Cm Math Evidence — Examples

The formula $h(T,l)=4{\pi }^{2}l{T}^{-2}$ is continuous on $U=\{T>0\}$, and its first partial derivatives in [L3] are continuous there. Thus $h$ is Borel measurable by [L7], and [L7] gives total differentiability at $\mu$ with Jacobian matrix $Dh(\mu )=(-8{\pi }^2{l}_0{T}_0^{-3},4{\pi }^2{T}_0^{-2})$; its entries agree with the partial derivatives of [L3] by [L4].

Differentiating [L3] once more gives $h_{TT}=24{\pi }^{2}l\mathrm{/}{T}^4$, $h_{Tl}=h_{lT}=-8{\pi }^2\mathrm{/}{T}^3$ and $h_{ll}=0$, hence the displayed Hessian by [L6]. On $U_B$, $T>t_{-}>0$ and $\mathrm{\mid }l\mathrm{\mid }<l_{+}$; the sum of squared entries is at most $(24{\pi }^2{l}_{+}\mathrm{/}{t}_{-}^4{)}^2+2(8{\pi }^2\mathrm{/}{t}_{-}^3{)}^2$, so [L6] gives $\mathrm{\parallel }{H}_h\mathrm{\parallel }\le B$. The formulas of [L8] verify continuity of every iterated partial derivative on $U$, as asserted in the Example.

By [L1] and [L2] the propagated covariance is the single entry ${\sum }_{k,l=1}^2{a}_{1k}{a}_{1l}{\mathrm{\Sigma }}_{kl}$ with $a_{11}=-8{\pi }^2{l}_0{T}_0^{-3}$, $a_{12}=4{\pi }^2{T}_0^{-2}$, ${\mathrm{\Sigma }}_{11}={\sigma }_T^2$, ${\mathrm{\Sigma }}_{22}={\sigma }_l^2$ and ${\mathrm{\Sigma }}_{12}={\mathrm{\Sigma }}_{21}={\sigma }_{Tl}$: $a_{11}^2{\sigma }_T^2+2a_{11}{a}_{12}{\sigma }_{Tl}+a_{12}^2{\sigma }_l^2=\frac{64{\pi }^4{l}_0^2}{T_0^6}{\sigma }_T^2-\frac{64{\pi }^4{l}_0}{T_0^5}{\sigma }_{Tl}+\frac{16{\pi }^4}{T_0^4}{\sigma }_l^2\mathrm{.}$

Under the additional box hypothesis, $\mathrm{\mid }X-\mu {\mathrm{\mid }}^2<{\rho }_T^2+{\rho }_l^2$ pointwise, so $M_4\le ({\rho }_T^2+{\rho }_l^2{)}^2<\mathrm{\infty }$. The box is open and convex and contains $\mu$ and all values of $X$, so [L5] with step 1.2 gives the pointwise, mean, mean-square and scalar covariance-error bounds displayed in the Example.

The three coefficients computed in step 2.1 are exactly those displayed in the Example: $a_{11}^2=64{\pi }^4{l}_0^2{T}_0^{-6}$, $2a_{11}{a}_{12}=-64{\pi }^4{l}_0{T}_0^{-5}$ and $a_{12}^2=16{\pi }^4{T}_0^{-4}$.

Steps 2.1 and 3.1 establish the exact propagated covariance of the linearization for the initial square-integrable model; steps 1.2 and 2.2 establish its explicit Hessian correction under the additional box hypothesis.

By [L1] the weights make $(\mathrm{\Omega },w)$ a probability space and the expectations below are the finite weighted sums of Expectation of a real random variable on a finite probability space: $E[Y]=\frac{1}{2}\cdot 0+\frac{1}{4}\cdot 1+\frac{1}{4}\cdot 3=1,E[Y^2]=\frac{1}{4}\cdot 1+\frac{1}{4}\cdot 9=\frac{5}{2},$ $E[Y^4]=\frac{1}{4}\cdot 1+\frac{1}{4}\cdot 81=\frac{82}{4}=\frac{41}{2},E[(Y-1{)}^2]=\frac{1}{2}\cdot 1+\frac{1}{4}\cdot 0+\frac{1}{4}\cdot 4=\frac{3}{2},$ and $E[(Y-1{)}^4]=\frac{1}{2}\cdot 1+\frac{1}{4}\cdot 0+\frac{1}{4}\cdot 16=\frac{1}{2}+4=\frac{9}{2}$.

