Comparing $\left\| A_n^{-1/2} B_n A_n^{-1/2}-I_k\right\|\xrightarrow{\ \ \mathbb P\ \ } 0$ and $\|B_n - A_n\|\xrightarrow{\ \ \mathbb P\ \ } 0$

Question

Let $(A_n)$ and $(B_n)$ be two sequences of random $k \times k$ positive definite matrices. Let $I_k$ be the identity matrix of size $k$, and $\left\|M\right\|=\sqrt{\mathrm{trace}(M^T M)}$ for any $M \in \mathbb R^{k \times k}$. I need to compare the two following conditions:

1. $\left\|A_n^{-1/2} B_n A_n^{-1/2} - I_k\right\| \xrightarrow{\ \ \mathbb P\ \ } 0$

2. $\left\|B_n - A_n\right\| \xrightarrow{\ \ \mathbb P\ \ } 0$

i.e.

1. $\mathbb P\!\left(\left\|A_n^{-1/2} B_n A_n^{-1/2} - I_k\right\| > \varepsilon\right) \to 0$ for every $\varepsilon >0$

2. $\mathbb P\!\left(\left\|B_n - A_n\right\| > \varepsilon\right) \to 0$ for every $\varepsilon >0$

What I am told is that the two conditions are different, “because $\left\|A_n\right\| \to 0$ in most applications” (I’m omitting the context because I’m fairly sure it isn’t relevant). But I fail to see if any of the two is stronger than the other, and what is the exact role of $\left\|A_n\right\|$ tending to $0$.

All I have done is noticing that, by the sub-multiplicativity of $\|\cdot\|$,

• $\left\|A_n^{-1/2} B_n A_n^{-1/2} - I_k\right\| = \left\|A_n^{-1/2} (B_n - A_n) A_n^{-1/2}\right\| \leq \left\|A_n^{-1/2}\right\|^2 \left\|B_n - A_n\right\|$

  $\implies \mathbb P\!\left(\left\|A_n^{-1/2} B_n A_n^{-1/2} - I_k\right\| > \varepsilon \right) \leq \mathbb P\!\left(\left\|A_n^{-1/2}\right\|^2 \left\|B_n - A_n\right\| > \varepsilon\right)$

• $\left\|B_n - A_n\right\| = \left\|A_n^{1/2} (A_n^{-1/2} B_n A_n^{-1/2} - I_k) A_n^{1/2}\right\| \leq \left\|A_n^{1/2}\right\|^2 \left\|A_n^{-1/2} B_n A_n^{-1/2} - I_k\right\|$

  $\implies \mathbb P\!\left(\left\|B_n - A_n\right\| > \varepsilon\right) \leq \mathbb P\!\left(\left\|A_n^{1/2}\right\|^2 \left\|A_n^{-1/2} B_n A_n^{-1/2} - I_k\right\| > \varepsilon\right)$

• $\left\|A_n^{-1/2}\right\|^2 \geq \left\| \left( A_n^{-1/2} \right)^2 \right\| = \left\|A_n^{-1}\right\| \geq \left\|A_n\right\|^{-1}$

• $\left\|A_n^{1/2}\right\|^2 \geq \left\| \left( A_n^{1/2} \right)^2 \right\| = \left\|A_n\right\|$

Any help is much appreciated.

Answer 1

Remark that $\Vert A_n \Vert \overset{ \mathbb P}{ \to} 0 \iff A_n \overset{ \mathbb P}{ \to} 0 $ (the null matrix). So, if $\Vert A_n \Vert \overset{ \mathbb P}{ \to} 0 $, then 1. implies 2. , using that $$ \Vert B_n - A_n \Vert \leq \Vert A_n^{1/2} \Vert^2 \Vert A_n^{-1/2} B_n A_n^{-1/2} - I \Vert $$ However, if $\Vert A_n \Vert \overset{ \mathbb P}{ \to} 0 $ then $$ \Vert B_n - A_n \Vert \leq \Vert B_n \Vert + \Vert A_n \Vert $$ tends to zero in probability for any $B_n$ such that $\Vert B_n \Vert \overset{ \mathbb P}{ \to} 0$. Thus, we can easily build a counter-example to show that 2. does not imply 1. in this case.

Conversely, $$ \Vert A_n^{-1/2} \Vert^2 \Vert A_n^{-1/2} B_n A_n^{-1/2} - I \Vert \leq \Vert B_n - A_n \Vert $$ so if $\Vert A_n^{-1} \Vert$ is bounded in probability, then so is $\Vert A_n^{-1/2} \Vert$ and 2. implies 1.