We consider the Berkson model of logistic regression with Gaussian and homoscedastic error in regressor. The measurement error variance can be either known or unknown. We deal with both functional and structural cases. Sufficient conditions for identifiability of regression coefficients are presented.
Conditions for identifiability of the model are studied. In the case where the error variance is known, the regression parameters are identifiable if the distribution of the observed regressor is not concentrated at a single point. In the case where the error variance is not known, the regression parameters are identifiable if the distribution of the observed regressor is not concentrated at three (or less) points.
The key analytic tools are relations between the smoothed logistic distribution function and its derivatives.
Logistic regressionbinary regressionerrors in variablesBerkson modelregression calibration model62J12Introduction
Statistical model. Consider logistic regression with Berkson-type error in the explanatory variable. One trial is distributed as follows. Xnobs is the observed (or assigned) surrogate regressor. The true regressor is Xn=Xnobs+Un, where the error Un∼N(0,τ2) is independent of Xnobs. The response Yn is a binary random variable and attains either 0 or 1 with
P(Yn=1|Xnobs,Xn)=exp(β0+β1Xn)1+exp(β0+β1Xn).
We consider both functional model and structural model. In the functional one, Xnobs are nonrandom variables, and in the structural one, Xnobs are i.i.d., and therefore in the latter model, (Xnobs,Xn,Yn) are i.i.d. random triples.
The couples (Xnobs,Yn), n=1,…,N, are observed. Vector β→=(β0,β1)⊤ is a parameter of interest.
The error variance τ2 can be either known or unknown, and we consider both cases. The conditions for identifiability of the model (or of the parameter β→) are presented.
Overview. Berkson models of logistic regression and probit regression were set up in Burr [1]. For probit regression, it is shown that the introduction of Berkson-type error is equivalent to augmentation of regression parameters. As a consequence, the Berkson model of probit regression is identifiable if τ2 is known and is not identifiable if τ2 is not known.
The identifiability of the classical model was studied by Küchenhoff [3]. He assumes that both the regressor and measurement error are normally distributed. Then univariate logistic regression is identifiable (here τ2 can be unknown), and multiple logistic regression is not identifiable. Our results can be proved similarly to [3] if we assume that the distribution of the surrogate regressor Xobs has an unbounded support.
For classification of errors-in-variables regression models and various estimation methods, see the monograph by Carroll et al. [2].
Identifiability of the statistical model can be used in the proof of consistency of the estimator. For known τ2, the strong consistency of the maximum likelihood estimator is obtained by Shklyar [4]. But if τ2 is not known, the maximum likelihood estimator seems to be unstable (see discussion in [2] or [3]).
Convolution of logistic function with normal density
Consider the function
L0(x,σ2)=Eexp(x−ξ)1+exp(x−ξ),ξ∼N(0,σ2),x∈R,σ2≥0,
that is, L0(x,0)=ex/(1+ex) and
L0(x,σ2)=12πσ∫−∞∞exp(x−t)1+exp(x−t)e−t2/(2σ2)dtforσ2>0.0.\]]]>
Denote the derivatives w.r.t. xLk(x,σ2)=∂k∂xkL0(x,σ2).
Differentiation of Lk(x,σ2) with respect to the second argument is described in Appendix A.
The distribution of Yi given Xiobs is
P[Yi=1|Xiobs]=E[P[Yi=1|Xiobs,Xi]|Xiobs]=E[exp(β0+β1Xi)1+exp(β0+β1Xi)|Xiobs]=L(β0+β1Xiobs,β12τ2)
since [β0+β1Xi∣Xiobs]∼N(β0+β1Xiobs,β12τ2).
Identifiability when <inline-formula id="j_vmsta27_ineq_025"><alternatives>
<mml:math><mml:msup><mml:mrow><mml:mi mathvariant="italic">τ</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msup></mml:math>
<tex-math><![CDATA[${\tau }^{2}$]]></tex-math></alternatives></inline-formula> is known
If in the functional model not allXobsare equal, then the model is identifiable.
Suppose that for two values of parameters β→(1)=(β0(1),β1(1)) and β→(2)=(β0(2),β1(2)), β→(1)≠β→(2), the distributions of observations are equal. Then for all i=1,2,…,N,
Pβ→(1)(Yi=1)=Pβ→(2)(Yi=1),L0(β0(1)+β1(1)Xiobs,(β1(1))2τ2)=L0(β0(2)+β1(2)Xiobs,(β1(2))2τ2).
