3
$\begingroup$

So I am really struggling with this problem and could use some help.

Consider the Epanechnikov kernel given by $$f_e(x)=\frac{3}{4}\left( 1-x^2 \right)$$

According to Devorye and Gyofri to generate a sample of a distribution having $f_e$ as its density function we can use the following method

  1. Generate iid $U_1,U_2,U_3 \sim \operatorname{Uniform}(-1,1)$.
  2. If $\left| U_3\right| \geq \left| U_2\right|$ and $\left| U_3\right| \geq \left| U_1\right|$, deliver $U_2$; otherwise deliver $U_3$.

I have to prove this works. I thought this was related to either the acceptance-rejection method or maybe order statistics. I have spent a fair bit of time trying both approaches but I am stuck. Any pointers will be greatly appreciated.

$\endgroup$
  • $\begingroup$ Please correct: Devroye and Györfi (and possibly Lugosi). $\endgroup$ – Did Sep 24 '15 at 8:54
0
$\begingroup$

$(U_1,U_2,U_3)$ is a random point in the cube $[-1,1]^3$. Let $V$ be the distribution generated by the given algorithm: it is obviously a symmetric distribution supported on $[-1,1]$. If we take some $\alpha\in[0,1]$, the probability that $|V|\leq \alpha$ is given by the probability that $\max(|U_1|,|U_2|)\leq|U_3|$ and $|U_2|\leq \alpha$, plus the probability that $\max(|U_1|,|U_2|)>|U_3|$ and $|U_3|\leq \alpha$, hence by:

$$\int_{0}^{\alpha}\left(\int_{0}^{u_2}(1-u_2)\,du_1+\int_{u_2}^{1}(1-u_1)\,du_1\right)\,du_2+ \int_{0}^{\alpha}(1-u_3^2)\,du_3$$ that is $\frac{3\alpha-\alpha^3}{2}$. That gives: $$ \mathbb{P}[0\leq V\leq \alpha] = \frac{3\alpha-\alpha^3}{4}$$ and by differentiating with respect to $\alpha$ we get that the PDF of $V$ is given by $\frac{3}{4}(1-x^2)$, as wanted.

$\endgroup$
  • 1
    $\begingroup$ Perfect! So it seems I should have been thinking about absolute basic probability instead of more advanced concepts! $\endgroup$ – cga007 Sep 22 '15 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.