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I couldn't find the right way to use the suggestions in the comments and answers but think about ways of using the Slutsky's theorem (sugested by @Glen_b) I came up with a proof using a nice properties fromproperty of the normal distribution.

Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have

$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$ $$b_n = \frac{1}{\sqrt{2\log n}}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$ (a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)

Now $$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$

and after some manipulation $$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$ We have $$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$ $$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$

And the result follows from Slutsky's theorem.

I couldn't find the right way to use the suggestions in the comments and answers but think about ways of using the Slutsky's theorem (sugested by @Glen_b) I came up with a proof using a nice properties from the normal distribution.

Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have

$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$ $$b_n = \frac{1}{\sqrt{2\log n}}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$ (a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)

Now $$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$

and after some manipulation $$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$ We have $$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$ $$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$

And the result follows from Slutsky's theorem.

I couldn't find the right way to use the suggestions in the comments and answers but think about ways of using the Slutsky's theorem (sugested by @Glen_b) I came up with a proof using a nice property of the normal distribution.

Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have

$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$ $$b_n = \frac{1}{\sqrt{2\log n}}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$ (a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)

Now $$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$

and after some manipulation $$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$ We have $$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$ $$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$

And the result follows from Slutsky's theorem.

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I couldn't find the right way to use the suggestions in the comments and answers but think about ways of using the Slutsky's theorem (sugested by @Glen_b) I came up with a proof using a nice properties from the normal distribution I was able to produce a proof.

Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have

$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$ $$b_n = \frac{1}{\sqrt{2\log n}}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$ (a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)

Now $$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$

and after some manipulation $$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$ We have $$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$ $$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$

And the result follows from Slutsky's theorem.

I couldn't find the right way to use the suggestions in the comments and answers but using a properties from the normal distribution I was able to produce a proof.

Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have

$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$ $$b_n = \frac{1}{\sqrt{2\log n}}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$ (a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)

Now $$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$

and after some manipulation $$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$ We have $$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$ $$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$

And the result follows from Slutsky's theorem.

I couldn't find the right way to use the suggestions in the comments and answers but think about ways of using the Slutsky's theorem (sugested by @Glen_b) I came up with a proof using a nice properties from the normal distribution.

Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have

$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$ $$b_n = \frac{1}{\sqrt{2\log n}}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$ (a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)

Now $$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$

and after some manipulation $$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$ We have $$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$ $$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$

And the result follows from Slutsky's theorem.

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source | link

I couldn't find the right way to use the suggestions in the comments and answers but using a properties from the normal distribution I was able to produce a proof.

Let $M_{n} = \max{X_1,\ldots,X_n}$. Because the $X_i\sim N(0,\sigma^2)$ we have

$$a_n = \sqrt{2\log n} - \frac{\log\log n + \log 4\pi}{2\sqrt{2\log n}}$$ $$b_n = \frac{1}{\sqrt{2\log n}}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s. as } n\rightarrow +\infty$$ (a proof of the last property can be found in Asymptotic theory of statistics and probability by Anirban DasGupta (2008, Springer Science & Business Media) in page 109 Example 8.13)

Now $$\frac{\frac{1}{\hat\sigma}M_n - a_n}{b_n} = \frac{\frac{1}{\sigma}M_n - a_n}{b_n} + \frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n}$$

and after some manipulation $$\frac{(\frac{1}{\hat\sigma}-\frac{1}{\sigma})M_n}{b_n} = \frac{1}{\sigma\hat\sigma}2(\log n)(\sigma -\hat\sigma)\frac{M_n}{\sqrt{2\log n}}$$ We have $$\frac{1}{\sigma\hat\sigma}\rightarrow \frac{1}{\sigma^2}\text{ a.s.}$$ $$(\log n)(\sigma -\hat\sigma)\rightarrow 0\text{ in probaility}$$ $$\frac{M_n}{\sqrt{2\log n}} \rightarrow 1\text{ a.s.}$$

And the result follows from Slutsky's theorem.