Deriving Local-to-Unity Matrix Limits with a Floored Time Index

# Question on the use of a limit in a proof


# Question on the use of a limit in a proof












I ran into a step in an argument that I can't quite figure out. It's basically how they use a limit that I don't seem to understand. The context is local-to-unity asymptotics in vector autoregressions, so I figured some people doing empirical work might know.

We have a $K \times K$ diagonal matrix $\Lambda := \text{diag}\left( \lambda_1, \dots, \lambda_K \right)$ and we define $h := \left[ \delta T \right]$ where $[.]$ refers to the integer part of a number and $\delta \in (0,1)$ is some fraction. We'd like to know what happens to $\Lambda^h$ as $T \rightarrow \infty$. The textbook notes that \begin{align*} \lim_{T \rightarrow \infty} \left( 1 + \frac{\delta \lambda_k}{T} \right)^T = e^{\delta \lambda_k} \; \forall k=1,\dots,K \end{align*} and infers that $\Lambda^h \rightarrow e^{\delta \Lambda} := \text{diag}\left( e^{\delta \lambda_1}, \dots, e^{\delta \lambda_K} \right)$. I don't quite follow what is going on. I see that the first limit is correct, but I fail to see how the conclusion follows. If it helps, later the authors actually use the reasoning to claim that $C^h \rightarrow e^{\delta C}$, but $C := I_K + \Lambda/T$. Now, in that case we have $\left( C^h \right)_{k,k} = \left( 1 + \frac{\lambda_k}{T} \right)^{[\delta T]}$. I still am having trouble seeing how the hell $\delta$ can be brought inside to invoke the above argument, but the expression is closer.

Anyone can help me out here?

## Answer by RRL (score 4, accepted)

https://quant.stackexchange.com/a/57896

It makes no sense to write $C^h \to e^{\delta C}$ as $T \to \infty$ when $C = I_K +\Lambda/T$ since $e^{\delta C}$ on the right-hand side depends on $T$.

What can be confirmed is $(C^h)_{k,k} \to e^{\delta \lambda_k}$ as $T \to \infty$ with $\delta$ fixed. Note that

$$\log (C^h)_{k,k} = \lfloor \delta T\rfloor \log\left(1 + \frac{\lambda_k}{T}\right) = \frac{\lfloor \delta T\rfloor}{T} \log\left(1 + \frac{\lambda_k}{T}\right)^T$$

As $T \to \infty$ we clearly have $\log\left(1 + \frac{\lambda_k}{T}\right)^T \to \log e^{\lambda_k} = \lambda _k$, since $\log(\cdot)$ is continuous.

Since $ \lfloor \delta T\rfloor \leqslant \delta T < \lfloor \delta T\rfloor +1$, it follows that $ \delta T -1 < \lfloor \delta T\rfloor \leqslant \delta T$, and $$\delta - \frac{1}{T} < \frac{\lfloor \delta T\rfloor}{T} \leqslant \delta$$.

Hence, by the squeeze theorem, $ \frac{\lfloor \delta T\rfloor}{T} \to \delta$ as $T \to \infty$, and

$$\lim_{T \to \infty}\log (C^h)_{k,k} = \lim_{T \to \infty}\frac{\lfloor \delta T\rfloor}{T} \lim_{T \to \infty}\log\left(1 + \frac{\lambda_k}{T}\right)^T = \delta \lambda_k$$

Therefore, by continuity of $\log(\cdot)$,

$$\lim_{T \to \infty} (C^h)_{k,k} = e^{\delta \lambda_k}$$