Weighted Euclidean Metrics

Euclidean Distance with Coordinate Weights

Ordinary Euclidean distance assigns the same role to every coordinate: a change of one unit in the first coordinate contributes just as much as a change of one unit in the second. In some problems, coordinates need not have equal influence. A weighted Euclidean metric incorporates that difference by assigning a positive weight to each coordinate. The points remain the same; it is the rule for measuring their separation that changes.

Throughout, \(n\in\mathbb{N}\), and \(\mathbb{R}^n\) has its usual coordinates and Euclidean distance. Fix positive real numbers \(w_1,\dots,w_n\), called the coordinate weights. For \(x=(x_1,\dots,x_n)\) and \(y=(y_1,\dots,y_n)\), define their weighted distance as follows.

The weights are coefficients inside the square root. A larger weight makes a fixed change in that coordinate contribute more to the squared distance. Positivity of every weight is essential: it ensures that no nonzero coordinate difference can be ignored.

Why This Formula Is a Metric

Proof. Define \(D_w:\mathbb{R}^n\to\mathbb{R}^n\) by \(D_w(x_1,\dots,x_n)=(\sqrt{w_1}x_1,\dots,\sqrt{w_n}x_n)\). For all \(x,y\in\mathbb{R}^n\), the ordinary Euclidean distance satisfies \(d_w(x,y)=\|D_w(x)-D_w(y)\|\), because the square of the right-hand side is \(\sum_{i=1}^n(\sqrt{w_i}x_i-\sqrt{w_i}y_i)^2=\sum_{i=1}^n w_i(x_i-y_i)^2\). We verify the metric axioms.

Each term \(w_i(x_i-y_i)^2\) is nonnegative, so \(d_w(x,y)\ge0\). If \(x=y\), every coordinate difference is zero and \(d_w(x,y)=0\). Conversely, if \(d_w(x,y)=0\), then \(\sum_{i=1}^n w_i(x_i-y_i)^2=0\). Every summand is nonnegative, so each must be zero. Since \(w_i>0\), this implies \((x_i-y_i)^2=0\) and hence \(x_i=y_i\) for every \(i\); therefore \(x=y\). Symmetry follows from \((x_i-y_i)^2=(y_i-x_i)^2\) for each coordinate.

For the triangle inequality, let \(x,y,z\in\mathbb{R}^n\). The map \(D_w\) gives \(D_w(x)-D_w(z)=(D_w(x)-D_w(y))+(D_w(y)-D_w(z))\). Applying the triangle inequality for ordinary Euclidean distance to these vectors gives \(\|D_w(x)-D_w(z)\|\le\|D_w(x)-D_w(y)\|+\|D_w(y)-D_w(z)\|\). Replacing each Euclidean distance by its equal weighted-distance expression yields \(d_w(x,z)\le d_w(x,y)+d_w(y,z)\). All four metric axioms hold, so \(d_w\) is a metric. \(\square\)

This proof uses the ordinary Euclidean metric and its triangle inequality. It also shows why the weights must be positive: \(D_w\) scales each coordinate by a positive number, so distinct points remain distinct after scaling.

Calculating and Comparing Weighted Distances

The effect of changing weights can be stated precisely. Because \(n\) is finite and each weight is positive, there is a smallest weight \(m=\min\{w_1,\dots,w_n\}>0\) and a largest weight \(M=\max\{w_1,\dots,w_n\}\).

Proof. For every coordinate \(i\), the definitions of \(m\) and \(M\) give \(m\le w_i\le M\). Since \((x_i-y_i)^2\ge0\), multiplication preserves these inequalities: \(m(x_i-y_i)^2\le w_i(x_i-y_i)^2\le M(x_i-y_i)^2\). Summing over \(i=1,\dots,n\) gives \(m\sum_{i=1}^n(x_i-y_i)^2\le\sum_{i=1}^n w_i(x_i-y_i)^2\le M\sum_{i=1}^n(x_i-y_i)^2\). All three quantities are nonnegative, so taking square roots preserves the inequalities. The left and right sums are the squares of \(\sqrt{m}\|x-y\|\) and \(\sqrt{M}\|x-y\|\), and the middle sum is \(d_w(x,y)^2\). This proves the stated bounds. \(\square\)

The same coordinate calculation also gives a useful monotonicity rule. If two positive weight lists \(w_i\) and \(v_i\) satisfy \(w_i\le v_i\) for every \(i\), then \(d_w(x,y)\le d_v(x,y)\) for every pair \(x,y\): multiply each inequality by \((x_i-y_i)^2\), sum, and take nonnegative square roots. For a particular pair, increasing a weight has no effect if the two points agree in that coordinate. If they differ there, increasing that weight strictly increases the squared distance and therefore the distance.

A Common Pitfall: Zero Weights

It is tempting to allow a coordinate weight to be zero and treat it as a metric weight. That fails the identity-of-indiscernibles axiom: distinct points can then have distance zero. For instance, the formula with weights \((1,0)\) gives, for \(p=(0,0)\) and \(q=(0,5)\), \(\sqrt{1(0-0)^2+0(5-0)^2}=\sqrt{0+0}=0\), although \(p\ne q\). Thus the formula is not a metric for these weights. A negative weight is also inadmissible: the expression under the square root can be negative for some pairs, so it need not define a real-valued distance. Require every coordinate weight to be strictly positive.

Weighted Euclidean metrics are useful when different coordinate changes should have different costs or scales. The definition gives a direct computation, the scaling map explains the triangle inequality, and the comparison bounds quantify how far the weighted distances can differ from ordinary Euclidean distances. In every case, the coordinates still describe the same points of \(\mathbb{R}^n\); only their measured separations have changed.