Laplace approximation and deterministic quadrature — Examples

The derivative is $h^{\mathrm{\prime }}(\theta )=-(\theta -\mu )$, so the unique maximizer is $\widehat{\theta }=\mu$; the second derivative is $h^{\mathrm{\prime }\mathrm{\prime }}(\mu )=-1$ and $h(\mu )=c$.

Use $\theta =\varphi (u):=\mu +\sqrt{2\mathrm{/}n}u$. Since $n\ge 1$, this is an increasing affine surjection of $\mathbb{R}$, with constant positive derivative $\sqrt{2\mathrm{/}n}$. The original integrand and ${\varphi }^{\mathrm{\prime }}$ are continuous, hence locally Riemann integrable by [F3], and on each compact truncation the map is differentiable through both endpoints, with integrable derivative and bounded continuous integrand, so the proper substitution hypotheses hold. By [F1], the transformed Gaussian converges at both infinite ends; applying [F2] separately at these ends proves convergence of $J_n$ and gives $J_n=e^{nc}\sqrt{2\mathrm{/}n}{\int }_{\mathbb{R}}{e}^{-u^2}du=e^{nc}\sqrt{2\pi \mathrm{/}n}$.

Substituting $h(\mu )=c$ and $-h^{\mathrm{\prime }\mathrm{\prime }}(\mu )=1$ into the one-dimensional quadratic expression gives $e^{nc}\sqrt{2\pi \mathrm{/}n}$, which equals step 1.2 for every $n\ge 1$.

The phase is $C^2$, satisfies $h(0)=0$, $h(x)<0$ for every $x\in (0,1]$, and $h^{\mathrm{\prime }\mathrm{\prime }}(0)=-2$. Thus its unique maximizer is the boundary point $0$, and the refuted formula predicts $\sqrt{\pi \mathrm{/}n}$.

For $x\in [0,1]$, $h(x)=-x-x^2\le -x$, so [F2] and [F4] give $0\le I_n\le {\int }_0^1{e}^{-nx}dx$. The antiderivative $G(x)=-e^{-nx}\mathrm{/}n$ has derivative $e^{-nx}$ by [F6]; [F3, F5] evaluate the latter integral as $(1-e^{-n})\mathrm{/}n\le 1\mathrm{/}n$.

On $[0,1\mathrm{/}n]$ with $n\ge 1$, $n(x+x^2)\le 1+1\mathrm{/}n\le 2$. Thus $e^{nh(x)}\ge e^{-2}$ there and [F3, F4] give $I_n\ge e^{-2}\mathrm{/}n$. Together with step 1.2, this proves $e^{-2}\mathrm{/}n\le I_n\le 1\mathrm{/}n$, so the boundary contribution has order $n^{-1}$.

Consequently $0\le I_n\mathrm{/}\sqrt{\pi \mathrm{/}n}\le 1\mathrm{/}\sqrt{\pi n}\to 0$. The ratio required by the refuted asymptotic formula therefore tends to $0$, not $1$.

Extend $h(x)=-x^4$ to $\mathbb{R}$. Repeated application of [F6] gives $h^{\mathrm{\prime }}(x)=-4x^3$, $h^{\mathrm{\prime }\mathrm{\prime }}(x)=-12x^2$, $h^{(3)}(x)=-24x$, $h^{(4)}(x)=-24$, and $h^{(j)}(x)=0$ for every $j\ge 5$. These derivatives and $h$ are continuous by [F7], so $h$ is $C^{\mathrm{\infty }}$ by [F1]. Since $x^4>0$ for $x\mathrm{\ne }0$, its unique maximum on $[-1,1]$ is $h(0)=0$ at the interior point $0$, with $h^{\mathrm{\prime }}(0)=h^{\mathrm{\prime }\mathrm{\prime }}(0)=0$. For each integer $m\ge 1$, the interval $[-1\mathrm{/}m,1\mathrm{/}m]$ lies in $[-1,1]$.

If $\mathrm{\mid }x\mathrm{\mid }\le 1\mathrm{/}m$, then $m^4{x}^4\le 1$, so [F2] gives $e^{-m^4{x}^4}\ge e^{-1}$. The polynomial $x\mapsto -m^4{x}^4$ is continuous by [F7], so its exponential is continuous by [F2] and [F8], and positive by [F2], hence integrable on each nondegenerate subinterval by [F3]. Split its integral at $-1\mathrm{/}m$ and $1\mathrm{/}m$ using [F5]. The two outer contributions are nonnegative by [F4] (and zero when $m=1$), while the central contribution is at least $2e^{-1}\mathrm{/}m$ by [F4]. Thus $I_{m^4}\ge 2e^{-1}\mathrm{/}m$. The Gaussian scale at $n=m^4$ is $n^{-1\mathrm{/}2}=m^{-2}$; therefore $I_{m^4}\mathrm{/}(m^4{)}^{-1\mathrm{/}2}\ge 2e^{-1}m⟶\mathrm{\infty }$. Hence these integrals are not $O(n^{-1\mathrm{/}2})$, which refutes the statement.

