Proof of Chernoff Bound for the Under-Likelihood Set of a Relatively Perturbed Causal Intensity: Elementary Exponential Inequalities, the Tilted Power-Product Exponent, and the Markov Step

Throughout, $\exp$ is given by its series $\exp(x)=\sum_{k\ge0}x^{k}/k!$ (The Real Exponential Function), and we use claims 1, 2 and 4 of Basic Properties of the Exponential Function (functional equation, positivity, monotonicity, and $\exp(u)\ge1+u$ for $u\ge0$). Termwise comparison of convergent series of nonnegative terms, and passage of inequalities to limits, are by claim 1 of Order Properties of Limits of Real Sequences, limits of sums and products by Arithmetic of Limits of Real Sequences, and limits of absolute values by claim 4 of Order Properties of Limits of Real Sequences.

Step 1 (claim 1). (a) $\exp(u)\ge1+u$ for all real $u$. For $u\ge0$ this is claim 4 of Basic Properties of the Exponential Function. For $u\le-1$ it holds since $\exp(u)>0\ge1+u$. For $-1<u<0$ put $v=-u\in(0,1)$; termwise $v^{k}/k!\le v^{k}$, and the partial sums $\sum_{k=0}^{n}v^{k}=(1-v^{n+1})/(1-v)\le1/(1-v)$, so $\exp(v)\le1/(1-v)$, whence $\exp(u)=1/\exp(v)\ge1-v=1+u$.

(b) $|\exp(z)-1-z|\le\tfrac12z^{2}\exp(|z|)$ for all real $z$. As $\exp(z)-1-z=\sum_{k\ge2}z^{k}/k!$ (the partial sums differ from those of $\exp(z)$ by the two initial terms), the absolute value is at most $\sum_{k\ge2}|z|^{k}/k!$ (triangle inequality for the partial sums and passage to the limit); for $k\ge2$, $k!\ge2(k-2)!$, so $|z|^{k}/k!\le\tfrac12z^{2}|z|^{k-2}/(k-2)!$, and comparing partial sums, $\sum_{k\ge2}|z|^{k}/k!\le\tfrac12z^{2}\sum_{j\ge0}|z|^{j}/j!=\tfrac12z^{2}\exp(|z|)$.

(c) The power inequality. Let $x$ be real and $\theta\ge1$ an integer with $2\theta|x|\le1$; then $|x|\le\tfrac12$ and $1+x\ge\tfrac12$, so $\frac{1}{1+x}=1-x+\frac{x^{2}}{1+x}\le1-x+2x^{2}=1+y,\qquad y:=-x+2x^{2},$ where $1+y\ge1-|x|\ge\tfrac12>0$. As $t\mapsto t^{\theta}$ is nondecreasing on $[0,\infty)$ (a product of $\theta$ nondecreasing nonnegative factors) and $1+y\le\exp(y)$ by (a), $(1+x)^{-\theta}=(1/(1+x))^{\theta}\le(1+y)^{\theta}\le\exp(y)^{\theta}=\exp(\theta y)$, the last step by the functional equation. Put $z=\theta y$; then $|z|\le\theta|x|(1+2|x|)\le2\theta|x|\le1$, so by (b) and the monotonicity of $\exp$, $\exp(z)\le1+z+\tfrac12z^{2}\exp(1)$. Now $z=-\theta x+2\theta x^{2}$ and $z^{2}\le\theta^{2}(|x|+2x^{2})^{2}\le\theta^{2}(2|x|)^{2}=4\theta^{2}x^{2}$ (as $2x^{2}\le|x|$), hence $(1+x)^{-\theta}\le1-\theta x+2\theta x^{2}+2\exp(1)\,\theta^{2}x^{2}\le1-\theta x+8\theta^{2}x^{2},$ using $\exp(1)\le3$ (termwise, $1/k!\le2^{-(k-1)}$ for $k\ge1$ since $k!\ge2^{k-1}$, so $\exp(1)=\sum_{k\ge0}1/k!\le1+\sum_{k\ge1}2^{-(k-1)}=3$, the geometric partial sums being at most $2$) and $2\theta+6\theta^{2}\le8\theta^{2}$ for $\theta\ge1$.

