"Efficiency" of a Kernel

Question

My understanding is that the Epanechnikov kernel is "efficient" in a mean squared error sense. Footnote 4 of Wikipedia's page defines the "efficiency" of a kernel as $$\sqrt{\int u^2 K(u)du} \int K(u)^2 du.$$

which is minimized by the Epanechnikov kernel. (I've seen this definition in places other than Wikipedia). I am struggling to map this definition directly into mean squared error.

For kernel density estimation of the density of $X$, the mean squared error I derived is,

$$\frac{1}{4} h^4 \left(∫_{-∞}^∞z^2 K(z)dz\right)^2 f^{''}(x)^2+\frac{1}{nh} f(x) \left(∫_{-∞}^∞ K(z)^2 dz\right)$$

For a local constant (Nadaraya-Watson) regression estimating $E[Y|X]$, my derivation of MSE results in: $$h^4 \left(\int u^2 K(u)du\right)^2 B^2 (x)+\frac{\sigma^2(x)(\int K(u)^2 du))}{nhf(x)} $$

Where $\sigma^2(x)$ is the variance of $Y$ at $X=x$.

Mean Integrated Squared Error would involve integrating over $x$.

Can anyone help me understand how the definition of "efficiency" as posted on Wikipedia is directly related to minimizing MSE? (or MISE/AMISE?) The components of Wikipedia's definition definitely appear in MSE, but it definitely seems distinct.

TIA!

Answer 1

Kernels can be normalized based on $\int K(u)^2 du$. i.e. one could normalize such that $\int K(u)^2 du =1$, and the only term that affects efficiency is $\int u^2 K(u) du $.

Thus, mulitplying by $\int K(u)^2 du$ is just scaling based on the variance of the kernel (which could be normalized).