# Kernel density estimator function explanation needed

I'm studying about kernel density estimation and from wikipedia I get this formula:

$$\hat{f}_h(x, h) = \frac{1}{n}\sum^{n}_{i=1}K_h(x-x_i) = \frac{1}{nh}\sum_{i=1}^nK(\frac{x-x_i}{h}).$$

I think I've got the basic idea of this formula, but the smoothing term $h$ is what troubles me. If my kernel function was the Gaussian:

$$K(u)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}u^2}$$

I would get my estimator to be:

$$\hat{f}_h(x, h) = \frac{1}{nh\sqrt{2\pi}}\sum_{i=1}^n e^{-\frac{1}{2}u^2},$$

where $\displaystyle u = \frac{x-x_i}{h}.$ Why is the $h$ term in this formula? What is its function? Why do we divide the sum by $h$ etc. Can someone clarify this a bit?

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Notice that your density function is an average of density functions of gaussians with standard deviation $h$ and different centers. Does that clarify why you're using this formula? –  eqperes May 7 at 7:45
+1 @eqperes yes I thought so that $h$ is related to the standard deviation. So does the sum $\sum e^{(-1/2) u^2} = nh\sqrt{2\pi}$ in the maximum case? Sry if my question is dumb, I'm bit mixed up with this. –  jjepsuomi May 7 at 7:52
Well, no. That will depend on all your $x_{i}$ points. But, for me to understand better your questions, why do you think it should be equal to $nh\sqrt{2\pi}$ in the maximum? –  eqperes May 7 at 8:13
+1 darmet x) sry @eqperes I thought about the function as a probability all the time and not as a probability density. That's why I thought the sum should equal $nh\sqrt{2\pi}$ in the maximun case so that the overall value of the function would (when divided with the normalizing term $1/nh\sqrt{2\pi}$) equal probability value 1. I need to further examine this function, I think I don't understand it yet well enough to make clear questions about it. Maybe you could explain about this function as an answer? I could then accept it as an answer? Thank you for your effort! :) –  jjepsuomi May 7 at 8:36
+1 @eqperes I got it =) no need to answer anymore, I integrated the density and it clarified it :) –  jjepsuomi May 7 at 9:10

$$K(y) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2}$$
If we have the $n$ observations: $\mu_1, ..., \mu_n$ and $y_i = \displaystyle \frac{x-\mu_i}{h}$, then
$$\int_{-\infty}^{\infty}\hat{f}_h(x)\;dx=\int_{-\infty}^{\infty}\frac{1}{nh}\sum_{i=1}^nK(y_i)\;dx = \int_{-\infty}^{\infty}\frac{1}{nh}\sum_{i=1}^nK(\frac{x-\mu_i}{h})\;dx$$
$$=\frac{1}{nh\sqrt{2\pi}}\int_{-\infty}^{\infty}\sum_{i=1}^n e^{-\frac{1}{2}(\frac{x-\mu_i}{h})^2}\;dx = \frac{1}{nh\sqrt{2\pi}}\left[ \int_{-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x-\mu_1}{h})^2}+\cdots+\int_{-\infty}^{\infty}e^{-\frac{1}{2}(\frac{x-\mu_n}{h})^2}\right] = \frac{1}{nh\sqrt{2\pi}}\cdot nh\sqrt{2\pi}=1.$$