simplifying an asymptotic expression

Question

I have this expression in a statistics book, namely $nh(f(x) +o(1)+O_p(1/\sqrt{nh}))$. Where $f$ is a density function. Now, this expression is equal to $nhf(x)\{1+o_p(1)\}$. Note, that $n\to \infty$, $h\to 0$, and $nh\to \infty$. I do appreciate any comment on how to get this final expression. Thanks in advance.

Answer 1

We are given an expression

$$E=nh\left(f(x)+o(1)+O_p\left(\sqrt{1/nh}\right)\right).$$

Provided that $f(x) \neq 0$ (a necessary assumption not given in your statement, but hopefully present in context), we obtain

$$E=nhf(x)\left(1+o(1)+O_p\left(f(x)^{-1} \sqrt{1/nh}\right)\right).$$

We are taking the limit independently of $x$, so the constant $f(x)^{-1}$ may be absorbed into the $O$-notation:

$$E= nhf(x)\left(1+o(1)+O_p\left(\sqrt{1/nh}\right)\right).$$

Finally, we use that $O_p(\sqrt{1/nh}) =o_p(1)$ as $nh \to \infty$, to get $$E=nhf(x)\left(1+o(1)+o_p(1)\right)=nhf(x)\left(1+o_p(1)\right),$$ in which we've used that $o(1)=o_p(1)$.