However, by Lemma 4.1 from [4] the equation
L0(β0(1)+β1(1)x,(β1(1))2τ2)=L0(β0(2)+β1(2)x,(β1(2))2τ2)
has no more than one solution x. Hence, all Xiobs are equal. □
By definition the degenerate distribution is the distribution concentrated at a single point. For the next theorem, see the proof of Theorem 5.1 in [4]. ([<xref ref-type="bibr" rid="j_vmsta27_ref_004">4</xref>]).
If in the structural model the distribution ofX1obsis not degenerate, then the parameterβ→is identifiable.
Identifiability when <inline-formula id="j_vmsta27_ineq_034"><alternatives>
<mml:math><mml:msup><mml:mrow><mml:mi mathvariant="italic">τ</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msup></mml:math>
<tex-math><![CDATA[${\tau }^{2}$]]></tex-math></alternatives></inline-formula> is unknown
For fixed σ2, the function L0(x,σ2) is a bijection R→(0,1). Hence, for fixed σ12 and σ22, the relation
L0(y,σ12)=L0(x,σ22)
sets the bijection R→R; see Fig. 1.
For fixedσ12≥0andσ22≥0, the sign of the second derivative of the implicit function (4) issign(d2ydx2)=sign(σ22−σ12)sign(x).
Differentiating (4), we get
L1(y,σ12)dy=L1(x,σ22)dx;dydx=L1(x,σ22)L1(y,σ12).
Then
d2ydx2=L2(x,σ22)L1(y,σ12)−L1(x,σ22)L2(y,σ12)dydxL1(y,σ12)2=L2(x,σ22)L1(y,σ12)2−L1(x,σ22)2L2(y,σ12)L1(y,σ12)3=(L2(x,σ22)L1(x,σ22)2−L2(y,σ12)L1(y,σ12)2)·L1(x,σ22)2L1(y,σ12).
Thus,
sign(d2ydx2)=sign(L2(x,σ22)L1(x,σ22)2−L2(y,σ12)L1(y,σ12)2).
The plot to equation L0(y,σ12)=L0(x,σ22) for σ12<σ22
Denote by μ(z,σ2) the solution to the equation L0(μ,σ2)=z. Note that as L0(x,σ2) is the cdf of a symmetric distribution, sign(L0(x,σ2)−0.5)=sign(x). Therefore, sign(μ(z,σ2))=sign(z−0.5). Find the derivative
ddv(L2(μ(z,v),v)L1(μ(z,v),v)2)
for fixed z. By the implicit function theorem,
dμ(z,v)dv=−L2(μ(z,v),v)2L1(μ(z,v),v);
also,
∂∂x(L2(x,v)L1(x,v)2)=L3(x,v)L1(x,v)−2L2(x,v)2L1(x,v)3,∂∂v(L2(x,v)L1(x,v)2)=L4(x,v)L1(x,v)−2L2(x,v)L3(x,v)2L1(x,v)3.
Then
ddv(L2(μ(z,v),v)L1(μ(z,v),v)2)=−L22L1·L3L1−2L22L13+L4L1−2L2L32L13=L4L12−3L3L2L1+2L232L14,
where Lk are evaluated at the point (μ(z,v),v). By Lemma 10,
sign(ddv(L2(μ(z,v),v)L1(μ(z,v),v)2))=sign(μ(z,v))=sign(z−0.5).
The function v↦L2(μ(z,v),v)L1(μ(z,v),v)2 is monotone (it is increasing for z>0.50.5$]]> and decreasing for z<0.5). For x and y satisfying (4),
x=μ(z,σ22)andy=μ(z,σ12)
with z=L0(y,σ12)=L0(x,σ22); note that sign(z−0.5)=sign(x). Then
sign(L2(x,σ22)L1(x,σ22)2−L2(y,σ12)L1(y,σ12)2)=sign(σ22−σ12)sign(x),
and with (5), we can obtain the desired equality
sign(d2ydx2)=sign(σ22−σ12)sign(x).