Factoring gives $h(x)=-(x-1{)}^2(x+1{)}^2\le 0$, with equality exactly at $x=-1$ and $x=1$. Direct differentiation gives $h^{\mathrm{\prime }}(x)=-4x^3+4x$ and $h^{\mathrm{\prime }\mathrm{\prime }}(x)=4-12x^2$, so each mode is interior, has height $0$, and has negative Hessian $h^{\mathrm{\prime }\mathrm{\prime }}(\pm 1)=-8$ by [F5].

On $K_{-}=[-2,0]$, the unique maximizer is $-1$. If $\mathrm{\mid }x+1\mathrm{\mid }\le 1\mathrm{/}2$, then $\mathrm{\mid }x-1\mathrm{\mid }\ge 1$ and $h(x)\le -(x+1{)}^2$; if $x\in K_{-}\setminus B(-1,1\mathrm{/}2)$, then $\mathrm{\mid }x+1\mathrm{\mid }\ge 1\mathrm{/}2$ and $\mathrm{\mid }x-1\mathrm{\mid }\ge 1$, so $h(x)\le -1\mathrm{/}4$. Thus the Laplace theorem's local quadratic and strict global gap conditions hold on $K_{-}$.

On $K_{+}=[0,2]$, the unique maximizer is $1$. If $\mathrm{\mid }x-1\mathrm{\mid }\le 1\mathrm{/}2$, then $\mathrm{\mid }x+1\mathrm{\mid }\ge 1$ and $h(x)\le -(x-1{)}^2$; if $x\in K_{+}\setminus B(1,1\mathrm{/}2)$, then $\mathrm{\mid }x-1\mathrm{\mid }\ge 1\mathrm{/}2$ and $\mathrm{\mid }x+1\mathrm{\mid }\ge 1$, so $h(x)\le -1\mathrm{/}4$. Both intervals are compact of positive length, the amplitude $a\equiv 1$ is integrable and positive at the mode, and the polynomial $h$ is $C^3$ by [F6].

Apply [F1] separately on $K_{-}$ and $K_{+}$. If $J_n^{-}:={\int }_{K_{-}}{e}^{nh}d{\lambda }_1$ and $J_n^{+}:={\int }_{K_{+}}{e}^{nh}d{\lambda }_1$, then $J_n^{-}\sim \sqrt{\pi }\mathrm{/}(2\sqrt{n})$ and $J_n^{+}\sim \sqrt{\pi }\mathrm{/}(2\sqrt{n})$, because each mode has height $0$ and $-h^{\mathrm{\prime }\mathrm{\prime }}=8$.

By [F2] and [F4], the Riemann and Lebesgue integrals on each half agree; by [F3], $I_n=J_n^{-}+J_n^{+}$. Therefore $I_n\sim \sqrt{\pi }\mathrm{/}\sqrt{n}$: dividing by $\sqrt{\pi }\mathrm{/}\sqrt{n}$ gives half the sum of two ratios tending to $1$.

A one-mode term is $\sqrt{\pi }\mathrm{/}(2\sqrt{n})$, while step 3.1 gives $I_n\mathrm{/}(\sqrt{\pi }\mathrm{/}(2\sqrt{n}))\to 2$. This ratio is not $1$, so the statement refuted fails.

The function is $C^2$, with $f^{\mathrm{\prime }\mathrm{\prime }}(x)=2$ and $\mathrm{\parallel }{f}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}=2$. By [F1], $T_1(f)=(f(0)+f(1))\mathrm{/}2=1\mathrm{/}2$.

By [F4], $f$ is integrable. The antiderivative $x^3\mathrm{/}3$ and [F5] give ${\int }_0^1{x}^{2}dx=1\mathrm{/}3$.

Therefore $\mathrm{\mid }{\int }_0^{1}f-T_1(f)\mathrm{\mid }=\mathrm{\mid }1\mathrm{/}3-1\mathrm{/}2\mathrm{\mid }=1\mathrm{/}6$, while the bound in [F2] is $1\cdot 1^2\cdot 2\mathrm{/}12=1\mathrm{/}6$. Equality holds, so this example attains the stated error bound.