Suppose that we use the kernel density estimate


With the kernel function $K\left(u\right)=\left(1-\left\vert{}u\right\vert{}\right)I\left(\left\vert{}u\right\vert{}\leq{}1\right)$and bandwidth h to estimate the probability density function f(x). Since the cumulative distribution function, $F\left(x\right)=\int_{-\infty{}}^xf\left(u\right)du$

F(x) can be estimated by ${\hat{F}}_h\left(x\right)=\int_{-\infty{}}^x{\hat{f}}_h\left(u\right)du$

Compute ${\hat{F}}_h\left(x\right)$

I tried to substitute all the information I have into the overall equation but I am having trouble integrating both the absolute and indicator function.

  • 1
    $\begingroup$ Please explain what you mean by "computing" $\hat F_h(x)$. Exactly what data or inputs would you begin with and what output do you expect? Although you refer to "all the information I have," you don't appear to have told us anything about it. $\endgroup$
    – whuber
    Commented Mar 1, 2016 at 19:33
  • $\begingroup$ @whuber Sorry if I wasn't clear. That was the full question given to me without any data. So the answer is supposed to be a general formula. $\endgroup$
    – Stongals
    Commented Mar 2, 2016 at 3:23

1 Answer 1


Please check, I am tired :


$=\frac{1}{nh}\sum_{i=1}^{n}\int_{-\infty}^{x}(1-|\frac{u−X_i}{h}|)\mathbf{1}_{|\frac{u−X_i}{h}|\leq 1}du$

$=\frac{1}{nh}\sum_{i=1}^{n}\int_{-\infty}^{\frac{x−X_i}{h}}(1-|y|)\mathbf{1}_{|y|\leq 1}h dy$

$=\frac{1}{n}\sum_{i=1}^{n}\int_{-\infty}^{+\infty}(1-|y|)\mathbf{1}_{-1\leq y\leq 1}\mathbf{1}_{y\leq\frac{x−X_i}{h}}dy$

$=\frac{1}{n}\sum_{i=1}^{n}[\int_{-1}^{\min(1,\frac{x−X_i}{h})}(1-|y|)dy] \mathbf{1}_{-1<\frac{x−X_i}{h}}$



  • $\begingroup$ Thank you! I have some parts that I do not understand though. Why did the h disappear from step 2 to 3? In the last 2 steps, I think it should be 3/2 instead of 1/2. The +1 was missing. Also, would it be possible to simplify the equation further? $\endgroup$
    – Stongals
    Commented Mar 2, 2016 at 3:32
  • $\begingroup$ In step 3, h disappears because of the variable change $y=(x-X_i)/h$. $\endgroup$
    – Jacky1
    Commented Mar 2, 2016 at 6:43
  • $\begingroup$ The last step is ok because the integral of |y| is y|y|/2, and I just factorize the result by $min$. I think it is maybe possible to simplify further. You can use indicator functions and distinguish the cases where $(x-X_i)/h\geq1$ and $(x-X_i)/h<1$. $\endgroup$
    – Jacky1
    Commented Mar 2, 2016 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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