0
$\begingroup$

Say we have constants $v \in \mathbb{R}$ and $w >0$. Assume that $a_n \to 0$ and $b_n \to 0$ as $n \to \infty$. I wish to show that

$$ \frac{v+O_p(a_n)}{w+O_p(b_n)}=\frac{v}{w}+O_p(a_n) $$

where we assume that $w+O_p(b_n)\neq 0$ for all $n$. Now, I am not 100% sure the statement is even correct, but so far I have tried the following:

Letting $Y_n =O_p(b_n)$ and $X_n = O_p(a_n)$ we show that $$ \frac{v+X_n}{w+Y_n}-\frac{v}{w}=\frac{wX_n-vY_n}{w(w+Y_n)} \in O_p(a_n) $$

Since $Y_n \in O_p(b_n)$ and $b_n \to 0$ we in particular have that $Y_n \in o_p(1)$, i.e. $Y_n \to 0$ in probability. Thus, picking $0 < \epsilon < w$

\begin{align} P\left( \left| \frac{wX_n-vY_n}{w(w+Y_n)} \right| > Ca_n \right) &\leq P\left( \left| \frac{wX_n}{w(w-\epsilon)} \right| +\left| \frac{vY_n}{w(w-\epsilon)} \right| > Ca_n,|Y_n| \leq \epsilon \right)+P(|Y_n| > \epsilon) \\ &\leq P\left( \left| X_n \right| > CK_1 a_n \right) + P\left( \left| Y_n \right| > CK_2 a_n \right)+P(|Y_n| >\epsilon) \end{align} We need the last expression to converge to $0$ as first $n \to \infty$ and then $C \to \infty$. The first and last term both go to zero - but the problem is the term $$ P\left( \left| Y_n \right| > CK_2 a_n \right) $$ which I believe will only converge to $0$ if $Y_n \in O_p(a_n)$. Since $Y_n \in O_p(b_n)$, we will have $Y_n \in O_p(a_n)$ only when $b_n = O(a_n)$. My question is then if it is possible in the case where we do not have $b_n = O(a_n)$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Your calculation seems correct to me! Notice that if in your calculation $X_n \equiv 0$, then clearly the rate is going to depend on $b_n$. I.e. the statement that the rate only depends on $a_n$ will not hold in general. You might see if you can show that the result can be established with rate $O_p( \max\{a_n,b_n\})$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .