# Nonparametric method - Parzen windows

I don't understand the main function of the Parzen Window

Let $u=[u_1, u_2,..., u_d]$ and define a window function

$φ(u)=\left\{ \begin{array}{l l} 1 & \quad \text{$|u_j|<\frac{1}{2}$,$j=1,2,...,d$}\\ 0 & \quad \text{otherwise} \end{array} \right.$

What exactly it means?

I found lots of presentation on the internet, so please do not direct me to one.

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When you edit a question such that parts of existing answers no longer make sense, please mark the edit as such. – joriki Dec 12 '12 at 18:14
Also, editing other people's posts while ignoring their comments is not exactly the best style. – joriki Dec 13 '12 at 6:05
My edit of your your answer is the sign that I didn't ignored your comment. I even wrote why I edited it. Take it easy man. You got your points, I got my answer. (I tried to answer my question an hour after I posted it but I couldn't because I'm new here, even though I accepted your answer and not my) – Bush Dec 13 '12 at 6:39
I am taking it easy :-) Relative to how annoying I find it that so many people on this site just edit around in posts (their own and others) without ever replying to comments, my reaction was quite mild :-) My annoyance was also perhaps furthered by the fact that you asked a question about what exactly something means but didn't bother to make sure that you quoted it exactly; that sort of thing often wastes a lot of time, and it would have been so easy for you to avoid. Feel free to accept your own answer if you find it more helpful; points are not so important. – joriki Dec 13 '12 at 8:44
I agree. Need to recheck anything I write, especially quote, so people will not waste their time on stupid mistakes. I learnt from this case, thank's :) – Bush Dec 13 '12 at 9:14

It means that $\varphi(u)$ is $1$ in a unit hypercube centred on the origin and $0$ outside (and on the boundary).
$u$ is a $d$ dimensional sample, we choose a "window" which is: a line with length h when $d=1$, a square with edge $h=2$ ($h^2$ space), an hypercube with edge $h=2$. The "window" is centered at the point $u$ and we normalize is so $h=1$.
Now our function assign 1 to each other sample that is close to $u$ and 0 to samples that are far.
Close means that the sample $x$ is in the range of $u-\frac{1}{2}h\leq{x}\leq{u+\frac{1}{2}h}$