Step 2 (claim 2). The lower bound $\mu'^\upsilon_s(r)\ge(1-\varepsilon)\mu^\upsilon_s(r)\ge\tfrac12\underline\mu>0$ of the statement follows from the perturbation hypothesis, so claim 1 of Integer Power Products of Causal-Intensity Likelihoods: Pathwise Identity, Integral Bounds, and the Pair Identity applies to $(\mu,\mu')$ with the exponents $(1+\theta,-\theta)$ (the only negative exponent being that of $\mu'$, bounded below by $\tfrac12\underline\mu$), and claim 3 of that lemma (applied with base $\mu$, and separately with base $\mu'$) gives $\ell>0$ and $\ell'>0$. Fix an integer $\theta\ge1$ with $2\theta\varepsilon\le1$. For $s,r,\upsilon$ put $x=x(s,r,\upsilon)=(\mu'^\upsilon_s(r)-\mu^\upsilon_s(r))/\mu^\upsilon_s(r)$, so that $|x|\le\varepsilon$, $2\theta|x|\le1$ and $\mu'^\upsilon_s(r)=\mu^\upsilon_s(r)(1+x)$. Then $\tilde\mu^\upsilon_{\theta,s}(r)=\mu^\upsilon_s(r)(1+x)^{-\theta}$, and the three integrated total intensities in $E_\theta$ are finite integrals over $[0,T]$ of functions which are measurable in $s$ and bounded (the total intensities of $\mu$, $\mu'$ and $\tilde\mu_\theta$ take values in $[0,\tilde{l}\bar\mu]$, $[0,\tilde{l}\bar\mu']$ and $[0,\tilde{l}\bar{\tilde\mu}_\theta]$ respectively, $\bar{\tilde\mu}_\theta$ being the bound of $\tilde\mu_\theta$ from claim 1 of Integer Power Products of Causal-Intensity Likelihoods: Pathwise Identity, Integral Bounds, and the Pair Identity, and their sections are measurable, by Causal Intensity on the Observation Record Space and Likelihood of a Causal Intensity on the Observation Record Space), so by the linearity of the integral for integrable functions (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) $E_\theta(r)=\int_{[0,T]}\sum_{\upsilon\in V}\mu^\upsilon_s(r)\Bigl((1+x)^{-\theta}-(1+\theta)+\theta(1+x)\Bigr)ds=\int_{[0,T]}\sum_{\upsilon\in V}\mu^\upsilon_s(r)\Bigl((1+x)^{-\theta}-1+\theta x\Bigr)ds .$ By Step 1(c) the bracket is at most $8\theta^{2}x^{2}\le8\theta^{2}\varepsilon^{2}$, and $0\le\mu^\upsilon_s(r)\le\bar\mu$, so by monotonicity the integrand is at most $\tilde{l}\bar\mu\cdot8\theta^{2}\varepsilon^{2}$ and $E_\theta(r)\le8\tilde{l}T\bar\mu\theta^{2}\varepsilon^{2}=\bar{E}_\theta$ for every $r$. Since $E_\theta$ is bounded (claim 1 of Integer Power Products of Causal-Intensity Likelihoods: Pathwise Identity, Integral Bounds, and the Pair Identity), there is a real $\underline{E}$ with $\underline{E}\le E_\theta\le\bar{E}_\theta$ on $\mathbf{R}$, and claim 2 of that lemma gives $\int_{\mathbf{R}}\ell^{1+\theta}\ell'^{-\theta}\,d\rho\le\exp(\bar{E}_\theta)$.

Step 3 (claim 3). Let $0<\delta'<1$ and $\theta$ as in Step 2. Since $\ell,\ell'>0$, $L>0$ on $\mathbf{R}$. On the set $\{L<1-\delta'\}$ one has $(1-\delta')/L>1$, hence $((1-\delta')/L)^{\theta}>1$; off the set the indicator is $0$. Therefore, pointwise on $\mathbf{R}$, $\ell\,\mathbf{1}\{L<1-\delta'\}\le(1-\delta')^{\theta}\,\ell\,L^{-\theta}=(1-\delta')^{\theta}\,\ell^{1+\theta}\ell'^{-\theta} .$ The set $\{L<1-\delta'\}$ is measurable ($L$ is measurable by claim 3 of Integer Power Products of Causal-Intensity Likelihoods: Pathwise Identity, Integral Bounds, and the Pair Identity), so the left side, a product of the measurable $\ell$ with a measurable indicator, is measurable and nonnegative, and the right side is measurable and nonnegative (claim 2 of that lemma); so monotonicity of the integral and the rule $\int cf\,d\rho=c\int f\,d\rho$ for constants $c\ge0$ (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) together with Step 2 give $\int\ell\mathbf{1}\{L<1-\delta'\}\,d\rho\le(1-\delta')^{\theta}\exp(\bar{E}_\theta)$. Finally $0<1-\delta'\le\exp(-\delta')$ by Step 1(a), so $(1-\delta')^{\theta}\le\exp(-\delta')^{\theta}=\exp(-\theta\delta')$ (monotonicity of $t\mapsto t^{\theta}$ on $[0,\infty)$, as in Step 1(c)), and $\exp(-\theta\delta')\exp(\bar{E}_\theta)=\exp(-\theta\delta'+\bar{E}_\theta)$ by the functional equation. $\blacksquare$