The function $h(y)=y^2$ is $C^2$ on the open convex set $U=(-1,4)$, with $h^{\mathrm{\prime }}(y)=2y$, so $h(1)=1$ and $h^{\mathrm{\prime }}(1)=2$; its Hessian is the constant matrix $(2)$, of operator norm $B=2$ by [L3] and [L4], and the values $0,1,3$ of $Y$ and the mean $\mu =1$ all lie in $U$.

By [L4], $\mathrm{Var}(Y)=E[Y^2]-E[Y{]}^2=\frac{5}{2}-1=\frac{3}{2}$ and $\mathrm{Var}(h(Y))=E[Y^4]-E[Y^2{]}^2=\frac{41}{2}-\frac{25}{4}=\frac{82-25}{4}=\frac{57}{4}$.

The linearization error is $R=h(Y)-1-2(Y-1)=Y^2-2Y+1=(Y-1{)}^2$.

By [L3] the propagated covariance is $h^{\mathrm{\prime }}(1{)}^2\mathrm{Var}(Y)=4\cdot \frac{3}{2}=6=\frac{24}{4}$, while step 2.1 gives $\mathrm{Var}(h(Y))=\frac{57}{4}$; these differ, so the propagated covariance is not the exact variance of $h(Y)$ in this model.

With $B=2$ and the moments of step 1.1, the bounds of [L2] read $\frac{B}{2}E[(Y-1{)}^2]=1\cdot \frac{3}{2}=\frac{3}{2}$ and $\frac{B^2}{4}E[(Y-1{)}^4]=1\cdot \frac{9}{2}=\frac{9}{2}$; since $R^2=(Y-1{)}^4$ by step 2.2, one has $E[R]=\frac{3}{2}$ and $E[R^2]=\frac{9}{2}$, so both bounds are attained.

The four outcomes have weight $1\mathrm{/}4$ each, so $\mathbb{P}(X_1=-1)=\mathbb{P}(X_1=1)=1\mathrm{/}2$ and the same for $X_2$; for every pair $(a,b)\in \{-1,1{\}}^2$ one has $\mathbb{P}(X_1=a,X_2=b)=1\mathrm{/}4=\mathbb{P}(X_1=a)\mathbb{P}(X_2=b)$. By [L4] the family is independent, the two distributions agree, and $E[X_1]=\frac{1}{2}(-1)+\frac{1}{2}(1)=0$, $E[X_1^2]=\frac{1}{2}(1)+\frac{1}{2}(1)=1$ by [L3].

Enumerating $\bar{X}=(X_1+X_2)\mathrm{/}2$ on the four outcomes gives the values $1,0,0,-1$, each with weight $1\mathrm{/}4$; hence $\mathbb{P}(\mid \bar{X}\mid \ge 1)=\frac{2}{4}=\frac{1}{2},\mathbb{P}(\mid \bar{X}\mid \ge \frac{1}{2})=\frac{2}{4}=\frac{1}{2}\mathrm{.}$

By the enumeration in step 1.2 and the expectation formula [L3], $E[\bar{X}]=0$ and $E\mathrm{\mid }\bar{X}{\mathrm{\mid }}^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}={\sigma }^2\mathrm{/}2$. For any $a>0$, the pointwise inequality $a^2{\mathbf{1}}_{\{\mathrm{\mid }\bar{X}\mathrm{\mid }\ge a\}}\le \mathrm{\mid }\bar{X}{\mathrm{\mid }}^2$ and summation with nonnegative weights give $\mathbb{P}(\mathrm{\mid }\bar{X}\mathrm{\mid }\ge a)\le E\mathrm{\mid }\bar{X}{\mathrm{\mid }}^2\mathrm{/}{a}^2={\sigma }^2\mathrm{/}(2a^2)$.

At $a=1$ the bound of step 2.1 is $1\mathrm{/}2$, equal to the exact probability from step 1.2; at $a=1\mathrm{/}2$ it is $2$, strictly larger than the exact probability $1\mathrm{/}2$. These establish the displayed finite-sample identity and bounds without requiring an infinite IID extension.

On $U=(-1,3)$ the second derivative of $h$ is the constant $2$, so $H_h=(2)$ and $\parallel H_h\parallel =2$ by [L2] and [L3]; $U$ is open and convex by [L4].

On $V=(-1,2)$ the second derivative of $g$ is $6y$, so $H_g(y)=(6y)$ and $\parallel H_g(y)\parallel =\mid 6y\mid \le 12$ for $\mid y\mid <2$; $V$ is open and convex by [L4].

