In statistics, the Horvitz–Thompson estimator, named after Daniel G. Horvitz and Donovan J. Thompson, is a method for estimating the total and mean of a superpopulation in a stratified sample. Inverse probability weighting is applied to account for different proportions of observations within strata in a target population. The Horvitz–Thompson estimator is frequently applied in survey analyses and can be used to account for missing data.
Formally, let
Y
i
,
i
=
1
,
2
,
…
,
n
be an independent sample from n of N ≥ n distinct strata with a common mean μ. Suppose further that
π
i
is the inclusion probability that a randomly sampled individual in a superpopulation belongs to the ith stratum. The Horvitz–Thompson estimate of the total is given by:
Y
^
H
T
=
∑
i
=
1
n
π
i
−
1
Y
i
,
and the estimate of the mean is given by:
μ
^
H
T
=
N
−
1
Y
^
H
T
=
N
−
1
∑
i
=
1
n
π
i
−
1
Y
i
.
In a Bayesian probabilistic framework
π
i
is considered the proportion of individuals in a target population belonging to the ith stratum. Hence,
π
i
−
1
Y
i
could be thought of as an estimate of the complete sample of persons within the ith stratum. The Horvitz–Thompson estimator can also be expressed as the limit of a weighted bootstrap resampling estimate of the mean. It can also be viewed as a special case of multiple imputation approaches.
For post-stratified study designs, estimation of
π
and
μ
are done in distinct steps. In such cases, computating the variance of
μ
^
H
T
is not straightforward. Resampling techniques such as the bootstrap or the jackknife can be applied to gain consistent estimates of the variance of the Horvitz–Thompson estimator. The Survey package for R conducts analyses for post-stratified data using the Horvitz–Thompson estimator.
The Horvitz–Thompson estimator can be shown to be unbiased when evaluating the expectation of the Horvitz–Thompson estimator,
E
X
¯
n
H
T
, as follows:
E
X
¯
n
H
T
=
E
1
N
∑
i
=
1
n
X
I
i
π
I
i
=
E
1
N
∑
i
=
1
N
X
i
π
i
1
i
∈
D
n
=
∑
b
=
1
B
P
(
D
n
(
b
)
)
[
1
N
∑
i
=
1
N
X
i
π
i
1
i
∈
D
n
(
b
)
]
=
1
N
∑
i
=
1
N
X
i
π
i
∑
b
=
1
B
1
i
∈
D
n
(
b
)
P
(
D
n
(
b
)
)
=
1
N
∑
i
=
1
N
(
X
i
π
i
)
π
i
=
1
N
∑
i
=
1
N
X
i
w
h
e
r
e
D
n
=
{
x
1
,
x
2
,
.
.
.
,
x
n
}
