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With $X_1,X_2,..$ as a sequence of i.i.d variables with F as a distribution function and $M_n=max\{X_m:m<=n\}$ for $n=1,2,..$

To prove $P(M_n<=x)=F^n(x)$, I did the following:

a) $P(M_n<=x)=P(max(X_1,X_2,...X_n)<=x)=\prod_{i=1}^{n}P(X_i<=x)=P(X_1<=x)\cdot P(X_2<=x)\cdot\cdot\cdot P(X_n<=x)=F\cdot F \cdot F.. = [F(x)]^n= F^n(x)$

b) Given $ F(x)=1-x^{-α}$ for $x>=1, α > 0$. Need to prove that as n approaches infinity $P(\frac{M_n}{n^{1/α}} <= y) -> exp(-y^{-α})$

I proved $P(\frac{M_n}{n^{1/α}} <= y)=P(M_n<=y \cdot n^{1/α})=(F(y \cdot n^{1/α}))^n=(1-\frac{1}{n \cdot y^α})^n -> e^{-y^{-α}}$ as n approaches $\infty$

Are proofs in a) and b) are done correctly?

Please let me know if something needs to be added to make it solid.

Thank you very much!

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marked as duplicate by BCLC, Math1000, Claude Leibovici, Mostafa Ayaz, Jimmy R. Mar 19 '18 at 9:06

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What are you doing in $(b)$?

$(a)$ is right:

  • Steps 1,3,5 and 6 are by definition.
  • Step 2 is the trick where one combines understanding of 'maximum' with utilising the independence of the random variables.
  • Step 4 utilises the identical distribution of the random variables.
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    $\begingroup$ Thank you. Edited the question by adding "Need to prove" part to the b). Is that better? $\endgroup$ – user629034 Mar 19 '18 at 17:59
  • $\begingroup$ @user629034 looks good to me $\endgroup$ – BCLC Mar 20 '18 at 0:19

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