Let $x,y \in \mathbb{R}$ where $y=x-t$. Translation-invariant (or shift-invariant) kernel $\kappa(\cdot,\cdot)$ is defined as $\kappa(x,y) = \kappa(x,x-t) = \kappa(t)$.
Can I say that the function $\kappa$ is symmetric ?
I think "yes", if I can define the translation-invariance as "DIFFERENCE between $x$ and $y$" so that $y=x+t$. And $\kappa(x,y)=\kappa(x,x+t)=\kappa(-t)$.
But, wanted make sure the relation between translation-invariance and symmetry (i.e., $\kappa(x,y)=\kappa(y,x)$).
Can I simply say that translation invariant kernel is $\kappa(x,y) = \kappa(x-y) = \kappa(y-x)$ ? translation-invariant $\to$ symmetric. Is this always correct??? Or, is the symmetry required for the second equality?
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$\begingroup$Unfortunately I still don't understand the parts of your question where you try to motivate the idea that translation invariance might imply symmetry, so I can only state that this implication doesn't hold. If a kernel $\kappa(x,y)$ is translation-invariant and is thus given by $\kappa(x,y)=g(x-y)$ for some function $g$, then $\kappa$ is symmetric in the sense $\kappa(x,y)=\kappa(y,x)$ if and only if $g$ is symmetric in the sense $g(t)=g(-t)$, since $\kappa(x,y)=g(x-y)$ and $\kappa(y,x)=g(y-x)=g(-(x-y))$. For instance, for symmetric $g(t)=t^2$, the kernel is symmetric with $\kappa(x,y)=\kappa(y,x)=(x-y)^2$, while for antisymmetric $g(t)=t$, the kernel is antisymmetric with $\kappa(x,y)=-\kappa(y,x)=x-y$.
$\endgroup$ $\begingroup$If you mean a kernel as in kernel method's literature, then k(x,y) = k(y,x) by definition as it corresponds to an inner product in some Hilbert space. If further k is translation invariant, then k(x-y) = k(y-x) implying that k(x) = k(-x) which is symmetric (even function). If you don't assume k(x,y)=k(y,x), I think @jriki is right.
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