Wavelet regression estimator (${\hat{m}}_J=P_J{\tilde{Y}}_n$)

At design sites $i/n$, let ${\tilde{Y}}_n$ be the step function taking response $Y_i$ on $((i-1)/n,i/n]$. Project it onto an interval-adapted wavelet basis through resolution $J$. Each coefficient is ${\hat{c}}_{\lambda }={\sum }_i{Y}_i{\int }_{I_i}{b}_{\lambda }(t)dt$. Its expectation is the corresponding coefficient of the step function of sampled regression means. For a Haar wavelet basis, dyadic sample size and resolution at most the sampling scale, this is exactly $n^{-1}{\sum }_i{Y}_i{b}_{\lambda }(i/n)$ with right-closed pointwise versions. Independent errors of variance ${\sigma }^2$ give coefficient variance at most ${\sigma }^2/n$: Cauchy-Schwarz inequality on each sampling cell yields ${\sum }_i({\int }_{I_i}{b}_{\lambda }{)}^2\le \Vert b_{\lambda }{\Vert }_2^2/n$.

Hard thresholding keeps detail coefficients whose magnitude exceeds a prescribed threshold and sets the rest to zero, while retaining the coarse approximation coefficients. Soft thresholding instead uses $\mathrm{sgn}(\hat{b})(|\hat{b}|-\tau {)}_{+}$. The simultaneous wavelet coefficient noise bound motivates thresholds of order $\sigma \sqrt{\log n/n}$ in a normalized Gaussian regression model with at most order $n$ details. This is a nonlinear selection rule: coefficients can be removed individually rather than discarding an entire finest resolution.

For coefficient weights $w_{\lambda i}$ and independent mean-zero sub-Gaussian random variables of parameter $\sigma$, put $q_{\lambda }^2={\sum }_i{w}_{\lambda i}^2$. Multiplying moment-generating-function bounds shows that the weighted noise has parameter $\sigma q_{\lambda }$. The exponential Markov inequality, optimized in its parameter, gives tail bound $2\exp (-t^2/(2{\sigma }^2{q}_{\lambda }^2))$. At the displayed thresholds, a union bound over $N$ coefficients gives simultaneous control with probability at least $1-\delta$. No coefficient independence is required. Integrated wavelet weights satisfy $q_{\lambda }\le n^{-1/2}$. Finite variance alone only supports the weaker Chebyshev choice ${\tau }_{\lambda }=\sigma q_{\lambda }\sqrt{N/\delta }$.