Boundary Points at Infinity and Compactification

Why Add Points at Infinity?

A risk function on a noncompact parameter space need not attain its supremum. A sequence of parameter values can move farther and farther away while its risks approach the worst-case value. Compactification makes this limiting behavior visible by adding boundary points that record how the parameters escape. The boundary is not part of the original statistical model; it is a device for studying limits of its risks.

For Euclidean parameter spaces, the useful choice is usually a radial compactification: distinct directions of escape remain distinct boundary points. That distinction matters whenever a risk has different limiting behavior in different directions.

Radial Compactification of Euclidean Space

Definition: For $d\geq 1$, define $T:\mathbb{R}^d\to B^d$, where $B^d=\{y\in\mathbb{R}^d:\|y\|<1\}$, by $$ T(\theta)=\frac{\theta}{1+\|\theta\|}. $$ The radial compactification identifies $\mathbb{R}^d$ with the open unit ball $B^d$ and uses the closed unit ball $\overline{B}^d$ as its compactification. Its boundary is the unit sphere $S^{d-1}=\{u:\|u\|=1\}$. The inverse map on the open ball is $T^{-1}(y)=y/(1-\|y\|)$.

The closed ball is compact. A sequence $\theta_n$ approaches the boundary point $u\in S^{d-1}$ precisely when its norms tend to infinity and its directions tend to $u$: $ \|\theta_n\|\to\infty$ and $ \theta_n/\|\theta_n\|\to u$. Thus a boundary point represents both unbounded magnitude and a limiting direction.

This differs from the one-point compactification, which adds a single point and treats every way of escaping as the same. For a real parameter, the one-point compactification can describe a continuous extension only if the limits at $+\infty$ and $-\infty$ agree. The radial compactification of $\mathbb{R}$ has two boundary points, one for each direction, so it can represent different limits.

Worked Example: A Supremum Attained Only at Infinity

Consider a decision problem with one available action and loss $L(\theta)=\theta^2/(1+\theta^2)$, so its risk is $R(\theta)=L(\theta)$. Compute its worst-case risk and its radial extension.

For every finite $\theta$, $0\leq \theta^2/(1+\theta^2)<1$, while $$ R(\theta)=1-\frac{1}{1+\theta^2}\longrightarrow 1 \qquad\text{as }|\theta|\to\infty. $$ Therefore $\sup_{\theta\in\mathbb{R}}R(\theta)=1$, and no finite parameter attains the supremum.

Under the radial compactification of $\mathbb{R}$, the boundary points are $+1$ and $-1$. The risk tends to 1 from either direction, so setting $\overline R(+1)=\overline R(-1)=1$ gives a continuous extension. Its boundary points are actual maximizers of the extended risk, even though the original risk has no maximizer on $\mathbb{R}$.

When Does a Risk Extend Continuously?

Fix a decision rule and write its risk as $R(\theta)$. For $\theta\neq 0$, write $\theta=rv$, where $r=\|\theta\|>0$ and $v=\theta/\|\theta\|\in S^{d-1}$. A natural candidate for the extended boundary values is a function $g:S^{d-1}\to\mathbb{R}$, where $g(u)$ is the limiting risk in direction $u$.

Theorem: Suppose $R:\mathbb{R}^d\to\mathbb{R}$ is continuous and $g:S^{d-1}\to\mathbb{R}$ is continuous. Suppose also that $$ \lim_{r\to\infty}\ \sup_{v\in S^{d-1}} |R(rv)-g(v)|=0. $$ Define $\overline R(y)=R(T^{-1}(y))$ for $y\in B^d$ and $\overline R(u)=g(u)$ for $u\in S^{d-1}$. Then $\overline R$ is continuous on $\overline B^d$.

The uniformity over directions is essential: it controls not only sequences on a fixed ray but also sequences whose direction changes as their norm grows.

Proof. Continuity at points in the open ball follows from continuity of $R$ and $T^{-1}$. It remains to check a boundary point $u\in S^{d-1}$. Take any sequence $y_n\in\overline B^d$ with $y_n\to u$. For terms in the open ball, put $\theta_n=T^{-1}(y_n)=r_nv_n$, with $r_n=\|\theta_n\|$ and $v_n=\theta_n/\|\theta_n\|$. Along these interior terms, $r_n\to\infty$ and $v_n\to u$. The assumed uniform condition and continuity of $g$ give $$ |R(\theta_n)-g(u)| \leq \sup_{v\in S^{d-1}}|R(r_nv)-g(v)|+|g(v_n)-g(u)| \longrightarrow 0. $$ For terms $y_n$ on the boundary, $\overline R(y_n)=g(y_n)\to g(u)$ by continuity of $g$. Thus the function values tend to $g(u)$ for both types of terms, including any sequence that contains a mixture of interior and boundary terms. The sequential criterion for continuity applies because $\overline B^d$ is a metric space. Hence $\overline R$ is continuous at $u$, and therefore everywhere. $\square$

Worked Example: Directional Boundary Values in Two Dimensions

Let the risk be $R(\theta_1,\theta_2)=1+\theta_1/\sqrt{1+\|\theta\|^2}$. Find its radial boundary values and the maximum risk on the compactification.

For a unit direction $u=(u_1,u_2)$, set $\theta=ru$. Then $$ R(ru)=1+\frac{ru_1}{\sqrt{1+r^2}} \longrightarrow 1+u_1 \qquad (r\to\infty). $$ The convergence is uniform in $u$, since $$ \sup_{\|u\|=1}\left| \frac{ru_1}{\sqrt{1+r^2}}-u_1 \right| \leq \left| \frac{r}{\sqrt{1+r^2}}-1\right| \longrightarrow 0. $$ Thus $g(u)=1+u_1$, a continuous boundary risk, and the theorem gives a continuous extension.

