Extreme Value Theorem on Logistic Distribution

Question

Let $_1,_2, … , _$ be a random sample of size from a distribution with cumulative distribution function $F() = \frac{1}{1 + \exp(-cx)}$, $−\infty < < \infty$, where $ > 0$. Let $_ = \max(X_1,X_2, \dots , X_n)$ and $Z_n = Y_n - $ log $n$. Show that $Z_$ converges in distribution to a random variable with cumulative distribution function of $(Z) = \exp(−\exp(−)) , −\infty < Z < \infty$.

I think that it is impossible to prove it unless $Z_n = Y_n - \frac{1}{c}$ log $n$ or if $c=1$?

Answer 1

We have that $$\mathbb{P}(Z_n \leq x) = \mathbb{P}(\max(X_1, \dots, X_n) + \log n \leq x) = \mathbb{P}(\max(X_1, \dots, X_n) \leq x + \log n) = \\ \mathbb{P}(X_1 \leq x + \log n, \dots, X_n \leq x + \log n) = (F(x + \log n))^n$$ since the $X_i's$ are iid. Thus the cdf of $Z_n$ is $$G_n(x) = (1 + \exp(-c(x + \log n)))^{-n} = (1 + n^{-c} \exp(-c x))^{-n} = \\ \big((1 + n^{-c} \exp(-c x))^{n^c \exp(c x)}\big)^{- \exp(- c x) n^{1 - c}}.$$ Of course, the quantity inside the big brackets will converge to $e$ as $n \to \infty,$ but we run into problems with the exponent. However, retracing our steps with your proposed formula for $Z_n = Y_n - \frac{1}{c} \log n,$ we find $$G_n(x) = \big((1 + \frac{1}{n} \exp(-c x))^{n \exp(c x)}\big)^{- \exp(- c x)},$$ which will of course converge to $G(x) = \exp(- \exp(- c x))$ as desired. I hope this helps. :)

Answer 2

I think the OP is in the right.

Let $Z_n=\max_{1\leq j\leq n}Y_j -\frac{1}{c}\log n$.

Notice that \begin{align} F_{Z_n}(x)&=\mathbb{P}\big[\max_{1\leq j\leq n}Y_n\leq x+\frac{\log n}{c}\big]=\Big(F_{Y_1}\big(x+\frac{\log n}{c}\big)\Big)^n=\frac{1}{(1+e^{-c(x+\frac{\log n}{c})})^n}\\ &=\frac{1}{\Big(1+\frac{e^{-cx}}{n}\Big)^n}\xrightarrow{n\rightarrow\infty}e^{-e^{-cx}} \end{align} Since $F(x)=e^{-{e^{cx}}}$ is the distribution of a Borel probability measure on $\mathbb{R}$ and is continuous in $\mathbb{R}$ , it follows that $Z_n$ converges in law (aka in distribution) to a random variable $Z$ whose distribution is given by $F$ (Helly-Bray theorem).

If $Z'_n:=\max_{1\leq j\leq n}Y_j - \log n$, then $$F_{Z'_n}(x)=\frac{1}{\big(1+\frac{e^{-cx}n^{1-c}}{n}\big)^n}$$ which converges to $1$ if $1-c<0$ and to $0$ if $c-1>0$. In the former case $Z'_n$ losses all mass at $-\infty$ and in the latter $Z'_n$ losses all mass at $+\infty$; in either case, $Z'_n$ does not converge in distribution to any probability measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$.