□
The equationL0(β0(1)+β1(1)x,σ12)=L0(β0(2)+β1(2)x,σ22)has no more than three solutions, unless eitherβ→(1)=β→(2)andσ12=σ22orβ1(1)=β1(2)=0andL0(β0(1),σ12)=L0(β0(2),σ22).In exceptional cases (7) and (8), equation (6) is an identity.
The proof has the following idea: if a twice differentiable function y(x) satisfies (4), then the plot of the function either is a straight line (if σ12=σ22) or intersects any straight line at no more than three points.
Consider four cases.
Case 1.σ12=σ22. Since the function L0(z,σ2) is strictly increasing in z, Eq. (6) is equivalent to
β0(1)+β1(1)x=β0(2)+β1(2)x.
Equation (6) has only one solution if β1(1)≠β1(2); it is an identity if β→(1)=β→(2), and it has no solutions if β1(1)=β1(2) but β0(1)≠β0(2).
Case 2.β1(2)=0 and β1(1)≠0. For any fixed σ2, the function z↦L0(z,σ2) is a bijection R→(0,1). Denote the inverse function μ(Z,σ2): L0(z,σ2)=Z if and only if z=μ(Z,σ2). Equation (6) has a unique solution
x=μ(L0(β0(2),σ22),σ12)−β0(1)β1(1).
Case 3.β1(2)=β1(1)=0. Neither side of (6) depends on x. Equation (6) becomes L0(β0(1),σ12)=L0(β0(2),σ22). Equation (6) either holds for all x or does not hold for any x.
Case 4.σ12≠σ22 and β1(2)≠0. Make a linear variable substitution: denote z2=β0(2)+β1(2)x. Then Eq. (6) becomes
L0(β0(1)+β1(1)β1(2)·(z2−β0(2)),σ12)=L0(z2,σ22).
Define the function z1(z2) from the equation
L0(z1(z2),σ12)=L0(z2,σ22).
The function z1(z2):R→R is implicitly defined by Eq. (4): there the equality holds if and only if y=z1(x). Hence, the function z1(z2) satisfies Lemma 3. Equation (9) is equivalent to
z1(z2)−β0(1)−β1(1)β1(2)·(z2−β0(2))=0.
By Lemma 3,
sign(d2dz22(z1(z2)−β0(1)−β1(1)β1(2)·(z2−β0(2))))=sign(d2z1(z2)dz22)=sign(σ22−σ12)sign(z2).
Then the derivative of the left-hand size of (10)
ddz2(z1(z2)−β0(1)−β1(1)β1(2)·(z2−β0(2)))
is strictly monotone on both intervals (−∞,0] and [0,+∞), and hence (11) attains 0 no more than at two points. Then the left-hand side of (10) has no more than three intervals of monotonicity, and Eq. (10) has no more than three solutions. Equation (6) has the same number of solutions. □
If in the functional model there are four differentXobs, then the parametersβ→andβ12τ2are identifiable.
Suppose that there are two sets of parameters (β→(1),(τ(1))2) and (β→(2),(τ(2))2) that for a given sample of the surrogate, the regressors {X0n,n=1,…,N} provide the same distribution of Yn, n=1,…,N. Then for all n=1,…,N,
Pβ→(1),(τ(1))2(Yn=1)=Pβ→(2),(τ(2))2(Yn=1);L0(β0(1)+β1(1)Xnobs,(β1(1))2(τ(1))2)=L0(β0(2)+β1(2)Xnobs,(β1(2))2(τ(2))2).
The equation
L0(β0(1)+β1(1)x,(β1(1))2(τ(1))2)=L0(β0(2)+β1(2)x,(β1(2))2(τ(2))2)
has at least four solutions. Then by Lemma 4 either
β→(1)=β→(2)and(β1(1))2(τ(1))2=(β1(2))2(τ(2))2,
or
β1(1)=β2(2)=0andL0(β0(1),(β1(1))2(τ(1))2)=L0(β0(2),(β1(2))2(τ(2))2).
In the latter alternative,
(β1(1))2(τ(1))2=(β1(2))2(τ(2))2=0andβ0(1)=β0(2)
since L0(b0,0)=11+e−b0 is a strictly increasing function in b0. □
If in the structural model the distribution ofX0is not concentrated at three (or less) points, then the parametersβ→andβ12τ2are identifiable.
