Continuous and strictly increasing norm distributions for density bridges

In either $E=M_0(w)$ or $E=C([0,1])$, the Borel sigma-algebra is generated by countably many continuous linear coordinates: the wavelet coordinates in the first case, and evaluations at rational times in the second. Indeed, the norm distance to a fixed point is their countable supremum, with the weights included in the first case; in the second case continuity makes the rational supremum equal the full supremum. Separability then gives a countable base of coordinate-measurable balls. Separability of $M_0(w)$ is [F1]; rational polygonal functions give a countable dense family in $C([0,1])$ by uniform continuity and finite rational approximation. Thus finite-coordinate laws determine Borel laws by [F4]. If $X$ is any centered Borel Gaussian random element in either space and $X^{\mathrm{\prime }}$ an independent copy, [F3] gives and : every finite coordinate vector on either side is centered Gaussian with the same covariance.

For the law $\mu$ of such an $X$, let $S$ be its support, the set of points whose every open neighborhood has positive probability. Its complement is the union of all zero-mass balls in a countable base, hence has measure zero by [F5]; $S$ is closed and nonempty. Symmetry from step 1.1 gives $S=-S$. The support of $X+X^{\mathrm{\prime }}$ is $\overline{S+S}$: independent positive-mass balls around $s,t\in S$ give positive mass to every neighborhood of $s+t$, proving one inclusion; the other follows because $X,X^{\mathrm{\prime }}\in S$ almost surely, so their sum lies almost surely in the closed set $\overline{S+S}$. Scaling by $\sqrt{2}$ is a homeomorphism, so step 1.1 gives $\sqrt{2}S=\overline{S+S}$. Taking any $s\in S$ and $-s\in S$ shows $0\in \sqrt{2}S$, hence $0\in S$. Now $x+0\in \overline{S+S}=\sqrt{2}S$ gives $S\subseteq \sqrt{2}S$, while $x+x\in \sqrt{2}S$ gives $\sqrt{2}x\in S$ and hence $\sqrt{2}S\subseteq S$. Therefore $\sqrt{2}S=S$ and $\overline{S+S}=S$. The set $S$ is additive and symmetric, and division by $2$ follows by applying division by $\sqrt{2}$ twice. It is closed under dyadic rational scalars, and closure gives all real scalars. Thus $S$ is a closed linear subspace.

Put $Z=B_1$ and $M={\max }_{0\le t\le 1}{\beta }_t$. Every finite vector of bridge values is jointly Gaussian with $Z$ and has zero cross-covariance, because $\mathrm{Cov}(B_t-t{B}_1,B_1)=t-t=0$. By [F8] these finite vectors are independent of $Z$. The coordinate generation in step 1.1 and finite-measure uniqueness [F4], applied to rectangles and then the generated sigma-algebra, show that the entire bridge path, and hence $M$, is independent of $Z$. Write $\varphi$ for the standard-normal density. For $c>\epsilon >0$, [F9] implies that the measure $P({\max }_{t\le 1}{B}_t<c,Z\in dz)$ on $[-\epsilon ,\epsilon ]$ has density $\varphi (z)-\varphi (2c-z)$: subtract the reflected-event measure from the terminal normal measure; the reflected-event CDF in [F9] differentiates to $\varphi (2c-z)$ for $z<c$. Thus $A_{\epsilon }(c):=\frac{P({\max }_{t\le 1}{B}_t<c,\mathrm{\mid }Z\mathrm{\mid }\le \epsilon )}{P(\mathrm{\mid }Z\mathrm{\mid }\le \epsilon )}=\frac{{\int }_{-\epsilon }^{\epsilon }[\varphi (z)-\varphi (2c-z)]dz}{{\int }_{-\epsilon }^{\epsilon }\varphi (z)dz}\mathrm{.}$ For fixed $a>0$, take $0<\epsilon <a\mathrm{/}3$. Since $\mathrm{\parallel }B-\beta {\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon$ on $\{\mathrm{\mid }Z\mathrm{\mid }\le \epsilon \}$, the inclusions with thresholds $a-\epsilon$ and $a+\epsilon$, together with independence, give $A_{\epsilon }(a-\epsilon )\le P(M<a)\le A_{\epsilon }(a+\epsilon )$. Continuity and positivity of $\varphi$ show that both bounds tend to $1-\varphi (2a)\mathrm{/}\varphi (0)=1-e^{-2a^2}$. Hence $P(M<a)=1-e^{-2a^2}$ for every $a>0$. By [F5], $P(M\le a)={\lim }_{k\to \mathrm{\infty }}P(M<a+1\mathrm{/}k)=1-e^{-2a^2}$ for $a>0$, and $M\ge 0$ together with the same formula as $a\downarrow 0$ gives $P(M=0)=0$. Therefore $M$ is atomless.

If $X$ has any nondegenerate scalar coordinate, then $P(X=0)=0$ and $S\mathrm{\ne }\{0\}$. For every $r>0$, $0\in S$ gives $P(\mathrm{\parallel }X\mathrm{\parallel }<r)>0$. A nonzero $s\in S$ can be scaled to have norm greater than $r$, so some neighborhood of that scaled point lies in $\{\mathrm{\parallel }X\mathrm{\parallel }>r\}$ and has positive probability. Likewise, for $0<a<b$, scale $s$ to have norm $(a+b)\mathrm{/}2$; a sufficiently small ball about it is contained in $\{a<\mathrm{\parallel }X\mathrm{\parallel }<b\}$ and has positive probability. Consequently its norm CDF is strictly between zero and one at every positive argument and strictly increasing there. This applies to $G$: any nonconstant wavelet has positive $P_0$-variance, because $f_0\ge \rho$ makes the $P_0$ and Lebesgue null sets identical and a nonconstant wavelet cannot equal a constant almost everywhere.

