= Discrete wavelet transform
{title2=$(a,b)=((u+v)/\sqrt2,(u-v)/\sqrt2)$}
A discrete wavelet transform computes multiscale approximation and detail coefficients from sampled data. For the <Haar wavelet>, replace each adjacent pair $(u,v)$ by
$$
a=\frac{u+v}{\sqrt2},\qquad d=\frac{u-v}{\sqrt2}.
$$
The approximation coefficient $a$ is $\sqrt2$ times the pair's <arithmetic mean>; it is not the ordinary mean. The detail coefficient $d$ records their signed contrast. Repeat on the approximation coefficients until only one remains, keeping every detail coefficient.
For a worked example, start with the vector $(2,4,6,8)$. The first pass gives approximation coefficients $(3\sqrt2,7\sqrt2)$ and detail coefficients $(-\sqrt2,-\sqrt2)$. Transforming the remaining pair gives the coarse coefficient $10$ and the half-to-half detail $-4$. In the order coarse value, coarse detail, then fine details, the transform is
$$
\boxed{(2,4,6,8)\longmapsto(10,-4,-\sqrt2,-\sqrt2).}
$$
This is an <orthogonal matrix> transformation. Each pair preserves $u^2+v^2=a^2+d^2$, so the full transform preserves the squared <Euclidean norm>:
$$
2^2+4^2+6^2+8^2=120=10^2+(-4)^2+(-\sqrt2)^2+(-\sqrt2)^2.
$$
The <inverse function> for each pair is $u=(a+d)/\sqrt2$, $v=(a-d)/\sqrt2$. Reversing the passes therefore recovers every original value. This squared-norm bookkeeping is often called signal energy; it need not represent physical <energy>. https://www.mathworks.com/help/wavelet/ug/haar-transforms-for-time-series-data-and-images.html[MathWorks: Haar transforms, approximation/detail coefficients and squared-norm conservation]{ref}
Continuous and discrete normalizations differ. If $n=2^J$ data values are interpreted as the heights of a <step function> on $n$ equal cells of $[0,1]$, its normalized <L2 inner product> coefficients are the discrete coefficients divided by $\sqrt n$. For our four values they are $(5,-2,-1/\sqrt2,-1/\sqrt2)$, as the <Haar projection> example shows. The same normalization is used by the <wavelet regression estimator> on $[0,1]$.
Under <independent> <Gaussian noise> with a common <variance>, <orthogonality> preserves that coefficient noise <variance> and independence. It does not make every arbitrary correlated <signal> independent: this statistical conclusion uses the Gaussian model and its equal-variance independent inputs.
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