What's the rate of $\log|1+\frac{O_p(M^{-1/2})}{fu}|$

Question

What's the rate of $\log|1+\frac{O_p(M^{-1/2})}{fu}|$, where f is a real valued constant, u follows standard normal distribution and hence $u=O_p(1)$. So $\log|1+\frac{O_p(M^{-1/2})}{fu}|=O_p(?)$. I do not know how to deal with the absolute values. If the case is $\log(1+\frac{O_p(M^{-1/2})}{fu})$, I think the result is $O_p(M^{-1/2})$ since $\log(1+x)\sim x$ when $x\rightarrow 0$.

Answer 1

The conclusion is right, but the argument isn't. First, you need to get the order of $1/u$, and you can't conclude anything useful about it just from $u=O_p(1)$. However, $1/u$ is an almost-surely finite random variable not depending on $M$, so $1/u=O_p(1)$. As you argued, $O_p(M^{-1/2})/fu$ is thus $O_p(M^{-1/2}$.

Now you have to worry about the absolute value. As $M\to\infty$, the probability of $1+O_p(M^{-1/2})<0$ goes to zero, so the absolute value is asymptotically unimportant, and as $\log(1+x)\sim x$ for small $x$ $$\log\left|1+\frac{O_p(M^{-1/2})}{fu}\right|=O_p(M^{-1/2})$$