Here is the question:
The PDF of $X_{1},X_{2}...X_{n}$ are $f_X(x)=1-e^{-x},x>0$. They are independent.
Let $M_{n}=max(X_{1},X_{2}...X_{n})$ and $G(z)=\lim_{n\to\infty}[P(\frac{M_{n}-b_{n}}{a_{n}})\leq z]$ and $a_{n}=1,b_{n}=n$. So what is G(z)?
My solution is:
First compute the CDF of $X$, $F_{X}(x)=\int_{0}^{x}f_{X}(t)dt=\int_{0}^{x}(1-e^{-t})dt=x+e^{-x}-1$
Then $F_{M_{n}}(x)=P(M_{n}\leq x)=P(X_{1}\leq x)P(X_{2}\leq x)...P(X_{n}\leq x)=F_{X}(x)^{n}=(x+e^{-x}-1)^{n}$
Finally $G(z)=\lim_{n\to\infty}[P(\frac{M_{n}-b_{n}}{a_{n}})\leq z]=\lim_{n\to\infty}[P(M_{n}\leq a_{n}z+b_{n})]=\lim_{n\to\infty}[P(M_{n}\leq z+n)]=\lim_{n\to\infty}F_{M_{n}}(z+n)=\lim_{n\to\infty}(z+n+e^{-(z+n)}-1)^{n}$
But how to compute that limit? Or did I make mistakes in the computation before? Could anyone help me? Thanks!!!!
Since your variables are exponentially distributed, G(z) will be the Gumbel Distribution. The link is for the Fisher-Tippet theorem, which shows how the Gumbel distribution is related to the Fisher-Tippet theorem. The $a_n$ and $b_n$ are special normalizing sequences that are specific to different distributions.
Also, this step in your derivation looks suspect:
$\lim_{n\to\infty}[P(M_{n}\leq a_{n}z+b_{n})]=\lim_{n\to\infty}[P(M_{n}\leq z+n)]=\lim_{n\to\infty}F_{M_{n}}(z+n)$
What made you conclude that the coefficients are 1 and $n$?
This paper may also be very helpful, as it shows how the exponential leads to the gumbel. You will note that $a_n\neq1$ and $b_n\neq n$