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!!!!

  • 1
    $\begingroup$ Probably you've mistaken X's CDF for its PDF. For $F(x)$ to be a CDF, at least it should satisfy $\lim_{x \rightarrow +\infty} F(x)= 1$ and for $f(x)$ to be a PDF, we should have $\int_{R}f(x)dx = 1$ $\endgroup$ – Petite Etincelle Nov 17 '13 at 12:15
  • $\begingroup$ Maybe the original question has a mistake, in which the PDF is $1-e^{-x},x>0$. Thanks for your help. $\endgroup$ – TonyLic Nov 17 '13 at 17:30
  • $\begingroup$ But if like you said that $1-e^{-x},x>0$ is the CDF, how to compute the $\lim_{x\to\infty}(1-e^{-(z+n)})^{n}$? $\endgroup$ – TonyLic Nov 17 '13 at 17:39
  • $\begingroup$ we know that $\lim_{n \rightarrow +\infty}(1-\frac{x}{n})^n = \frac{1}{e^x}$, so $\lim_{n \rightarrow +\infty}(1 - \frac{e^{-z}}{e^n})^{e^n} = \frac{1}{e^{e^{-z}}}$, so your limit is equal to $\lim_{n \rightarrow +\infty}(\frac{1}{e^{e^{-z}}})^{\frac{n}{e^n}} = 1$ $\endgroup$ – Petite Etincelle Nov 17 '13 at 18:04

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 $b_n\neq n$

  • $\begingroup$ Those coefficients are given by the question. Thanks for your help. $\endgroup$ – TonyLic Nov 17 '13 at 17:27
  • $\begingroup$ But in that paper, how could I conclude that $a_{n}\neq $ and $b_{n}\neq n$??? $\endgroup$ – TonyLic Nov 17 '13 at 17:34

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