Problem on spherical kernel and spherical distribution

Question

Suppose, $X$ is a random variable which follows $f$ (i.e., $X$ $\sim$ $f$), such that $f(x)$ is the probability density function (p.d.f.) of a spherical distribution.

Here, $X$ is multi-dimensional ! By spherical distribution, I mean any distribution defined on its unit sphere $S^{n-1}$.

Also, $X$ and $HX$ follow the same distribution, where $H$ is an orthogonal vector.

Let $k$ be the spherical kernel and $h$ be the bandwidth of the kernel function.

Let, $$\lambda_y \; = \; \int \; k\, \Big(\frac{y - x}{h} \Big) \; f(x) \; dx$$

and $$\lambda_z \; = \; \int \; k\, \Big(\frac{z - x}{h} \Big) \; f(x) \; dx.$$

Then how do I show that : $$ ||y|| \; = \; ||z|| \; \; \; \; \; \; \Rightarrow \; \; \; \; \lambda_y \; = \; \lambda_z \; ?$$

Here, $||y||$ indicates the norm of $y$.

Answer 1

Preliminary definitions and observations

What "spherical" means

That $f$ is "spherical" (often called "elliptical") means

$$f(Hx) = f(x)\tag{1}$$

for all $x\in\mathbb{R}^n$ and all orthogonal matrices $H$.

That $k$ is "spherical" means the same thing. Applying it to the argument in the integral gives

$$k\left(\frac{Hy - Hx}{h}\right) = k\left(H \frac{y-x}{h}\right) = k\left(\frac{y-x}{h}\right)\tag{2}$$

for all $x,y\in\mathbb{R}^n$, all positive real numbers $h$, and all orthogonal matrices $H$. (The first equality is a consequence of the linearity of $H$.)

Invariance of the volume element

Any orthogonal matrix $H$ has unit determinant due to the defining relation $H^\prime H = I_n$ because

$$1 = |I_n| = |H^\prime H| = |H^\prime| |H| = |H|^2,$$

whence the absolute value of $|H|$ must equal $1$. This implies the volume element $|dx|$ is invariant under orthogonal transformations, because

$$|\pm 1\, dx| = |\,|H|\,dx| = |d(Hx)| = |dx|.\tag{3}$$

Transitivity of the orthogonal group

Finally, whenever $||y||=||z||$, there is an orthogonal matrix $H$ for which

$$H y = z.\tag{4}$$

For instance, the matrix $H$ representing the reflection through the midpoint of the line segment from $z$ to $y$ will work.

The derivation

After finding an $H$ by virtue of $(4)$, substitute $z=Hy$ to produce

$$\lambda_z = \int_{\mathbb{R}^n} k\left(\frac{z-x}{h}\right) f(x) dx = \int_{\mathbb{R}^n} k\left(\frac{Hy-x}{h}\right) f(x)dx.$$

Make the change of variable $x = H x^\prime$. The invariance of $|dx^\prime|$ under orthogonal transformations $(3)$ yields

$$\int_{\mathbb{R}^n} k\left(\frac{Hy-x}{h}\right) f(x) dx = \int_{\mathbb{R}^n} k\left(\frac{Hy-Hx^\prime}{h}\right) f(Hx^\prime) dx^\prime.$$

This is the point where we require the spherical invariance of both $f$ (equation $(1)$) and $k$ (equation $(2)$), for they allow us to simplify this expression as

$$\lambda_z = \int_{\mathbb{R}^n} k\left(\frac{Hy-Hx^\prime}{h}\right) f(Hx^\prime) dx^\prime = \int_{\mathbb{R}^n} k\left(\frac{y-x^\prime}{h}\right) f(x^\prime) dx^\prime = \lambda_y,$$

QED.