Suppose that there are two sets of parameters (β→(1),(τ(1))2) and (β→(2),(τ(2))2) for which the same bivariate distribution of (X1obs,Y1) is obtained. The random variable P[Y1=1∣X1obs] satisfies Eq. (3) almost surely for each set of parameters. Hence, the equality
L0(β0(1)+β1(1)X1obs,(β1(1))2(τ(1))2)=L0(β0(2)+β1(2)X1obs,(β1(2))2(τ(2))2)
holds almost surely. The rest of the proof is the same as in Theorem 5. □
Consider the sum of two independent random variables ζ=λ+ξ, where λ has the logistic distribution
P(λ≤x)=exp(x)1+exp(x),x∈R,
and ξ∼N(0,σ2). We allow σ2=0, and then ξ=0 almost surely.
The function L0(x,σ2) defined in (1) is the cdf of ζ, and the function L1(x,σ2) defined in (2) is the pdf of ζ.
The partial derivatives of Lk(x,v) are
∂∂xLk(x,v)=Lk+1(x,v),∂∂vLk(x,v)=12Lk+2(x,v);
see the proof in [4, Section 2]. The functions Lk(x,v) are infinitely differentiable and bounded on R×[0,+∞).
Since the distribution of ζ is symmetric,
Lk(−x,σ2)=(−1)k−1Lk(x,σ2),k≥1,
that is, L1(x,σ2) and L3(x,σ2) are even functions in x, and L2(x,σ2) and L4(x,σ2) are odd functions in x.
The key inequality
The next lemma is similar to Lemma 2.1 in [4]. Hence, the proof is brief; see [4] for details.
Let ξ and η be two independent random variables, whereξ∼N(0,1). Denoteζ=ξ+ηand letpζ(z)be the pdf of ζ. Thend3dz3(lnpζ(z))=μ3[η|ζ=z],whereμ3[η∣ζ=z]is the third conditional central moment,μ3[η|ζ=z]=E[(η−E[η|ζ=z])3|ζ=z].
We have
pζ(z)=Epξ(z−η)=12πEe−12(z−η)2.
Then
pζ′(z)=12πE[(η−z)e−12(z−η)2],ddz(lnpζ(z))=pζ′(z)pζ(z)=E[(η−z)e−12(z−η)2]Ee−12(z−η)2=Eηe−12(z−η)2Ee−12(z−η)2−z,d2dz2(lnpζ(z))=Eη2e−12(z−η)2Ee−12(z−η)2−(Eηe−12(z−η)2)2(Ee−12(z−η)2)2−1,d3dz3(lnpζ(z))=(Ee−12(z−η)2)−3×(E[η2(η−z)e−12(z−η)2](Ee−12(z−η)2)2+Eη2e−12(z−η)2E[(η−z)e−12(z−η)2]Ee−12(z−η)2−2E[η(η−z)e−12(z−η)2]Eηe−12(z−η)2Ee−12(z−η)2−2Eη2e−12(z−η)2Ee−12(z−η)2E[(η−z)e−12(z−η)2]+2(Eηe−12(z−η)2)2E[(η−z)e−12(z−η)2])=(Ee−12(z−η)2)−3×(Eη3e−12(z−η)2(Ee−12(z−η)2)2−3Eη2e−12(z−η)2Eηe−12(z−η)2Ee−12(z−η)2+2(Eηe−12(z−η)2)3).
If η has a pdf, the conditional pdf of η given ζ=z is equal to
pη|ζ=z(y)=pη(y)e−12(z−y)2Ee−12(z−η)2;
otherwise, we can use the conditional density of η w.r.t. marginal density
dcdfη|ζ=z(y)dcdfη(y)=e−12(z−y)2Ee−12(z−η)2.
Anyway, the conditional moments of η given ζ=z are equal to
E[ηk|ζ=z]=Eηke−12(z−η)2Ee−12(z−η)2.
From (13) and (14) it follows that
d3dz3(lnpζ(z))=E[η3|ζ=z]−3E[η2|ζ=z]E[η|ζ=z]+2(E[η|ζ=z])3=μ3[η|ζ=z].
