Semiparametric Models and Tangent Spaces — Examples

Every allowed score is in $L_0^2(P)$ by [F2]; because $L_0^2(P)$ is closed in $L^2(P)$ by [F7] under [A2], its closed tangent space is contained in $L_0^2(P)$.

Let $M={\sup }_{x\in \mathcal{X}}\mathrm{\mid }s(x)\mathrm{\mid }<\mathrm{\infty }$ for the bounded representative in Given, and choose $0<ϵ<(1+M{)}^{-1}$. For $0\le t<ϵ$, let $a_t=1+ts\mathrm{/}2$, $c_t=P(a_t^2)=1+t^{2}P{s}^2\mathrm{/}4$, and $p_t=a_t^2\mathrm{/}{c}_t$. The bound on $t$ makes $a_t\ge 0$; $c_t>0$ and $P{p}_t=1$, so $p_t$ is a probability density with respect to $P$.

Let $f\in L_0^2(P)$ and choose a measurable representative. Define $u_m=(-m)\vee (f\wedge m)$. Then $\mathrm{\mid }{u}_m-f{\mathrm{\mid }}^2\le f^2{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>m\}}$, whose right side tends pointwise to zero and is dominated by the integrable $f^2$. By [F5], $\mathrm{\parallel }{u}_m-f{\mathrm{\parallel }}_2\to 0$. Put $s_m=u_m-P{u}_m$; it is bounded and $P{s}_m=0$ by [F8] and $P1=1$. Also [F6] and [F8] give $\mathrm{\mid }P{u}_m\mathrm{\mid }=\mathrm{\mid }P(u_m-f)\mathrm{\mid }\le \mathrm{\parallel }{u}_m-f{\mathrm{\parallel }}_2$. Therefore $\mathrm{\parallel }{s}_m-f{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }{u}_m-f{\mathrm{\parallel }}_2\to 0$.

Since $\mathrm{\parallel }{a}_t{\mathrm{\parallel }}_2=\sqrt{c_t}$, $\mathrm{\parallel }\sqrt{p_t}-1-(t\mathrm{/}2)s{\mathrm{\parallel }}_2=\mathrm{\parallel }{a}_t\mathrm{/}\sqrt{c_t}-a_t{\mathrm{\parallel }}_2=\mathrm{\mid }1-\sqrt{c_t}\mathrm{\mid }=O(t^2)=o(t)$. The density family is explicit, so it fits [A1]'s general convention without any selection. Thus $P_t=p_{t}P$ is an allowed DQM path with score $s$, and every bounded mean-zero score lies in the tangent set.

Each $s_m$ is a tangent score by step 2.1, and the sequence converges to the arbitrary $f\in L_0^2(P)$ by step 1.3. Hence the tangent space contains $L_0^2(P)$ by [F4]; step 1.1 gives the reverse inclusion. Thus the tangent space is exactly $L_0^2(P)$.

Every subset of $\{0,1\}$ and its complement lie in the full power set, and every countable union of such subsets is still a subset; thus the power set is a sigma-algebra by [F4], and with this sigma-algebra the sample set is a measurable space by [F4]. Counting measure on this two-point set is a measure of total mass two by [F5, F6], hence sigma-finite by the constant sequence of covers $E_n=\{0,1\}$ in [F6]. Its only null set is empty, so each $P_t$ is absolutely continuous with respect to it. The choice of $\delta$ ensures $p+t>0$ and $1-p-t>0$ throughout $\mathrm{\mid }t\mathrm{\mid }<\delta$; the displayed masses sum to one, so each $P_t$ is a probability measure.

Since $p,1-p>0$, [F7] and the definition of derivative give $\sqrt{p+t}=\sqrt{p}+t\mathrm{/}(2\sqrt{p})+o(t)$ and $\sqrt{1-p-t}=\sqrt{1-p}-t\mathrm{/}(2\sqrt{1-p})+o(t)$. These are exactly the two coordinates of $\sqrt{p}+(t\mathrm{/}2)s\sqrt{p}+o(t)$ for the stated $s$; because counting-measure $L^2$ on two points is the Euclidean norm, the combined remainder is $o(t)$. The density versions are explicit, so the calculation uses [A1]'s convention without a selection. Thus the path is DQM with this score by [F1].

Direct calculation gives $Ps=p(1\mathrm{/}p)+(1-p)(-1\mathrm{/}(1-p))=0$ and $P{s}^2=p\mathrm{/}{p}^2+(1-p)\mathrm{/}(1-p{)}^2=1\mathrm{/}[p(1-p)]<\mathrm{\infty }$. Hence the score is in $L_0^2(P)$ and has the stated variance.

Every subset of $\{-1,+1\}$ and its complement lie in the full power set, and every countable union of such subsets is still a subset; thus the power set is a sigma-algebra by [F4], and with this sigma-algebra the sample set is a measurable space by [F4]. Counting measure on the two-point set is a measure of total mass two by [F5, F6], hence sigma-finite by the constant sequence of covers $E_n=\{-1,+1\}$; its only null set is empty. For every $\theta \in [0,1]$, the masses $(1+\theta )\mathrm{/}2$ and $(1-\theta )\mathrm{/}2$ are nonnegative and sum to one, so $P_{\theta }$ is a probability measure dominated by counting measure. At $\theta =0$ both masses are $1\mathrm{/}2$, so $\sqrt{p_0(x)}=1\mathrm{/}\sqrt{2}$.

If a one-sided DQM path has score $s$, [F1] on each atom gives $\sqrt{P_{\theta (t)}\{x\}}=1\mathrm{/}\sqrt{2}+ts(x)\mathrm{/}(2\sqrt{2})+o(t)$. Squaring each expansion yields $P_{\theta (t)}\{x\}=1\mathrm{/}2+ts(x)\mathrm{/}2+o(t)$.

Conversely, for each $a\ge 0$, take $\theta (t)=at$ for sufficiently small $t\ge 0$. Then $P_{\theta (t)}\{x\}=(1+atx)\mathrm{/}2$, and [F7] at $1\mathrm{/}2$ gives $\sqrt{P_{\theta (t)}\{x\}}=1\mathrm{/}\sqrt{2}+tax\mathrm{/}(2\sqrt{2})+o(t)$ on each atom. The two-coordinate counting-measure $L^2$ remainder is $o(t)$, so this path has score $aX$. Its densities are explicit, so [A1] requires no selection in this construction. Thus every vector on the nonnegative ray is an allowed score.

Subtract the two mass expansions from step 2.1. Since $\theta (t)=P_{\theta (t)}\{+1\}-P_{\theta (t)}\{-1\}$, we get $\theta (t)\mathrm{/}t\to (s(+1)-s(-1))\mathrm{/}2$. By [F2], $P_{0}s=(s(+1)+s(-1))\mathrm{/}2=0$, so this limit equals $s(+1)$. The constraints $\theta (t)\ge 0$ and $t>0$ force the limit $a:=s(+1)$ to be nonnegative, and $s(-1)=-a$; hence every score is $s(x)=ax$ with $a\ge 0$.

The real span of $\{aX:a\ge 0\}$ is $\{bX:b\in \mathbb{R}\}$, and every mean-zero function on the equiprobable two-point space has values $(b,-b)$, so this span is $L_0^2(P_0)$. This space is closed in $L^2(P_0)$ by [F8], hence its closure—and therefore the tangent space by [F3]—is $L_0^2(P_0)$. It strictly contains the non-linear score cone because $X$ is in the cone but $-X$ is not.