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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 $y$.

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  • $\begingroup$ Just to clarify a possible ambiguity: by "spherical distribution" do you mean a distribution on $\mathbb{R}^n$ that is invariant under rotations about the origin? (Or do you mean any distribution defined on its unit sphere $S^{n-1}$?) And by "spherical kernel" do you mean any kernel that is also invariant under rotations about the origin, or do you mean some particular kernel (as indicated by your use of "the spherical kernel")? $\endgroup$ – whuber May 27 '16 at 16:07
  • $\begingroup$ I have edited my question above. $\endgroup$ – Dwaipayan Gupta May 27 '16 at 16:46
  • $\begingroup$ I have assumed that being a "distribution defined on [the] unit sphere S" is not what you mean, for otherwise your conclusion is obviously false (and generally true only when $f$ is the uniform distribution on $S^{n-1}$). In fact, that characterization of "spherical" would render the condition "$||y||=||z||$" superfluous, because by definition the norms of all points on the sphere are the same. $\endgroup$ – whuber May 27 '16 at 20:27
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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.

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