□
Let ξ and η be independent random variables such thatξ∼N(μ,σ2). Denoteζ=ξ+η, and denote the pdf of ζ bypζ(z). Thend3dz3(lnpζ(z))=1σ6μ3[η∣ζ=z].
Assume that the distribution of a random variable X satisfies the following conditions:
X has a continuously differentiable densitypX(x).
X is unimodal in the following sense: there exists a modeM∈Rsuch that for allx∈R, we have the equalitysign(pX′(x))=sign(M−x).
1) EX>MM$]]>. Denote by x1(z) and x2(z) the solutions to the equation pX(x)=z (see Fig. 2):
x1(z)<M<x2(z)if0<z<max(pX);x1(z)=M=x2(z)ifz=max(pX);pX(x1(z))=pX(x2(z))=zif0<z≤max(pX).
Represent the expectation as a double integral and change the order of integration:
EX=M+∫−∞∞(x−M)pX(x)dx=M+∬{(x,z)|0≤z≤pX(x)}(x−M)dxdz=M+∫0max(pX)(∫x1(z)x2(z)(x−M)dx)dz=M+∫0max(pX)(x2(z)−M)2−(M−x1(z))22dz.
To proof of Lemma 9, part 1). Sample pX(x) and definition of x1(z) and x2(z)
For all x2>MM$]]>, by the implicit function theorem,
ddx2x1(pX(x2))=pX′(x2)pX′(x1(pX(x2)))>−1-1\]]]>
because pX(x1(pX(x2)))=pX(x2) implies pX′(x1(pX(x2)))>−pX′(x2)>0-{p^{\prime }_{X}}(x_{2})>0$]]>. Note that x1(pX(M))=M. By the Lagrange theorem,
x1(pX(x2))=M+(x2−M)·ddx3x1(pX(x3))|x3=M+(x2−M)θ
for some θ∈(0,1);
x1(pX(x2))>M−(x2−M)forx2>M;x1(z)>M−(x2(z)−M)for0<z<max(pX);x2(z)−M>M−x1(z)>0;(x2(z)−M)22>(M−x1(z))22;M-(x_{2}-M)\hspace{1em}\text{for}\hspace{2.5pt}x_{2}>M;\\{} \displaystyle x_{1}(z)& \displaystyle >M-\big(x_{2}(z)-M\big)\hspace{1em}\text{for}\hspace{2.5pt}0M-x_{1}(z)>0;\\{} \displaystyle \frac{{(x_{2}(z)-M)}^{2}}{2}& \displaystyle >\frac{{(M-x_{1}(z))}^{2}}{2};\end{array}\]]]>
the last integrand in (15) is positive, and then (15) implies EX>MM$]]>.
2) Consider the function
f(t)=pX(EX+t)−pX(EX−t),
which is odd and strictly decreasing on the interval [−(EX−M),EX−M]. Therefore, f(t) attains 0 only once on this interval, that is, at the point 0 (see Fig. 3).
To proof of Lemma 9, part 2)
If t>EX−M\operatorname{\mathsf{E}}X-M$]]> (more generally, |t|>EX−M\operatorname{\mathsf{E}}X-M$]]>) and f(t)=0, then f′(t)=pX′(EX+t)+pX′(EX−t)>00$]]> by condition 3) of Lemma 9. Therefore, f(t) can attain 0 only once on (EX−M,+∞), and if it attains 0 (say, at a point t1>EX−M>0\operatorname{\mathsf{E}}X-M>0$]]>), it is increasing in the neighborhood of t1.
Hence, there may be two cases of sign changing of f(t) (Fig. 3). Either
∃t1>0∀x∈R:sign(f(t))=sign(t)sign(|t|−t1),0\hspace{2.5pt}\forall x\in \mathbb{R}\hspace{0.2778em}:\hspace{0.2778em}\operatorname{sign}\big(f(t)\big)=\operatorname{sign}(t)\operatorname{sign}\big(|t|-t_{1}\big),\]]]>
or
∀x∈R:sign(f(t))=−sign(t).
3) We have
0=E[X−EX]=∫−∞∞(x−EX)pX(x)dx=∫−∞∞tpX(EX+t)dt=∫0∞tpX(EX+t)dt+∫0∞(−t)pX(EX−t)dt=∫0∞tf(t)dt,
where f(t) is defined in the second part of the proof.
