Wavelet Multiresolution and Coefficient Decomposition

Why Multiresolution Decomposes a Regression Function

A series representation describes a function by its coefficients in a chosen collection of basis functions. Wavelets organize those coefficients by both location and resolution: coarse-scale terms capture broad averages, while finer-scale terms describe changes within smaller regions. This organization is useful in nonparametric regression because a regression function may be smooth in some regions and change rapidly in others.

Here the domain is $[0,1]$, and the target regression function $m(x)=E[Y\mid X=x]$ is assumed to belong to $L^2([0,1])$, meaning $\int_0^1 m(x)^2\,dx<\infty$. We use the Haar system to make the definitions and calculations explicit. It gives an exact orthonormal multiresolution decomposition without requiring boundary conventions for a more general wavelet family.

Scaling Functions, Wavelets, and Resolution Levels

A scaling function generates a family of functions used to represent a signal at a given resolution. A wavelet generates the corresponding detail functions. In the usual whole-line notation, a mother scaling function $\phi$ and mother wavelet $\psi$ generate translated and dilated functions

\phi_{j,k}(x)=2^{j/2}\phi(2^jx-k), \qquad \psi_{j,k}(x)=2^{j/2}\psi(2^jx-k),

where $j$ is the resolution level and $k$ is the location index. The factor $2^{j/2}$ normalizes the $L^2$ norm under dilation: if $\|\phi\|_2=1$, then $\|\phi_{j,k}\|_2=1$, and likewise for $\psi$. For an orthonormal multiresolution analysis, the scaling functions at level $j$ form an orthonormal basis for a space $V_j$. These spaces are nested, $V_j\subset V_{j+1}$. The detail space $W_j$ is the part of $V_{j+1}$ orthogonal to $V_j$, so $V_{j+1}=V_j\oplus W_j$.

For the Haar system on $[0,1]$, define the scaling function and wavelet by

\phi(x)=\mathbf{1}_{[0,1)}(x), \qquad \psi(x)= \begin{cases} 1, & 0\leq x<1/2,\\ -1, & 1/2\leq x<1,\\ 0, & \text{otherwise}. \end{cases}

For integer $j\geq 0$ and $k=0,\ldots,2^j-1$, set $\phi_{j,k}(x)=2^{j/2}\phi(2^jx-k)$ and $\psi_{j,k}(x)=2^{j/2}\psi(2^jx-k)$. At level $j$, each scaling function is constant on one dyadic interval $[k2^{-j},(k+1)2^{-j})$. The space $V_j$ consists of functions constant on each of the $2^j$ such intervals. The wavelet $\psi_{j,k}$ compares the left and right halves of the interval indexed by $k$.

Definition: For $f\in L^2([0,1])$, its Haar scaling coefficient at level $j$ and location $k$ is $\alpha_{j,k}=\int_0^1 f(x)\phi_{j,k}(x)\,dx$. Its wavelet detail coefficient is $\beta_{j,k}=\int_0^1 f(x)\psi_{j,k}(x)\,dx$. The projection onto $V_j$ is $P_j f=\sum_{k=0}^{2^j-1}\alpha_{j,k}\phi_{j,k}$. The coefficients describe the function’s components in an orthonormal system; they are not, by themselves, empirical estimates from a regression sample.

How One Resolution Splits Into the Next

For Haar functions, each interval at level $j$ divides into two intervals at level $j+1$. The two scaling functions on those child intervals can be replaced by one parent scaling function and one parent wavelet. In symbols, for each $k$,

\phi_{j,k} =\frac{\phi_{j+1,2k}+\phi_{j+1,2k+1}}{\sqrt{2}}, \qquad \psi_{j,k} =\frac{\phi_{j+1,2k}-\phi_{j+1,2k+1}}{\sqrt{2}}.

The first combination has the same sign on both children, so it represents their shared average. The second has opposite signs, so it represents their contrast. These identities establish that the space at the finer level contains both the coarser representation and precisely the new detail.

