Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know empirical distribution function is defined as $F_n(x)=\frac{1}{n}\sum\limits_{i=1}^nI(X_i \leq x)$. Then define empirical density function as $ f_n(x) = \frac{F_n(x+b_n)-F_n(x-b_n)}{2b_n} $ .

I can show by using definition that $2nb_nf_n(x)$ is distributed binomial as $B(n,F(x+b_n)-F(x-b_n))$. Further, $E(f_n(x))=\frac{F(x+b_n)-F(x-b_n)}{2b_n}$. Hence as $b_n\rightarrow 0$, $E(f_n(x))\rightarrow f(x) $.

Now, I wanna show that $Var(f_n(x))\rightarrow0$ if $b_n\rightarrow 0$ and $nb_n\rightarrow\infty $. However, I cant find the exact form of $Var(f_n(x))$ by explicit calculation. Can anyonehelp me with it ?Thanks in advance

share|cite|improve this question
up vote 0 down vote accepted

Since $2nb_nf_n(x)$ is binomial $(n,p_n)$ with $p_n=F(x+b_n)-F(x-b_n)$, its expectation and variance are $np_n$ and $np_n(1-p_n)$ respectively. Thus, $\mathbb E(f_n(x))=np_n/(2nb_n)=p_n/(2b_n)$, as you said, and $\mathrm{var}(f_n(x))=np_n(1-p_n)/(2nb_n)^2=p_n(1-p_n)/(4nb_n^2)$.

Assume that $b_n\to0$. If $f$ is regular enough, then $p_n/b_n\to2f(x)$ hence $\mathbb E(f_n(x))\to f(x)$, as you said. Likewise, $p_n\to0$ hence $\mathrm{var}(f_n(x))\sim p_n/(4nb_n^2)\sim f(x)/(2nb_n)$. Thus, $\mathrm{var}(f_n(x))\to0$ if and only if $nb_n\to\infty$.

This shows that the regime of interest is indeed $1\ll nb_n\ll n$, for example one can use $b_n=n^{-\beta}$ for any $\beta$ in $(0,1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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