On the boundary, $u_1\leq 1$, so $g(u)\leq 2$, with equality at $u=(1,0)$. At every finite $\theta$, $\theta_1/\sqrt{1+\|\theta\|^2}<1$, so $R(\theta)<2$. The extended maximum is 2 at the boundary point $(1,0)$, approached, for example, along $\theta=(r,0)$ as $r\to\infty$.

Boundary Maxima and Worst-Case Risk

Theorem: If a risk $R$ has a continuous extension $\overline R$ to $\overline B^d$, then $$ \sup_{\theta\in\mathbb{R}^d}R(\theta) =\max_{y\in\overline B^d}\overline R(y). $$ In particular, if the extended maximum is attained at a boundary point $u$, then there is a sequence of original parameter values escaping in direction $u$ whose risks approach that maximum.

Proof. The closed ball is compact, so continuity of $\overline R$ implies that it has a maximum $M$ there. Since the open ball is a subset of the closed ball, $\sup_{y\in B^d}\overline R(y)\leq M$. Conversely, the open ball is dense in the closed ball. If a maximizer $y^*$ is interior, its value is already included in the supremum over the open ball. If $y^*$ is on the boundary, choose $y_n=(1-1/n)y^*\in B^d$ for $n\geq 2$. Then $y_n\to y^*$, so continuity gives $\overline R(y_n)\to M$. In either case, $\sup_{y\in B^d}\overline R(y)=M$. Since $T$ maps $\mathbb{R}^d$ onto $B^d$, this is the claimed equality.

If $y^*=u$ is a boundary maximizer, the sequence $y_n=(1-1/n)u$ corresponds under $T^{-1}$ to $$ \theta_n=\frac{y_n}{1-\|y_n\|}=(n-1)u. $$ Thus $\|\theta_n\|\to\infty$, $\theta_n/\|\theta_n\|=u$, and $R(\theta_n)=\overline R(y_n)\to M$. This proves the escaping-sequence claim. $\square$

The boundary need not be the only location of a maximizer. If a finite parameter also attains the maximum, then the original worst-case risk is attained, even if the extension has additional boundary maximizers. The boundary interpretation is most informative when no finite parameter realizes the supremum.

Worked Example: Fixed-Ray Limits That Do Not Give Continuity

A pointwise limit along every fixed direction may look reassuring but fail to define a continuous boundary extension. Consider the bounded continuous risk on $\mathbb{R}^2$ $$ R(x,y)=\frac{1}{1+e^{-x}}\exp\!\left[-\left(y-\sqrt{x_+}\right)^2\right], \qquad x_+=\max(x,0). $$ For any fixed unit direction $u=(u_1,u_2)$, examine $R(ru_1,ru_2)$ as $r\to\infty$.

If $u_1<0$, the factor $1/(1+e^{-ru_1})$ tends to zero, while the exponential factor is at most 1. If $u_1=0$, then $x_+=0$ and $|ru_2|=r$, so the exponential tends to zero. If $u_1>0$ and $u_2\neq 0$, the difference $ru_2-\sqrt{ru_1}$ has magnitude tending to infinity, so again the exponential tends to zero. If $u_1>0$ and $u_2=0$, that difference is $-\sqrt{ru_1}$, which also has magnitude tending to infinity. Thus the limit along every fixed ray is zero.

However, along $x=t^2,\ y=t$, the direction $(x,y)/\sqrt{x^2+y^2}$ tends to $(1,0)$, while $$ R(t^2,t)=\frac{1}{1+e^{-t^2}}\longrightarrow 1. $$ A continuous extension with boundary value zero at $(1,0)$ would have to give risks tending to zero along every sequence approaching that point. This sequence contradicts that requirement. The ray limits are pointwise, not uniform over directions, and they do not establish continuity.

Common Pitfall and Practical Use

The most common error is to compute the limit along each fixed ray and then declare the boundary extension continuous. Continuity at a boundary point concerns all sequences approaching it, including sequences whose directions vary. The uniform radial condition in the theorem rules out the moving-direction failure illustrated above. When uniform convergence is inconvenient to prove, the direct alternative is to verify the sequential condition for every sequence approaching every boundary point.

Compactification is useful for interpreting nonattainment: a boundary maximizer identifies a direction in which the worst-case risk is approached. It does not turn the boundary into a genuine parameter value, nor does it by itself prove that any particular risk has a continuous extension. Those conclusions require the relevant limiting conditions to be checked for that risk.

Key takeaway: Radial compactification preserves the direction of escape. A continuous risk extension turns the original supremum into a maximum on the closed ball; a boundary maximizer then describes a limiting worst case, not a parameter value in the original model.

Check Your Understanding

Use the definitions and results above to answer the following questions.

What boundary point does the sequence $\theta_n=(-3n,4n)$ approach in the radial compactification of $\mathbb{R}^2$?
Why must the limits of a risk at $+\infty$ and $-\infty$ agree for a continuous extension to the one-point compactification of $\mathbb{R}$?
For the risk $R(\theta)=\theta^2/(1+\theta^2)$, state its supremum and explain why it is not attained at a finite parameter.
In the continuity theorem, what role does uniformity over directions play?
If the maximum of an extended risk occurs at a boundary point $u$, give an explicit sequence of original parameter values whose risks approach that maximum.