Suppose $\hat{\theta}_1 = O_p(n^{-1/2})$ and $\hat{\theta}_2 = O_p(n^{-1/2})$, what is $\sqrt{\hat{\theta}_1\hat{\theta}_2}$? Suppose $\hat{\theta}_1 = O_p(n^{-1/2})$ and $\hat{\theta}_2 = O_p(n^{-1/2})$, what is the big $O_p$ for $\sqrt{\hat{\theta}_1\hat{\theta}_2}$?
I think $\hat{\theta}_1\hat{\theta}_2 = O_p(n^{-1/2})O_p(n^{-1/2})=O_p(n^{1/4})$. However, what is the big $O_p$ for $\sqrt{\hat{\theta}_1\hat{\theta}_2}$?
 A: More generally, if you have $X=O_p(a_n)$ and $Y=O_p(b_n)$ for two independent random variables $X$ and $Y$, you can show that $XY = O_p(a_nb_n)$.
Indeed, by definition, for any $\epsilon>0$, you can find $M_X,M_Y$ and $N_X,N_Y$ such that for all $n>N_X$ (respectively $N_Y$), you have
$$\mathbb P(X/a_n>M_X)\le\epsilon\ \text{ and }\ \mathbb P(Y/b_n>M_Y)\le\epsilon$$
Now for any $\epsilon>0$, if you take $M:=M_X\times M_Y$, and let $N:=\max\{N_X,N_Y\}$, it follows that for all $n\ge N$
$$\begin{align}\mathbb P(XY/a_nb_n>M)&=\mathbb P((X/a_n)\times(Y/b_n)>M) \\
&\le\mathbb P\left(\{X/a_n>M_X\}\cap\{Y/b_n>M_Y\}\right)\\
&=\mathbb P(X/a_n>M_X)\times\mathbb P(Y/b_n>M_Y)\\
&\le \epsilon^2\end{align}$$
So letting $\epsilon'=\epsilon^2$ you get the desired statement.
By a similar argument, you can show that if $\varphi$ is a monotonically increasing function and $Z$ a random variable such that $Z=O_p(c_n)$, then $\varphi(Z) = O_p(\varphi(c_n))$.
With all that you should be able to conclude that $\sqrt{\hat{\theta}_1\hat{\theta}_2} = O_p(n^{-1/2})$, as Christoph Hanck rightfully pointed out.
