# Rate of convergence and asymptotic dominance in $\Vert x \Vert \gg \Vert(\hat\beta-\beta)\cdot u\Vert$

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 $$\sqrt N$$?

• "...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 $\sqrt N$..."? – Michael Jul 13 '19 at 14:07
• 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. – econ86 Jul 13 '19 at 14:20
• 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. – Michael Jul 13 '19 at 14:34
• $f(s) \ll g(s)$ is equivalent to $f(s) = O(g(s))$ and $\cdot$ is just the dot product – econ86 Jul 13 '19 at 15:00
• 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. – Michael Jul 13 '19 at 15:06