$ \frac{(1+o_p(1) )A_n}{(1+o(1) )E[A_n]}=\frac{A_n}{E[A_n]}+ o_p(\frac{A_n}{E[A_n]}) ??$

Question

I was reading a paper and in it, they stated for a random sequence $ A_n=\Theta(1)$,

$$ \frac{(1+o_p(1) )A_n}{(1+o(1) )E[A_n]}=\frac{A_n}{E[A_n]}+ o_p(\frac{A_n}{E[A_n]}) ??$$

without any proof. Although seemingly intuitive, I couldnt come up with a proof, and this actually is a crucial step, so I wanted to make sure.
Also, if possible, can you tell me if there is a easy way to estimate the precise order of $ o_p(\frac{A_n}{E[A_n]}) $ ,other than actually calculating $ \frac{(1+o_p(1) )A_n}{(1+o(1) )E[A_n]}-\frac{A_n}{E[A_n]}$,which is quite tedious in this case.
Thanks beforehand.

Answer 1

The term $ \frac{(1+o_p(1) )A_n}{(1+o(1) )E[A_n]}$ refers to something that can be expressed as $$ \frac{(1+Y_n )A_n}{(1+\delta_n )E[A_n]}, $$ where $(Y_n)_n$ is a sequence of random variables that converges in probability to $0$ and $(\delta_n)_n$ is a sequence of real numbers converging to 0$.

One has $$ \frac{(1+Y_n )A_n}{(1+\delta_n )E[A_n]}=\frac{A_n}{E[A_n] }+\frac{A_n}{E[A_n] }\underbrace{\left(\frac{1+Y_n }{1+\delta_n }-1\right)}_{o_p(1)}. $$ The term $o_p\left(\frac{A_n}{E[A_n]}\right)$ can therefore be expressed as $$ \frac{A_n}{E[A_n] }+\frac{A_n}{E[A_n] } \frac{Y_n-\delta_n }{1+\delta_n } . $$