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Let $X=(X_1,X_2,...,X_n)$ be a random sample from the logistic distribution with cdf $F(x)=1/(1+e^{-x})$, $x \in \Bbb R.$ Let $Y_n$ be the $n$-th order statistic. Find the limiting distribution for $Y_n-\ln(n)$ as $n \to \infty $.

Here's what I've done so far:

$$F_{Z_n}=P(Z_n\le z) \\=P(Y_n\le z+\ln(n)) \\=F_{Y_n}(z+\ln(n)) =F^n(z+\ln(n));\\ F(z+\ln(n))=1/({1+e^{-(z+\ln(n))}})$$

Now how do I find $F^n(z+\ln(n))$?

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  • $\begingroup$ Hint: $F(z+\ln n)^n =(1 + e^{-z}/n)^{-n}$. $\endgroup$ – zhoraster May 4 '17 at 5:31
  • $\begingroup$ I see. Using the hint I got $F^n(z+\ln n) \to e^{-e^{-z}}$ as $n \to \infty$, Is this correct? $\endgroup$ – ViC May 4 '17 at 10:18
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    $\begingroup$ Correct! This is so-called double exponential (or Gumbel) distribution, one of the few distributions which arise as weak limits of rescaled maxima of iid random variables. $\endgroup$ – zhoraster May 4 '17 at 12:45

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