we have at our disposal the auxiliary information ‘number of
employees’, denoted by x, for each business. The data coming from the sample
are:
• X = 50 employees (true mean for xk ),
• Y = 5.2 × 106 Euros (average sales in the sample),
, 212 6 Calibration with an Auxiliary Variable
• X = 45 employees (sample mean),
• s2y = 25 × 1010 (corrected sample variance of yk ),
• s2x = 15 (corrected sample variance of xk ),
• ρ = 0.80 (linear correlation coefficient between x and y calculated in the
sample).
1. What is the ratio estimator? (We denote this as Y R .) Is this estimator
biased?
2. Recall the ‘true’ variance formula for this estimator.
3. Calculate an estimate of the true variance. Is the variance estimator used
biased?
4. Give a 95% confidence interval for Y .
Solution
1. By definition:
Y 5.2 × 106
Y R = X = 50 × ≈ 5.8 × 106 Euros.
X 45
We have Y R > Y because the sample contains businesses that are on
average too small (in terms of employees), and thus with sales that are
a little bit too small. A priori, the estimator is biased: the 1/n term
appearing in the bias is null when
Sx Sy
=ρ .
X Y
None of the terms of this equality can be estimated without bias, but a
calculation of magnitudes (bias 1/n) compares:
sx sy
≈ 0.086 and ρ̂ ≈ 0.077.
X Y
Numerically, they are close values, which lets us think that the bias must
be very small.
2. For n ‘large’, we have:
1−f 2 1−f
var(Y R ) ≈ Su = [Sy2 + R2 Sx2 − 2RSxy ].
n n
Su2 is the population variance of ui , where ui = yi − Rxi with R = Y /X.
3. We have
1 − f 2 2 2
Y R ) =
var( [sy + R sx − 2Rs xy ].
n
Calibration with an Auxiliary Variable
6.1 Calibration with a qualitative variable
We assume that the sizes Nh , where h = 1, ..., H, of H types of a qualitative
variable are known in the population. The qualitative variable specifies H
parts Uh , where h = 1, ..., H, called post-strata in the population. If the sample
S is selected in accordance with a simple design without replacement, then
the size of the sample intersecting post-strata h, being nh = #(Uh ∩ S) has a
hypergeometric distribution. If we denote Yh as the true total of a variable y
over Uh , we can construct the post-stratified estimator of the total
H
Ypost = Nh Y h ,
h=1
nh >0
where Y h = Yh /N
h . With a simple design without replacement,
Y h =
1
yk .
nh
k∈Uh ∩S
With a simple design without replacement, the post-stratified estimator is
unbiased as soon as we keep to the conditions of nh non-null for all h, and it
is all the more precise since the auxiliary variable is ‘linked’ to the variable of
interest. If n is ‘large enough’, the variance of Ypost is approximately, for the
simple design without replacement:
var(Ypost )
H
n1 n 1
H
Nh 2 Nh
≈ N2 1− S + 1− 1− 2
Syh ,
N n N yh N n2 N
h=1 h=1
,210 6 Calibration with an Auxiliary Variable
and is estimated by
Ypost )
var(
H
n1 n 1
H
Nh 2 Nh
=N 2
1− s + 1− 1− s2yh ,
N n N yh N n2 N
h=1 h=1
where
1
2
Syh = (yk − Y h )2 ,
Nh − 1
k∈Uh
and
(yk − Y h )2 .
1
s2yh =
nh − 1
k∈Uh ∩S
6.2 Calibration with a quantitative variable
If the total X of a quantitative variable x is known, we can use this information
to construct a more precise estimator. If X π and Yπ designate respectively the
Horvitz-Thompson estimators of the totals of variables x and y, then we can
construct
• the difference estimator:
YD = Yπ + X − X
π ,
• the ratio estimator:
X
YR = Yπ ,
π
X
• the regression estimator:
Yreg = Yπ + (X − X
π )b̂,
where b̂ is an estimator of the affine regression coefficient of y over x:
Sxy
b= ,
Sx2
and
1
Sxy = (xk − X)(yk − Y ).
N −1
k∈U
We can choose, to estimate b:
1 Xπ Yπ
xk − yk −
πk Nπ Nπ
k∈S
b̂ = 2 .
1 π
X
xk −
πk π
N
k∈S
, Exercise 6.1 211
All of these estimators satisfy a fundamental property of calibration, as they
estimate with null variance the total X (we are speaking about estimators
calibrated on x):
XD = X R = X
reg = X.
We can show that:
yk − xk
• var(YD ) = var ,
πk
⎛k∈S
⎞
Y
yk − xk
⎜ X ⎟
• var(YR ) ≈ var ⎝ ⎠ (n ‘large enough’),
πk
k∈S
(yk − Y ) − b(xk − X)
• var(Yreg ) ≈ var (n ‘large enough’),
πk
k∈S
which comes back to using the general expressions of Chapter 3 with new
individual variables. Thus, with simple random sampling, we estimate these
variances with:
n1 1
N2 1 − (yk − α − βxk )2 ,
N nn−1
k∈S
by holding:
• α = 0, β = 1 with YD ;
Y
• α = 0, β = with YR ;
X
(xk − X)(y k −Y)
α = Y − b X,
β = b = with Yreg .
k∈S
• 2
(xk − X)
k∈S
EXERCISES
Exercise 6.1 Ratio
In a population of 10 000 businesses, we want to estimate the average sales
Y . For that, we sample n = 100 businesses using simple random sampling.
Furthermore, we have at our disposal the auxiliary information ‘number of
employees’, denoted by x, for each business. The data coming from the sample
are:
• X = 50 employees (true mean for xk ),
• Y = 5.2 × 106 Euros (average sales in the sample),
, 212 6 Calibration with an Auxiliary Variable
• X = 45 employees (sample mean),
• s2y = 25 × 1010 (corrected sample variance of yk ),
• s2x = 15 (corrected sample variance of xk ),
• ρ = 0.80 (linear correlation coefficient between x and y calculated in the
sample).
1. What is the ratio estimator? (We denote this as Y R .) Is this estimator
biased?
2. Recall the ‘true’ variance formula for this estimator.
3. Calculate an estimate of the true variance. Is the variance estimator used
biased?
4. Give a 95% confidence interval for Y .
Solution
1. By definition:
Y 5.2 × 106
Y R = X = 50 × ≈ 5.8 × 106 Euros.
X 45
We have Y R > Y because the sample contains businesses that are on
average too small (in terms of employees), and thus with sales that are
a little bit too small. A priori, the estimator is biased: the 1/n term
appearing in the bias is null when
Sx Sy
=ρ .
X Y
None of the terms of this equality can be estimated without bias, but a
calculation of magnitudes (bias 1/n) compares:
sx sy
≈ 0.086 and ρ̂ ≈ 0.077.
X Y
Numerically, they are close values, which lets us think that the bias must
be very small.
2. For n ‘large’, we have:
1−f 2 1−f
var(Y R ) ≈ Su = [Sy2 + R2 Sx2 − 2RSxy ].
n n
Su2 is the population variance of ui , where ui = yi − Rxi with R = Y /X.
3. We have
1 − f 2 2 2
Y R ) =
var( [sy + R sx − 2Rs xy ].
n
