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We have to show:

Let $\theta_0$ be a k-dim vector. Show, that the following statements are equivalent:

(1) $\hat{\theta}_n$ is consistent for $\theta_0$.

(2) For each component $i= 1,...,k$: $\hat{\theta}_{n,i}$, is $\hat{\theta}_{n,i}$ consistent for $\theta_{0,i}$, where $\hat{\theta}_{n,i}$ and $\theta_{0,i}$ are the i'th component of the respective vector.

Hint: $||x||_\infty \leq ||x||_2 \leq \sqrt{k}||x||_\infty$ as well as: $P(A+B \geq \varepsilon) \leq P(A\geq \frac{\varepsilon}{2}) + P(B\geq \frac{\varepsilon}{2})$

Any suggestions would be greatly appreciated.

So an estimator is consistent if:

$\lim_{n \rightarrow \infty} P(|\hat{\theta}_n - \theta| > \varepsilon) = 0$, however using the inequality: $P(A+B \geq \varepsilon) \leq P(A\geq \frac{\varepsilon}{2}) + P(B\geq \frac{\varepsilon}{2})$ doesn't make much sense here (from my understanding).

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    $\begingroup$ Could you give some context about what you have tried so far? $\endgroup$ Apr 13 '21 at 18:53
  • $\begingroup$ Just edited my question. $\endgroup$ Apr 13 '21 at 18:55
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    $\begingroup$ I think there is a small typo - it should be $i = 1,\ldots,k$. Based on the hint, I am going to assume that $\hat \theta_n$ is consistent for $\theta_0$ means specifically that for fixed $\varepsilon, \delta$, there eixsts $N$ such that $\mathbb P(||\hat\theta_n - \theta_0||_2 > \varepsilon) < \delta$ for $n \geq N$. Can you show that (2) is the same as the above statement holding, but with the $L-\infty$ norm in place of the $L2$ norm? $\endgroup$ Apr 13 '21 at 18:59
  • $\begingroup$ you mean like: $P(||\hat{\theta}_n - \theta||_\infty \geq \frac{\varepsilon}{\sqrt{k}})$? $\endgroup$ Apr 13 '21 at 19:07
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    $\begingroup$ Ok, so the argument is, that if $max(|\hat{\theta}_{n,i} -\theta_{n,i}| > \frac{\varepsilon}{\sqrt{k}})$ the so are all the others, and hence the statement is proved, correct? $\endgroup$ Apr 13 '21 at 19:15

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