Simple example of Parzen window (kernel density estimation)
In kernel density estimation, rectangle, triangle, or Gaussian  kernels assign weight to positions around query point $x$. For rectangle and square, the weight is steady, for triangle, weight drops linearly with distance, and for Gaussian, weight drops exponentially with distance. Here is an image for square, rectangle, and triangle kernels (one dimensional).



Here is the formula for kernel density estimation in one dimension:

$$\hat{f}_{h}(x) =\frac{1}{nh}\sum_{i=1}^{n}K\left(\frac{x-x_i}{h}\right)=\frac{1}{n}\sum_{i=1}^{n}K_h\left(x-x_i\right)$$

where scaled kernel $K_h$ is defined as:
$$K_h(x-x_i) := \frac{1}{h}K(\frac{x-x_i}{h})$$

Example

We have two training points located at $0.5$, and $0.7$ in one dimension.

If we assume query point is $x=0.5$, and $h$ is 0.3 (for rectangle and triangle, the scaled width is 0.6 from -0.3 to 0.3), for rectangle kernel we have:
$$\begin{align*}
\hat{f}_{0.3}(0.5) &amp;=\frac{1}{2*0.3}\left(rect \left(\frac{0.5-\color{blue}{0.5}}{0.3} \right)+rect \left(\frac{0.5-\color{blue}{0.7}}{0.3} \right)\right) \\
&amp;= \frac{1}{0.6}(0.5+0.5) = 1.67
\end{align*}$$
and for triangle kernel we have:
$$\begin{align*}
\hat{f}_{0.3}(0.5) &amp;=\frac{1}{2*0.3}\left(tri \left(\frac{0.5-\color{blue}{0.5}}{0.3} \right)+tri \left(\frac{0.5-\color{blue}{0.7}}{0.3} \right)\right) \\
&amp; = \frac{1}{0.6}(1+0.33) = 2.22
\end{align*}$$

For square kernel $sqr$, kernel outputs 1 for each point inside the hyper-cube, and $h$ would be the width of scaled hyper-cube (from $-h/2$ to $h/2$). For the square kernel, estimation in $d$ dimension is:
$$\hat{f}_{h}(x) =\frac{1}{nV}\sum_{i=1}^{n}sqr\left(\frac{x-x_i}{h}\right)=\frac{1}{nV}k=\frac{\frac{k}{n}}{V}$$

where $k$ is the number of points inside a hyper-cube centered at $x$ with width $h$, and $V=h^d$ is the size of hyper-cube in $d$ dimensions (length of a segment in one dimension).

The previous example for $h=0.3$ would be:
$$\begin{align*}
\hat{f}_{0.3}(0.5) &amp;=\frac{1}{2*0.3}\left(sqr \left(\frac{0.5-\color{blue}{0.5}}{0.3} \right)+sqr \left(\frac{0.5-\color{blue}{0.7}}{0.3} \right)\right) \\
&amp; = \frac{1}{0.6}(1+0) = \frac{\frac{1}{2}}{0.3} = 1.67
\end{align*}$$
Note that these are estimations for density not probability, thus, they are allowed to be larger than one (imagine a rectangle with width 0.5 and height 2 as pdf of a continuous random variable).