Why Smoothness Across Coordinates Matters
A multivariate density need not be equally smooth in every direction. For example, a density may change gradually with one coordinate but have sharper variation with another. A single bandwidth applied to all coordinates ignores this difference: it may smooth too little in a rough direction or too much in a smooth one. An anisotropic estimator instead assigns a bandwidth to each coordinate and chooses those bandwidths in light of their separate smoothness orders.
The central calculation balances two contributions to integrated squared error (ISE). Smoothing creates bias, whose size depends on both each bandwidth and its coordinate’s smoothness. Sampling creates variance, which grows as the product of the bandwidths shrinks. The resulting rate shows how dimension and directional smoothness jointly determine the cost of estimating a density.
Directional Smoothness and the Estimator
For a precise rate calculation, consider a density \(f\) on the \(d\)-dimensional unit torus, so that boundary effects do not enter the calculation. Assume observations \(X_1,\ldots,X_n\) are independent and identically distributed from \(f\). Let \(\beta_j>0\) be the smoothness order in coordinate \(j\), for \(j=1,\ldots,d\). We use noninteger \(\beta_j\) here. Write \(m_j=\lfloor\beta_j\rfloor\) and \(\alpha_j=\beta_j-m_j\), so \(0<\alpha_j<1\).
For orders above one, the corresponding Taylor remainder in direction \(j\) is bounded in \(L_2\) by a constant times \(|t|^{\beta_j}\). This is the feature of the smoothness condition used below. Take a one-dimensional kernel \(K_j\) for each coordinate, with integral one, finite \(L_2\) norm and finite absolute \(\beta_j\)-moment. Require its moments of orders \(1,\ldots,m_j\) to vanish. The product kernel estimator, with bandwidths \(h_1,\ldots,h_d\), is
where kernels are periodized on the torus. The moment conditions remove the lower-order terms in the directional Taylor expansion. They can require a signed kernel when higher-order moments must vanish; the estimator need not be nonnegative. That issue is distinct from the bandwidth-rate calculation.
Bias and Variance with Separate Bandwidths
Let \(T_j\) denote convolution with the \(j\)th scaled kernel in coordinate \(j\). The directional Taylor remainder and the kernel moment conditions imply \(\|(T_j-I)f\|_2\le C_j h_j^{\beta_j}\), where \(I\) is the identity operator. The product smoothing operator is \(T_1\cdots T_d\). Subtracting \(f\) and telescoping the product gives a sum of directional smoothing differences. Each other convolution operator has \(L_2\) operator norm at most the \(L_1\) norm of its kernel, so these factors change only the constant. Consequently, for a constant \(C_1\) independent of the bandwidths,
Squaring and using \((\sum_j a_j)^2\le d\sum_j a_j^2\) yields a squared-bias bound proportional to \(\sum_j h_j^{2\beta_j}\). For variance, independence of the observations gives a factor \(1/n\). Integrating the pointwise variance and dropping its nonnegative squared-mean term bounds it by \(n^{-1}\) times the squared \(L_2\) norm of the product kernel. For sufficiently small bandwidths, that norm squared is a constant times \(\prod_j h_j^{-1}\). Thus:
To see the decomposition, write \(\hat f_h-f=(\hat f_h-\mathbb{E}\hat f_h)+(\mathbb{E}\hat f_h-f)\). The cross term has expectation zero at every \(x\); integration therefore gives the sum of integrated variance and squared bias. Applying the two bounds above proves the result. This derivation makes clear why every coordinate affects the rate: the bias terms are directional, while the variance depends on the volume \(\prod_j h_j\).
Balancing the Terms
The bias contribution for coordinate \(j\) is \(h_j^{2\beta_j}\), while making any bandwidth smaller increases the variance. For a rate calculation, set \(h_j=n^{-a_j}\). If all the squared-bias terms and the variance have the same order \(n^{-r}\), then \(2\beta_j a_j=r\) for every \(j\), and \(1-\sum_j a_j=r\). Solving these equations gives the bandwidth allocation and rate.
For all \(j\), set \(2\beta_j a_j=r\), so \(a_j=r/(2\beta_j)\).
The variance has order \(n^{-1+\sum_j a_j}\), so its exponent must satisfy \(r=1-\sum_j a_j\).
Substitution gives \(r=1-r\sum_j1/(2\beta_j)\), hence \(r=2/(2+\sum_j1/\beta_j)=2/D\).