For $h$ with $\mu =1$ and $B=2$, the difference in the Example equals $(y-1{)}^2$ by expansion, and the bound of [L1] is $\frac{2}{2}(y-1{)}^2$, so equality holds for every $y\in U$.

For $g$ with $\mu =1\mathrm{/}2$ and $B=12$, the difference at $y=1$ is $1-\frac{1}{8}-\frac{3}{8}=\frac{1}{2}$, while the bound of [L1] is $\frac{12}{2}\cdot \frac{1}{4}=\frac{3}{2}$, and $\frac{1}{2}<\frac{3}{2}$.

Step 2.1 exhibits equality in the bound, and step 2.2 exhibits strict inequality, so both alternatives occur under the lemma's hypotheses.

By [L1] each $\theta \in \mathrm{\Theta }$ determines a probability measure ${\mathbb{P}}_{\theta }$ on $(S,\mathcal{S})$ with ${\mathbb{P}}_{\theta }(\{s\})=1\mathrm{/}4$ for $s\in D_{\theta }$ and ${\mathbb{P}}_{\theta }(\{s\})=0$ otherwise. Define $C(s):=\{\begin{array}{ll}\{s_1-1\}\cap \mathrm{\Theta },& s_1=s_2,\\ \left\{{\displaystyle \frac{s_1+s_2}{2}}\right\}\cap \mathrm{\Theta },& s_1\mathrm{\ne }{s}_2\mathrm{.}\end{array}$ Each $C(s)$ is a subset of $\mathrm{\Theta }$, and each set $\{\theta \in C\}$ is a subset of $S$ and therefore an element of the full power set $\mathcal{S}=2^S$, so $C$ is a confidence set by [L2].

For fixed $\theta$ the four points of $D_{\theta }$ evaluate under $C$ as follows: $(\theta +1,\theta +1)$ gives $C=\{(\theta +1)-1\}=\{\theta \}$, which contains $\theta$; $(\theta -1,\theta -1)$ gives $C=\{\theta -2\}\cap \mathrm{\Theta }$, which is empty for $\theta =14$ and equals $\{\theta -2\}$ for $\theta =16,18$; in every case it does not contain $\theta$; and the two points with distinct coordinates, $(\theta +1,\theta -1)$ and $(\theta -1,\theta +1)$, both give $C=\{\theta \}$. Hence ${\mathbb{P}}_{\theta }(\theta \in C)=3\mathrm{/}4$ for every $\theta \in \mathrm{\Theta }$, and the coverage level of $C$ is $3\mathrm{/}4$ by [L2].

Let $s^{\ast }:=(15,17)$. Since $s^{\ast }$ has distinct coordinates, step 1.1 gives $C(s^{\ast })=\{(15+17)\mathrm{/}2\}=\{16\}$. Moreover, $s^{\ast }\in D_{\theta }$ holds exactly for $\theta =16$, so ${\mathbb{P}}_{\theta }(\{s^{\ast }\})$ equals $1\mathrm{/}4$ for $\theta =16$ and $0$ for $\theta \in \{14,18\}$.

Fix weights $\pi (14),\pi (16),\pi (18)\ge 0$ summing to $1$ with $\pi (16)>0$, and put $W:=\{(\theta ,s):\theta \in \mathrm{\Theta },s\in D_{\theta }\}$, a twelve-element set, and give $(\theta ,s)\in W$ weight $\pi (\theta )\mathrm{/}4$. These nonnegative weights sum to $1$, so [L3] defines a probability measure $Q$ on the finite space $W$. In events below, $\theta$ and $s$ denote the two coordinate maps on $W$. The event $\{s=s^{\ast }\}$ has $Q$-probability $\pi (16)\cdot \frac{1}{4}>0$, so the conditional probability given $s^{\ast }$ is defined by [L3].

The event $\{\theta \in C(s),s=s^{\ast }\}$ equals $\{\theta =16,s=s^{\ast }\}$, because $C(s^{\ast })=\{16\}$ by step 2.2; its $Q$-probability is $\pi (16)\cdot \frac{1}{4}$, the same as $Q(\{s=s^{\ast }\})$ by step 2.2. Hence $Q(\theta \in C(s)\mid s=s^{\ast })=\frac{\pi (16)\mathrm{/}4}{\pi (16)\mathrm{/}4}=1.$

The confidence set $C$ has level $3\mathrm{/}4$ by step 2.1, while for the prior $\pi$ and the datum $s^{\ast }$ the conditional probability that $\theta \in C(s)$ is $1$ by step 4.1; since $1\mathrm{\ne }3\mathrm{/}4$, the refuted statement is false.