Affine invariance conjecture for the Poisson-series function class

The unrestricted condition $a\ne0$ is false: negative slopes do not preserve the Poisson-series class. The corrected positive-slope assertion holds, with a complete convergence criterion.

Let $N$ be a rate-$\lambda$ Poisson process on $[0,\infty)$, with arrival times $S_j$. Consider the smooth bounded function on the whole real line

$$f(x)=\frac1{1+e^x}.$$

Campbell's formula gives

$$\mathbb E\sum_{j\ge1}f(S_j) =\lambda\int_0^\infty\frac{dx}{1+e^x} =\lambda\log2<\infty.$$

Hence

$$\sum_{j\ge1}f(S_j)<\infty \qquad\text{almost surely}.$$

For the permitted affine reflection $x\mapsto-x$, however,

$$f(-S_j)=\frac1{1+e^{-S_j}}\longrightarrow1 \qquad\text{almost surely}.$$

Thus the transformed series diverges, since its terms do not even tend to zero. Therefore invariance under all $a\ne0$ fails.

In the primary source, functions are defined on $[0,\infty)$, so negative slopes are not affine self-maps of the stated domain. The intended assertion for $a>0$ is true. More precisely, for any measurable $f$, put

$$B=\{x:|f(x)|>1\}, \qquad g=f\mathbf1_{\{|f|\le1\}}.$$

Then

$$\sum_{j\ge1}f(S_j)\text{ converges almost surely}$$

if and only if

$$|B|<\infty, \qquad \int_0^\infty|g(x)|^2\,dx<\infty, \qquad \lim_{T\to\infty}\int_0^Tg(x)\,dx \text{ exists}.$$

For sufficiency, the contribution from $B$ has finitely many terms, while

$$M_T=\int_0^Tg(x)(dN_x-\lambda\,dx)$$

is an $L^2$-bounded martingale because

$$\mathbb E|M_T-M_S|^2 =\lambda\int_S^T|g(x)|^2\,dx.$$

Adding the convergent deterministic drift proves convergence.

Conversely, convergence forces finitely many large jumps, hence $|B|<\infty$. After removing them, the independent-increment characteristic function gives

$$\left| \mathbb E e^{i\int_S^Tg\,dN} \right| = \exp\left( -\lambda\int_S^T(1-\cos g(x))\,dx \right) \longrightarrow1.$$

Since $|g|\le1$ and $1-\cos g\ge c|g|^2$, this forces $g\in L^2$. The compensated martingale therefore converges, and subtracting it from the convergent Poisson integral forces convergence of the deterministic drift.

For $a>0$, $b\ge0$, the substitution $u=ax+b$ preserves all three conditions:

$$|\{|f(a\,\cdot+b)|>1\}| =\frac1a|B\cap[b,\infty)|,$$ $$\int_0^\infty|g(ax+b)|^2\,dx =\frac1a\int_b^\infty|g(u)|^2\,du,$$ $$\int_0^Tg(ax+b)\,dx =\frac1a\int_b^{aT+b}g(u)\,du.$$

Thus the source's well-defined positive-orientation conjecture is true, while its unrestricted negative-slope reformulation is false.