I am trying to understand why $lim_{||u|| \rightarrow+\infty}{\varphi(u)}\prod_{i=1}^{d}u_{i} = 0$ is necessary for convergence of Parzen density estimates. Similar question has been asked here

Intuitively we estimate density using parzen windows by squeezing a distribution into smaller windows as the number of observations increases. When $n=\infty$ this distribution should converge to spike at $0$. Thus we design density kernel in a way that probability is 0 everywhere except origin after squeezing infinitely. If the density function has a finite value at someplace close to infinity it will be difficult to squeegee it to a spike at 0. This can be understood in terms of convergence of kernel density function to delta dirac function. I have tried to understand this mathetically as follows:

Convergence in mean square for parzen window assumes that limit of sequence of function in $\mathbf{R}^m$ given by $\epsilon^{-m} \varphi(x/\epsilon)$ converges to Dirac Delta function due to which we can say $\bar{p}(x)=\int \frac{1}{V} \varphi(\frac{x-v}{\epsilon}) p(v) dx$ converges to $p(x)$ as $\epsilon \to 0$.

Coming to $lim_{||u|| \rightarrow+\infty}{\varphi(u)}\prod_{i=1}^{d}u_{i} = 0$.

Substituting $x_i/\epsilon=u_i $ for any $x >0$ in $L=lim_{||u|| \rightarrow+\infty}{\varphi(u)}\prod_{i=1}^{d}u_{i}$ we get

$L=lim_{||\epsilon|| \rightarrow 0}(\prod_{i=1}^{d}x_{i})\epsilon^{-m} \varphi(x/\epsilon)$

Since $lim_{||\epsilon|| \rightarrow 0}\epsilon^{-m} \varphi(x/\epsilon)$ is delta dirac function its value should be be $0$ everywhere except $x=0$, so $L=0$

Please let me know if my understanding is correct.


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