I want to understand whether the following holds as $N \rightarrow \infty$:
Given: $a_N = O_p \left(N^{-\frac{1-\rho}{2}} \right)$, $b_N = o_p(1)$, $1/2 \leq \rho < 1$
Then: $a_N + b_N = o_p(1)$.
Using the following property:
$o_p(a_n) + O_p(b_n) = O_p(a_n+b_n)$
I have that:
$a + b = O_p \left(N^{-\frac{1-\rho}{2}} + 1\right)$
But does this mean that the quantity is also $o_p(1)$? My reasoning is as follows:
By definition, $a N^{\frac{1-\rho}{2}} = O_p(1)$. Since the exponent is positive, $N^{\frac{1-\rho}{2}} \rightarrow \infty$. So if $a$ is bounded when multiplied by a number that goes to infinity, then $a$ should converge to zero when not multiplied by infinity. Is this correct? And is there an explicit property I should refer to?
You have to prove that $a_N+b_N$ (it is better to keep indices) converges in probability to zero. Since the sum of two sequence that converge to $0$ in probability also converges in probability, it suffices to show that $a_N\to 0$ in probability. Letting $p=(1-\rho)/2$, which is positive, we have that $N^{p}a_N=O_{\mathbb P}(1)$, which means that for each positive $\eta$, there exists $M(\eta)$ such that $\sup_{N\geqslant 1}\mathbb P\left(N^{p}\lvert a_N\rvert>M(\eta)\right)<\eta$.
For a fixed positive $\varepsilon$, take $N_0$ such that $N_0^{-p}M(\eta)<\varepsilon$. Then for each $N\geqslant N_0$, $\mathbb P(\lvert a_N\rvert>\varepsilon)<\eta$, which shows the convergence in probability.