Note that the case (17) is impossible because otherwise the last integrand in (18) would be negative and thus the integral could not be equal to 0.
4) Similarly to (18),
E(X−EX)3=∫0∞t3f(t)dt.
Subtract t12 times Eq. (18), where t1 comes from (16):
E(X−EX)3=∫0∞t(t2−t12)f(t)dt.
The integrand is positive for t>00$]]>, t≠t1, and hence μ3[X]=E(X−EX)3>00$]]>. □
For allx∈Randσ2≥0,sign(L4(x,σ2)L1(x,σ2)2−3L3(x,σ2)L2(x,σ2)L1(x,σ2)+2L2(x,σ2)3)=sign(x).
Lemma 11 is needed to prove Lemma 10. The notation F(y) and y0 is common for Lemmas 10 and 11.
For fixed x>00$]]> and σ2, consider the function
F(y)=ln(ey(ey+1)2)−(y−x)22σ2.
Its derivative
F′(y)=1−2eyey+1−y−xσ2
is strictly decreasing, and
limy→−∞F′(y)=+∞,limy→+∞F′(y)=−∞.
Hence, F′(y) attains 0 at a unique point. Denote this point by y0, and then
sign(F′(y))=−sign(y−y0).
For the functionF(y)defined in (19), fory0satisfying (20), and fory3andy4such thatF′(y3)+F′(y4)=0andy3<y4, we have the following inequalities:
y3<y0<y4andF′(y3)=−F′(y4)>00$]]>.
y3+y4>00$]]>.
F″(y3)<F″(y4)<0.
F(y3)>F(y4)F(y_{4})$]]>.
1) The inequalityy3<y0<y4 is a consequence of (20), and (20) implies F′(y3)>00$]]>.
2) y3+y4>00$]]>. For all y∈R,F′(y)+F′(−y)=2xσ2>0.0.\]]]>
Since F′(y3)+F′(−y3)>00$]]> and F′(y3)+F′(y4)=0, we have F′(−y3)>F′(y4){F^{\prime }}(y_{4})$]]>, and then −y3<y4 because the derivative F′(y) is decreasing.
3) F″(y3)<F″(y4)<0. The second derivative
F″(y)=−2ey(ey+1)2−1σ2
is an even function strictly increasing on [0,+∞) and attaining only negative values.
The inequalities y3<y4 and y3+y4>00$]]> can be rewritten as |y3|<y4, and then
F″(y3)=F″(|y3|)<F″(y4)<0.
4) F(x3)>F(x4)F(x_{4})$]]>. Consider the inverse function
(F′)−1(t),t∈R.
Its derivative is
ddt((F′)−1(t))=1F″((F′)−1(t))<0.
Then
ddt(F((F′)−1(t)))=F′((F′)−1(t))F″((F′)−1(t))=tF″((F′)−1(t));ddt(F((F′)−1(t))−F((F′)−1(−t)))=tF″((F′)−1(t))+−tF″((F′)−1(−t)).
Apply already proven part 3) of Lemma 11. If t>00$]]>, then (F′)−1(t)<(F′)−1(−t) (because (F′)−1(t) is a decreasing function) and F′((F′)−1(t))+F′((F′)−1(−t))=t−t=0. Then by part 3)
F″((F′)−1(t))<F″((F′)−1(−t))<0,t>0.0.\]]]>
Hence,
ddt(F((F′)−1(t))−F((F′)−1(−t)))>0,t>0.0,\hspace{1em}t>0.\]]]>
Note that
F((F′)−1(0))−F((F′)−1(−0))=0.
By the Lagrange theorem, for t>00$]]>,
F((F′)−1(t))−F((F′)−1(−t))=t·ddt1(F((F′)−1(t1))−F((F′)−1(−t1)))>0,0,\]]]>
where the derivative is taken at some point t1∈(0,t).
