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Let $\Vert A \Vert$ denote the spectral norm of a random matrix. Let $x$ and $u_k$ be N$\times$T matrices. Denote $\beta \cdot u = \sum_{k=1}^K\beta_ku_k $, where $\beta$ is a K-vector and $\beta_k$ a scalar. Further, let $\Vert x \Vert = O_p(\sqrt N)$ and $\Vert u_k \Vert = O_p(\sqrt{NT})$ for all $k=1,...,K$ as N and T grow at the same rate.

Why do we have $\Vert x \Vert \gg \Vert(\hat\beta-\beta)\cdot u\Vert $ asymptotically if the convergence rate of $\hat\beta$ to $\beta$ is faster than $\sqrt N$?

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  • $\begingroup$ "...convergence rate of $\hat\beta$ is faster than $\sqrt N$ (that is, if $\Vert\hat\beta-\beta\Vert=o_p(\sqrt N)$..." So according to you, if $\Vert\hat\beta-\beta\Vert=O(N^{ \frac{1}{4} }) = o_p(\sqrt N)$, i.e. $\Vert\hat\beta-\beta\Vert$ diverges to $\infty$ at the rate $N^{ \frac{1}{4} }$, then "...convergence rate of $\hat\beta$ is faster than $\sqrt N$..."? $\endgroup$ – Michael Jul 13 at 14:07
  • $\begingroup$ Thanks for pointing it out, @Michael. The "if" in the parenthesis should not have been there. Now should be correct, I was just trying to clarify that it is the rate of convergence of the estimator to its true value $\beta$ what I referred to. $\endgroup$ – econ86 Jul 13 at 14:20
  • $\begingroup$ What does $\gg$ mean? What is $\cdot$ ? In general, for any operator norm, one has the formula, $\| A\xi \| \leq \|A\| \|\xi\|$, where $\|A\|$ is the operator norm of $A$, and $\| \xi \|$ and $\| A\xi \|$ are vector norms. If you get your formulation straight, the claim probably follows immediately. $\endgroup$ – Michael Jul 13 at 14:34
  • $\begingroup$ $f(s) \ll g(s)$ is equivalent to $f(s) = O(g(s))$ and $\cdot$ is just the dot product $\endgroup$ – econ86 Jul 13 at 15:00
  • $\begingroup$ dot product between the (presumably) vector β^−β and the matrix u ("Let x and u be N by T matrices...")? As I said, try applying the standard operator norm estimate on the quantity ||u(β^−β)||, and your claim follows. $\endgroup$ – Michael Jul 13 at 15:06

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