Theorem: For every integer $j\geq0$, the Haar spaces satisfy $V_{j+1}=V_j\oplus W_j$, where $W_j$ is spanned by $\{\psi_{j,k}:0\leq k<2^j\}$. Consequently, for every $f\in L^2([0,1])$, $$ P_{j+1}f=P_jf+\sum_{k=0}^{2^j-1}\beta_{j,k}\psi_{j,k}. $$

Proof. On the two child intervals of a fixed parent, $\phi_{j+1,2k}$ and $\phi_{j+1,2k+1}$ are orthonormal indicator functions with the $L^2$-normalization $2^{(j+1)/2}$. The two displayed combinations are also orthonormal: each has squared norm $1$, and their inner product is $0$. They span the same two-dimensional space as the child functions because the transformation between the two pairs is invertible. The first combination is $\phi_{j,k}$, and the second is $\psi_{j,k}$. Repeating this on each parent interval gives an orthogonal decomposition of $V_{j+1}$ into $V_j$ and the span of all level-$j$ wavelets. Projecting $f$ onto this orthogonal sum gives its projection onto $V_j$ plus its wavelet coefficients at level $j$, proving the formula. $\square$

Worked Example: A Single Step and Its Coefficients

Let $f(x)=1$ on $[0,1/2)$ and $f(x)=0$ on $[1/2,1)$. At level $0$, there is one scaling function, $\phi_{0,0}=1$, and one wavelet, which is $1$ on the left half and $-1$ on the right half.

The scaling coefficient is $\alpha_{0,0}=\int_0^{1/2}1\,dx=1/2$. The detail coefficient is $\beta_{0,0}=\int_0^{1/2}1\,dx-\int_{1/2}^1 0\,dx=1/2$. Thus the level-zero projection is $(1/2)\phi_{0,0}$, equal to $1/2$ on both halves. Adding the detail gives $(1/2)\phi_{0,0}+(1/2)\psi_{0,0}$, which equals $1$ on the left and $0$ on the right, exactly $f$.

The parent average is $(1+0)/2=1/2$; the contrast is $1-0=1$, and the normalized Haar detail coefficient is half that contrast, $1/2$. These are distinct quantities. In particular, the average of child values $1$ and $0$ is not $1/4$.

The Full Multiresolution Expansion

Iterating the one-level decomposition gives, for any integer $J>0$,

P_J f=P_0 f+\sum_{j=0}^{J-1}\sum_{k=0}^{2^j-1}\beta_{j,k}\psi_{j,k}.

This is an exact finite-level identity: the right side is the projection of $f$ onto $V_J$. To obtain an infinite expansion, the projections must approach $f$ in $L^2$. For the Haar system this follows because the union of the dyadic step-function spaces is dense in $L^2([0,1])$. One way to see the density is that continuous functions are dense in $L^2([0,1])$, and uniform continuity lets a continuous function be approximated uniformly by a function constant on sufficiently short dyadic intervals. Since $P_J$ is the orthogonal projection onto $V_J$, it gives at least as accurate an $L^2$ approximation as any element of $V_J$. Hence $\|P_Jf-f\|_2\to0$.

Theorem: Every $f\in L^2([0,1])$ has the Haar expansion $$ f=\alpha_{0,0}\phi_{0,0} +\sum_{j=0}^{\infty}\sum_{k=0}^{2^j-1}\beta_{j,k}\psi_{j,k}, $$ with convergence in $L^2([0,1])$. Moreover, $$ \|f\|_2^2=|\alpha_{0,0}|^2+\sum_{j=0}^{\infty}\sum_{k=0}^{2^j-1}|\beta_{j,k}|^2. $$

Proof. The finite decomposition above shows that $P_Jf$ is the partial expansion containing the scaling term and all detail levels $0,\ldots,J-1$. Density of the dyadic step functions gives $\|P_Jf-f\|_2\to0$, so these partial expansions converge to $f$ in $L^2$. For each finite $J$, all included Haar functions are orthonormal, and therefore $\|P_Jf\|_2^2=|\alpha_{0,0}|^2+\sum_{j=0}^{J-1}\sum_k|\beta_{j,k}|^2$. Taking $J\to\infty$, the left side tends to $\|f\|_2^2$, because norm convergence implies convergence of norms. The finite sums on the right increase to the infinite sum of nonnegative squared coefficients. This proves the energy identity. $\square$

Worked Example: Details for a Function Constant on Quarter-Intervals

Let $f$ equal $2$ on $[0,1/2)$ and $0$ on $[1/2,1)$. At level zero, $\alpha_{0,0}=\int_0^1 f(x)\,dx=2(1/2)=1$, and $\beta_{0,0}=\int_0^{1/2}2\,dx-\int_{1/2}^1 0\,dx=1$. Therefore $P_1f=\phi_{0,0}+\psi_{0,0}$: its value is $1+1=2$ on the left half and $1-1=0$ on the right half.

