7.4. Completeness and Uniqueness 431Let us consider the family{g 1 (y 1 ;θ):θ> 0 }of probability mass functions. Suppose
that the functionu(Y 1 )ofY 1 is such thatE[u(Y 1 )] = 0 for everyθ>0. We shall
show that this requiresu(y 1 )tobezeroateverypointy 1 =0, 1 , 2 ,....Thatis,
E[u(Y 1 )] = 0 forθ>0requires
0=u(0) =u(1) =u(2) =u(3) =···.We have for allθ>0that
0=E[u(Y 1 )] =∑∞y 1 =0u(y 1 )(nθ)y^1 e−nθ
y 1!= e−nθ[
u(0) +u(1)nθ
1!+u(2)(nθ)^2
2!+···]
.Sincee−nθdoes not equal zero, we have shown that0=u(0) + [nu(1)]θ+[
n^2 u(2)
2]
θ^2 +···.However, if such an infinite (power) series converges to zero for allθ>0, then each
of the coefficients must equal zero. That is,
u(0) = 0,nu(1) = 0,n^2 u(2)
2=0,... ,and thus 0 =u(0) =u(1) =u(2) =···, as we wanted to show. Of course, the
conditionE[u(Y 1 )] = 0 for allθ>0 does not place any restriction onu(y 1 )wheny 1
is not a nonnegative integer. So we see that, in this illustration,E[u(Y 1 )] = 0 for all
θ>0 requires thatu(y 1 ) equals zero except on a set of points that has probability
zero for each pmfg 1 (y 1 ;θ), 0 <θ. From the following definition we observe that
the family{g 1 (y 1 ;θ):0<θ}is complete.
Definition 7.4.1.Let the random variableZof either the continuous type or the
discrete type have a pdf or pmf that is one member of the family{h(z;θ):θ∈Ω}.If
the conditionE[u(Z)] = 0, for everyθ∈Ω, requires thatu(z)be zero except on a set
of points that has probability zero for eachh(z;θ),θ∈Ω, then the family{h(z;θ):
θ∈Ω}is called acomplete familyof probability density or mass functions.
Remark 7.4.1.In Section 1.8, it was noted that the existence ofE[u(X)] implies
that the integral (or sum) converges absolutely. This absolute convergence was
tacitly assumed in our definition of completeness and it is needed to prove that
certain families of probability density functions are complete.
In order to show that certain families of probability density functions of the
continuous type are complete, we must appeal to the same type of theorem in anal-
ysis that we used when we claimed that the moment generating function uniquely
determines a distribution. This is illustrated in the next example.