Substituting t=F′(y3)>00$]]> (then −t=F′(y4)), we obtain F(y3)−F(y4)>00$]]>. □
Case 1.x>00$]]>andσ2>00$]]>. Recall that for fixed σ2, L1(x,σ2) is the pdf of η+ξ, where η and ξ are independent variables, P(η<y)=eyey+1 and ξ∼N(0,σ2) (see Appendix A). By Corollary 8,
d3dx3(lnL1(x,σ2))=1σ6μ3[η|η+ξ=x],
but
d3dx3(lnL1(x,σ2))=L4L12−3L3L2L1+2L23L13,
where Lk are evaluated at the point (x,σ2). Since L1(x,σ2)>00$]]>, we have to prove that μ3[η|η+ξ=x]>00$]]>. Therefore, we apply Lemma 9.
The pdf of the conditional distribution of η given η+ξ=x is equal to
pη|η+ξ=x(y)=1Ee−(η−x)22σ2·ey(1+ey)2e−(y−x)22σ2.
The pdf pη|η+ξ=x(y) is continuously differentiable. The conditional distribution has a finite kth moment because yke−(y−x)22σ2 is bounded for any k∈N. Hence, conditions 1) and 4) of Lemma 9 are satisfied.
Evaluate
lnpη|η+ξ=x(y)=ln(ey(ey+1)2)−y−x2σ2−ln(Ee−(η−x)22σ2)=F(y)+C,
where the function F(y) is defined in (19), and C=−ln(Eexp(−(η−x)22σ2)) depends only on x and σ2 and does not depend on y.
We check condition 2) of Lemma 9:
pη|η+ξ=x(y)=eF(y)+C;ddypη|η+ξ=x(y)=F′(y)eF(y)+C;sign(ddypη|η+ξ=x(y))=sign(F′(y))=−sign(y−y0),
and condition 2) holds with M=y0, where y0 is defined just above (20).
Now check condition of 3) of Lemma 9. The proof is illustrated by Fig. 4. Assume that pη|η+ξ=x(y1)=pη|η+ξ=x(y2) and y1<y0<y2. Then F(y1)=F(y2).
Denote
y4=(F′)−1(−F′(y1)).
Then F′(y1)+F′(y4)=F′(y1)−F′(y1)=0, and by (20), as y1<y0, we have F′(y1)>00$]]>, F′(y4)<0, y4>y0>y1y_{0}>y_{1}$]]>. By Lemma 11, F(y1)>F(y4)F(y_{4})$]]>.
To proof of Lemma 10. Checking condition 3) of Lemma 9
Hence, F(y2)=F(y1)>F(y4)F(y_{4})$]]>. Because the function F(y) is decreasing on (y0,+∞) (see (20)), we have y2<y4. Since the function F′(y) is decreasing, F′(y2)>F′(y4)=−F′(y1){F^{\prime }}(y_{4})=-{F^{\prime }}(y_{1})$]]>, which implies F′(y1)+F′(y2)>00$]]>. By (24) we have pη|η+ξ=x′(y1)+pη|η+ξ=x′(y2)>00$]]>.
All the conditions of Lemma 9 are satisfied. By Lemma 9, μ3[η|η+ξ=x]>00$]]>, and by (22)–(23),
L4(x,σ2)L1(x,σ2)2−3L3(x,σ2)L2(x,σ2)L1(x,σ2)+2L2(x,σ2)>00\]]]>
for all x>00$]]> and σ2>00$]]>.
Case 2.x≤0andσ2>00$]]>. The distribution of η+ξ is symmetric. Hence, L1(x,σ2) and L3(x,σ2) are even functions in x, and L2(x,σ2) and L4(x,σ2) are odd functions in x. Then
L4(x,σ2)L1(x,σ2)2−3L3(x,σ2)L2(x,σ2)L1(x,σ2)+2L2(x,σ2)3
is an odd function in x. It is equal to 0 for x=0, and it is negative for x<0 by Case 1; see (25).
Case 3.σ2=0. The function L1(x,0) is the pdf of the logistic distribution, and Lk+1(x,0) is its kth derivative:
L1(x,0)=ex(1+ex)2;L2(x,0)=ex(1−ex)(1+ex)3;L3(x,0)=ex(1+ex)4(1−4ex+e2x);L4(x,0)=ex(1−ex)(1+ex)5(1−10ex+e2x).
Then
L4L12−3L3L2L1+2L23=e3x(1−ex)(1+ex)9(−2ex);sign(L4L12−3L3L2L1+2L23)=sign(x),
where Lk are evaluated at the point (x,0).
Lemma 10 is proven. □
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