At level one, the two scaling coefficients are $\alpha_{1,0}=2^{1/2}\int_0^{1/2}2\,dx=\sqrt{2}$ and $\alpha_{1,1}=2^{1/2}\int_{1/2}^1 0\,dx=0$. The detail coefficients are $\beta_{1,0}=0$, because the function is constant on the first half, and $\beta_{1,1}=0$, because it is constant on the second half. There is no further detail: $P_1f=f$. The parent average is $1$, the child averages are $2$ and $0$, and their average is $(2+0)/2=1$, consistent with the level-zero scaling coefficient.

Worked Example: Wavelet Coefficients of a Linear Function

Take $f(x)=x$. Its level-zero scaling coefficient is $\alpha_{0,0}=\int_0^1x\,dx=1/2$. The level-zero detail coefficient is $\beta_{0,0}=\int_0^{1/2}x\,dx-\int_{1/2}^1x\,dx=1/8-3/8=-1/4$. Thus the coarsest approximation equals $1/4$ on the left half and $3/4$ on the right half.

For any $j\geq0$, write $h=2^{-j}$ and $a=k2^{-j}$. The wavelet is $2^{j/2}$ on $[a,a+h/2)$, and $-2^{j/2}$ on $[a+h/2,a+h)$. Direct integration gives $\int_a^{a+h/2}x\,dx-\int_{a+h/2}^{a+h}x\,dx=-h^2/4$. Therefore $\beta_{j,k}=-2^{j/2}h^2/4=-2^{-3j/2}/4$, for every $k$ at that level.

As a check, there are $2^j$ coefficients at level $j$, so their total squared energy is $2^j(2^{-3j/2}/4)^2=2^{-2j}/16$. Summing over levels gives $(1/16)\sum_{j=0}^{\infty}2^{-2j}=1/12$. Adding the scaling energy $(1/2)^2=1/4$ gives $1/3$, equal to $\int_0^1x^2\,dx$. This verifies Parseval’s identity for this example.

Reading the Coefficients in Regression

For a regression function $m\in L^2([0,1])$, the coefficient $\alpha_{j,k}$ summarizes its component aligned with a scaling function at a particular location and resolution. The coefficient $\beta_{j,k}$ measures a signed local contrast at that location. A zero detail coefficient does not mean the function is zero on the interval; it means that its component in that particular Haar contrast is zero. For example, any function constant across the two halves has zero Haar detail there.

Orthogonality also makes coefficient energy interpretable: the squared $L^2$ norm is partitioned into the squared scaling coefficient and the squared details. In regression, one may estimate coefficients from data and decide how many levels or details to retain. Those are estimation and selection questions; the expansion here describes the population function and its basis coefficients, not how to estimate or threshold them.

Common Mistake: Confusing Averages With Normalized Coefficients

Wrong reasoning: “The left and right child values are $1$ and $0$, so their average is $(1/2+0)/2=1/4$.” This has silently replaced the left child value $1$ by $1/2$, so it is not averaging the stated child values.
Correction: The parent average is $(1+0)/2=1/2$. For the Haar example with values $1$ and $0$ on the two halves, the normalized wavelet coefficient is also $1/2$, since it is the integral of the function against the unit-norm wavelet. The scaling coefficient and detail coefficient have specific integral definitions; verify them from those definitions rather than treating them as interchangeable averages or contrasts.
Another common confusion: A coefficient at a fine level is not a coefficient for the whole domain. Its wavelet is localized to one dyadic interval, and its index $k$ identifies that location.

Key takeaway: A multiresolution expansion starts with a coarse scaling component and adds orthogonal wavelet details level by level. For Haar wavelets, the scaling functions encode dyadic averages, the details encode left-right contrasts, and every square-integrable function on $[0,1]$ is recovered as the $L^2$ limit of these terms.

Check Your Understanding

Use the definitions and verify each calculation before interpreting it.

For the function equal to $3$ on the left half of $[0,1]$ and $1$ on the right half, calculate its level-zero scaling and detail coefficients.
Explain why $V_{j+1}=V_j\oplus W_j$ means that a finer projection adds detail without changing the coarser component.
For $f(x)=x$, what is the sign of every Haar detail coefficient, and what feature of the function determines that sign?
What convergence is asserted by the infinite Haar expansion, and what does the assertion not necessarily claim about pointwise values?
Why does a zero Haar detail coefficient on an interval not imply that the function is zero on that interval?