For each nonzero $x\in M_0(w)$, its norm is attained by some coordinate: if $r=\mathrm{\parallel }x{\mathrm{\parallel }}_{M(w)}>0$, the normalized level maxima are below $r\mathrm{/}2$ after finitely many levels, and the maximum over the remaining finite set equals $r$. Enumerate all coordinates by $j$, and write $b_j>0$ for their norm weights: $b_j=w_l$ on detail level $l$, and $b_j=1$ for any additional Haar scaling coordinate. Hence, for $r>0$, $\{\mathrm{\parallel }G{\mathrm{\parallel }}_{M(w)}=r\}$ is contained in the countable union of events $\{\mathrm{\mid }{G}_j\mathrm{\mid }=r{b}_j\}$. Each has probability zero: a positive-variance normal has a density, and a zero-variance centered normal is zero whereas $r{b}_j>0$. By [F5] the union is null. At $r=0$, the nondegenerate coordinate from step 3.1 again gives probability zero; negative values are impossible. Absence of atoms and continuity of probability in [F5] show that this norm CDF is continuous on $\mathbb{R}$.

First verify the bridge used in [F7] directly. Every finite linear combination of ${\beta }_{t_j}=B_{t_j}-t_j{B}_1$ is a finite linear combination of Brownian values with time $1$ appended, so it is normal by [F7], including repeated times and singular combinations. Finite expectation linearity makes its mean zero. Covariance bilinearity gives $\mathrm{Cov}({\beta }_s,{\beta }_t)=\min (s,t)-ts-st+st=\min (s,t)-st$ for $s,t\in [0,1]$. On the single continuous zero-start Brownian path event supplied in [F7], subtraction of the continuous function $t{B}_1$ makes $\beta$ continuous, with ${\beta }_0={\beta }_1=0$. Thus it is the needed centered continuous Gaussian bridge, without an example as a logical supplier. Write $Y=JG$. For every continuous real linear functional $L$ on $C([0,1])$, $L\circ J$ is continuous linear on $M_0(w)$, so $L(Y)$ is centered normal by [F3]; thus $Y$ is centered Gaussian. For $s\in [0,1]$, let $p_N{1}_{[0,s]}$ be the orthogonal wavelet projection through level $N$. Its Lebesgue $L^2$ limit is $1_{[0,s]}$ by [F2], and $f_0\le D$ gives the same convergence in $L^2(P_0)$. Put $U_N(s)=J(P_{N}G)(s)$. The coefficient covariance in [F3] gives $\mathrm{Cov}(U_N(s),U_M(t))=P_0[(p_N{1}_{[0,s]}-P_0{p}_N{1}_{[0,s]})(p_M{1}_{[0,t]}-P_0{p}_M{1}_{[0,t]})]$. In particular $U_N(s)$ is Cauchy in $L^2$, and it converges almost surely to $Y(s)$ by [F1]. Its $L^2$ limit therefore equals $Y(s)$ almost surely, since convergence in $L^2$ implies convergence in probability and two probability limits agree. The means are zero; applying [F6] to the product differences passes the covariances to the limit. Thus $Y$ is a centered continuous Gaussian process with covariance $F_0(s\wedge t)-F_0(s)F_0(t)$. The function $F_0$ is a continuous strictly increasing bijection of $[0,1]$ onto itself. By [F3], $Y$ and $t\mapsto {\beta }_{F_0(t)}$ from [F7] have the same finite Gaussian laws; step 1.1 then gives equality of their Borel laws on $C([0,1])$. In particular , and $Y$ has a nondegenerate evaluation coordinate, for example at any $s$ with $0<F_0(s)<1$. Step 3.1 therefore applies to its norm as well.

By symmetry of the centered Gaussian bridge, $-{\min }_t{\beta }_t$ has the same atomless distribution as $M$. For $r\ge 0$, continuity on a compact interval gives $\{\mathrm{\parallel }\beta {\mathrm{\parallel }}_{\mathrm{\infty }}=r\}\subseteq \{M=r\}\cup \{-{\min }_t{\beta }_t=r\}$. Both events on the right are null by step 2.2 and symmetry; [F5] gives no atom at $r$. Negative norm values are impossible. By step 4.2, the CDF of $\mathrm{\parallel }JG{\mathrm{\parallel }}_{\mathrm{\infty }}$ is therefore continuous on $\mathbb{R}$.

Both norm CDFs are continuous by steps 4.1 and 5.1, vanish at zero, and are strictly increasing on $(0,\mathrm{\infty })$ with values strictly between zero and one by steps 3.1 and 4.2. They tend to one at infinity because both random norms are finite almost surely. For each $u\in (0,1)$, continuity therefore gives a positive finite point $q$ with CDF value $u$, strict increase gives its uniqueness, and monotonicity makes it the lower $u$-quantile. Any sufficiently small open interval around $q$ is contained in $(0,\mathrm{\infty })$, where strict increase holds. These are all the asserted quantile conditions.