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!


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.
  • 1
    $\begingroup$ 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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