Indeed, the proposed bandwidth has \(a_j=1/(\beta_jD)\), so \(h_j^{2\beta_j}=n^{-2/D}\). Also, \(\sum_j a_j=(D-2)/D\); hence the variance term is \(n^{-1+(D-2)/D}=n^{-2/D}\). The bias sum has only \(d\) terms, so its order is also \(n^{-2/D}\). Substitution into the risk bound proves the stated upper rate. This is a rate-optimal allocation for that bias-variance bound: any power allocation balancing all terms must satisfy the equations above, which uniquely determine \(r\) and the \(a_j\). \(\square\)
Worked Bandwidth Allocations
Worked Example: One Rough and One Smooth Coordinate
Suppose \(d=2\), \(\beta_1=1/2\), and \(\beta_2=3/2\). Then
The rate exponent is \(2/D=3/7\), so the ISE upper bound has order \(n^{-3/7}\). For the first coordinate, the bandwidth exponent is \(1/(\beta_1D)=1/((1/2)(14/3))=3/7\). For the second, it is \(1/(\beta_2D)=1/((3/2)(14/3))=1/7\). Thus \(h_1=n^{-3/7}\) and \(h_2=n^{-1/7}\).
At \(n=128=2^7\), these bandwidths are \(h_1=1/8\) and \(h_2=1/2\). The two squared-bias terms scale as \(h_1^{2\beta_1}=h_1=1/8\) and \(h_2^{2\beta_2}=h_2^3=1/8\). The variance factor is \(1/(nh_1h_2)=1/(128\cdot(1/8)\cdot(1/2))=1/8\). All three terms in the rate calculation agree, as the allocation requires.
Worked Example: Equal Smoothness in Three Dimensions
Let \(d=3\) and let every coordinate have smoothness order \(\beta_j=3/2\). Then
The ISE rate is \(n^{-2/4}=n^{-1/2}\). Directly checking the bound, each squared-bias term is \(h_j^3=n^{-1/2}\). The product of the three bandwidths is \(n^{-1/2}\), so the variance is \(1/(n n^{-1/2})=n^{-1/2}\). This recovers the equal-bandwidth calculation when all directions have the same smoothness.
Worked Example: Five Coordinates with Three Smoothness Levels
Take \((\beta_1,\ldots,\beta_5)=(1,1,2,2,4)\). The reciprocal smoothness sum is \(1+1+1/2+1/2+1/4=13/4\), giving \(D=2+13/4=21/4\). Therefore the risk exponent is \(2/D=8/21\).
For each of the first two coordinates, the bandwidth exponent is \(1/(1\cdot21/4)=4/21\). For coordinates three and four it is \(1/(2\cdot21/4)=2/21\); for the fifth it is \(1/(4\cdot21/4)=1/21\). The squared-bias exponents are respectively \(2(1)(4/21)=8/21\), \(2(2)(2/21)=8/21\), and \(2(4)(1/21)=8/21\). The sum of all bandwidth exponents is \(8/21+4/21+1/21=13/21\). Thus the variance exponent is \(1-13/21=8/21\), matching the bias and giving rate \(n^{-8/21}\).
Dimension, Interpretation, and a Common Pitfall
The sum \(\sum_j1/\beta_j\) is the key measure of anisotropic complexity. A low smoothness order contributes a large reciprocal and slows the rate; a smoother coordinate contributes less. If all coordinates have common order \(\beta\), then \(D=2+d/\beta\), giving rate \(n^{-2\beta/(2\beta+d)}\). Thus the familiar dimension penalty is recovered, while unequal smoothness replaces \(d/\beta\) with the sum of the coordinate-specific reciprocals. One can summarize this using the effective smoothness \(\beta_{\mathrm{eff}}=d/(\sum_j1/\beta_j)\), for which \(D=2+d/\beta_{\mathrm{eff}}\).
Common mistake: use one bandwidth exponent for every coordinate. A tempting argument is that there are \(d\) coordinates, so the optimal bandwidth should have the same exponent in each, regardless of smoothness. That allocation generally fails to balance the directional bias terms: if \(h_j=n^{-a}\), then coordinate \(j\)'s squared bias has exponent \(2\beta_j a\), which differs across coordinates when the \(\beta_j\) differ. The correction is to choose \(a_j=1/(\beta_jD)\). Rough coordinates then receive smaller bandwidths, while smoother coordinates can use larger ones. The product of all bandwidths must still be included when calculating variance.
Check Your Understanding
Use the bias-variance bound and bandwidth balance to answer these questions.
- For two coordinates with smoothness orders \(1\) and \(2\), calculate \(D\), the rate exponent, and both bandwidth exponents.
- Explain why a rougher coordinate receives a smaller bandwidth under the allocation derived here.
- For three coordinates with smoothness orders \(1,1,2\), verify that the proposed bandwidth exponents make the variance and each squared-bias term have the same order.
- In the equal-smoothness case, how does increasing the dimension affect the ISE rate?
- Why does the variance depend on the product of the coordinate bandwidths rather than on a separate bandwidth term for each coordinate?