Why do we divide the kernel estimate function by h and also the entire equation by h?

Question

I believe the explanation is that the kernel estimate function would integrate to 1. But I do not quite understand the intuition behind it. How would dividing by h help to make the function integrate to 1. Also, what is the output of the kernel k(.) if it were Uniform. EXPLANATION

Answer 1

How would dividing by h help to make the function integrate to 1.

Like this: $$\begin{align} &\int_{-\infty}^\infty\hat{f_h}(x)dx\\ &=\int_{-\infty}^\infty \frac{1}{n}\sum_1^n\frac{1}{h}K\left(\frac{x-x_i}{h}\right) dx\\ &=\frac{1}{n}\frac{1}{h}\sum_1^n\int_{-\infty}^\infty K\left(\frac{x-x_i}{h}\right) dx\\ &=\frac{1}{n}\frac{1}{h}\sum_1^n\int_{-\infty}^\infty K\left(\frac{x-x_i}{h}\right)h\,d\left(\frac{x-x_i}{h}\right)\\ &=\frac{1}{n}\sum_1^n\int_{-\infty}^\infty K\left(\frac{x-x_i}{h}\right)\,d\left(\frac{x-x_i}{h}\right)\\ &=\frac{1}{n}\sum_1^n1\\ &=1 \end{align} $$ where we've used the fact that $K$ is a probability density function, i.e.

$$\int_{-\infty}^\infty K(t)dt = 1. $$

what is the output of the kernel k(.) if it were Uniform

If uniform on interval $(-a,a)$, then the kernel is $$K(t) = \frac{1}{2a}[-a<t<a] $$ and $$\begin{align} \hat{f_h}(x)&=\frac{1}{n}\sum_1^n\frac{1}{h}K\left(\frac{x-x_i}{h}\right)\\ &=\frac{1}{n}\frac{1}{h}\frac{1}{2a}\sum_1^n\left[-a<\frac{x-x_i}{h}<a\right]\\ &=\frac{1}{2ah}\left(\frac{1}{n}\sum_1^n\left[x-ah<x_i<x+ah\right]\right)\\ &=\frac{1}{2ah}\hat{P}(x-ah<X<x+ah) \end{align}$$ where $\hat{P}(.)$ is an estimate of the probability of the "small" interval $(x-ah,x+ah)$ of width $2ah$.

NB: Note how this gives $$\hat{f_h}(x)\,2ah\approx P(x-ah<X<x+ah) $$ compared to the actual density function $f(x)$, which has the property that for "small" $dx$, $$f(x)\,dx\approx P\left(x-\frac{dx}{2}<X<x+\frac{dx}{2}\right).$$