Consider $$\zeta_n=\max_{1\le k \le n} X_k ,$$ where $X_k\sim \exp(1)$. Let $\eta_n=\frac{\zeta_n}{\ln(n)}$. What is the limit of $\eta_n$ as $n\to \infty$? (Convergence in distribution).
Convergence in Distribution
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probability
probability-theory
probability-distributions
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1Well, knowing that $\eta_n$ is a maximum, you could try to write the event $[\eta_n\le x]$ for a fixed $x$ using only events $[X_k\le x]$. // Re your approach of MSE, let me venture two guesses: First, the present question is homework. Second, you ask questions here at a pace such that you do not have the time to accept the answers you get, even less to **study and understand** them. Am I right on both counts? – 2011-09-12
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0I am thinking of $P[\zeta_n\le \ln(n)x]$, so we will have $F(\ln(n) x)$, where $\zeta_n$ has a distribution of $F(x)=(1-\exp(-\lambda x))^n$, then find the distribution for one n and take the limit. i have $\lim_{n\rightarrow\infty} (1-\exp(-x\ln(n)))^n=\lim_{n\rightarrow\infty}(1-(\exp(-\ln(n))^x)^n=\lim_{n\rightarrow\infty} (1-(\frac{1}{n})^x)^n$, but i am not even sure if this limit exists. for x>0, we have a limit of 1 x=0,we have a limit of 0. can we just definite x<0 to have a limit of 0 as well? – 2011-09-12
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0The logarithm of the quantity of interest is $n\log(1-n^{-x})$. Perhaps you know the limit of $n\log(1-an^{-1})$ for every given $a$. Then you might want to show that the limit of $n\log(1-n^{-x})$ is $-\infty$ if $x<1$ and $0$ if $x>1$. Once you know *that*, what does that tell you on the limit in distribution you are after? – 2011-09-12
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0we have F(x)=1 for x>1, and F(x)=0, for x<1 – 2011-09-12
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0Sorry? Do you mean the limits? (At the moment, you consider F(log(n)x) which depends on n and is certainly not 1 for x>1 nor 0 for x<1.) Anyway, how do you prove this? – 2011-09-12
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0sorry, i mean in the limit. what i computed was x>0, we have 1 x=0, limit of 0, – 2011-09-12
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0@bear: I have merged your two accounts. Please consider registering your account. Currently, because your account is unregistered, everytime you switch computers or IP addresses you are treated as a new user and your previous history is "lost". By registering your account you will be able to keep track of your own questions and answers, post comments on your own questions, and enjoy other benefits. – 2011-09-12
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0I agree that for x=0, the limit of F(log(n)x)=F(0) is 0 but for x>0, the limit of F(log(n)x) is not always 1. In fact, the limit depends on x and this is precisely this behaviour that you have to determine to solve the exercise. (You might want to reread my previous comments.) – 2011-09-12