Estimation, Evaluation, and Selection of Actuarial Models

3.3. VARIANCE AND CONFIDENCE INTERVALS 49
When data are censored or truncated the matter becomes more complex. Counts no longer
have the binomial distribution and therefore the distribution of the estimator is harder to obtain.
While there are proofs available to back up the results presented here, they will not be provided.
Instead, an attempt will be made to indicate why the results are reasonable.
Consider the Kaplan-Meier product-limit estimator of S(t). It is the product of a number of
terms of the form (r j − s j )/r j where r j was viewed as the number available to die at age y j and s j
is the number who actually did so. Assume that the death ages and the number available to die are
fixed, so that the value of s j is the only random quantity. As a random variable, S j has a binomial
distribution based on a sample of r j lives and success probability [S(y j−1 ) − S(y j )]/S(y j−1 ). The
probability arises from the fact that those available to die were known to be alive at the previous
death age. For one of these terms,
µ
rj − S j
E
= r j − r j [S(y j−1 ) − S(y j )]/S(y j−1 )
= S(y j)
r j r j S(y j−1 ) .
That is, this ratio is an unbiased estimator of the probability of surviving from one death age to
the next one. Furthermore,
h
i
µ S(y
rj − S r j−1 )−S(y j )
j
j S(y j−1 )
1 − S(y j−1)−S(y j )
S(y j−1 )
Var
=
r j
r 2 j
= [S(y j−1) − S(y j )]S(y j )
r j S(y j−1 ) 2 .
Now consider the estimated survival probability at one of the death ages. Its expected value is
E[Ŝ(y j)] = E
=
jY
i=1
" jY
i=1
µ
ri − S i
r i
# =
S(y i )
S(y i−1 ) = S(y j)
S(y 0 )
jY
i=1
µ
ri − S i
E
r i
where y 0 is the smallest observed age in the sample. In order to bring the expectation inside the
product, it was assumed that the S-values are independent. The result demonstrates that at the
death ages, the estimator is unbiased.
With regard to the variance, we first need a general result concerning the variance of a product of
independent random variables. Let X 1 ,...,X n be independent random variables where E(X j )=µ j
and Var(X j )=σ 2 j . Then,
Var(X 1 ···X n ) = E(X 2 1 ···X 2 n) − E(X 1 ···X n ) 2
= E(X 2 1) ···E(X 2 n) − E(X 1 ) 2 ···E(X n ) 2
= (µ 2 1 + σ 2 1) ···(µ 2 n + σ 2 n) − µ 2 1 ···µ 2 n.