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As a follow up to this question, concerning this paper:

Given random variables $X_1,\ldots,X_N,X_q:\Omega\rightarrow\mathbb{R}^d$, where $X_1,\ldots,X_N$ are independent and identical distributed. Draw an $\omega\in\Omega$. We now want to pick the $k$ points from $X_1(\omega),\ldots,X_N(\omega)$ that are closest to $X_q(\omega)$. Wlog assume these are $X_1(\omega),\ldots,X_k(\omega)$. Define $$B_k(X_q(\omega))=\{x\in\mathbb{R}^d\mid D(x,X_q(\omega)\leq D(X_k(\omega),X_q(\omega))\}$$ where $D(x,y)$ is some dissimilarity measure (think of a metric). The volume of this ball is $$v_k(X_q(\omega))=\int_{B_k(X_q(\omega))}1\:dx.$$ Let $\delta(x)$ be the probability density function of the $X_i$, then let $$u_k(X_q(\omega))=\int_{B_k(X_q(\omega))}\delta(x)\:dx$$ be the probability content of the ball $B_k(X_q(\omega))$. Obviously we have $0\leq u_k(X_q(\omega))\leq 1$.

Now assume $N$ large enough such that $\delta(x)$ can be approximated by a constant $\overline{p}(X_q(\omega))$ on $B_k(X_q(\omega))$.

Fianlly, the authors of the paper state that $\overline{p}(X_q(\omega))\neq 0$, but I do not see why this should hold. I though that $$\overline{p}(X_q(\omega))\approx\int_{B_k(X_q(\omega))}\delta(x)\:dx.$$ which can be $0$ for example if $\delta(x)$ is not continuous, but discrete. Can someone explain, why $\overline{p}(X_q(\omega))\neq 0$?

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  • $\begingroup$ The very fact that they're asking to approximate a function ($\delta(x)$) by a constant in a small ball already assumes that the function is continuous. $\endgroup$ – Joe Neeman Apr 3 '15 at 17:42
  • $\begingroup$ Thanks for your comment, this already answers my question! $\endgroup$ – Skrodde Apr 9 '15 at 10:30
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As Joe Neeman pointed out in the comment on the OP:

The very fact that they're aksing to approximate a function $\delta(x)$ by a constant in a small ball already assumes that the function is continuous.

We now approximate $\delta(x)$ on $B_k(X_q(\omega))$ by $\overline{p}(X_q(\omega))=\int_{B_k(X_q(\omega))}\delta(x)\:dx$. Then $\overline{p}(X_q(\omega))\neq0$ since $v_k(X_q(\omega))\neq0$, $\delta(x)$ continuous, and $X_1(\omega),\ldots,X_k(\omega)\in B_k(X_q(\omega))$. That is, there is a small neighborhood $U_i$ around $X_i(\omega)$, $i=1,\ldots,k$, such that $\delta(x)>0$ for all $x\in U_i$.

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