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Let $X_1, ..., X_n$ be random variables independent and identically distributed on $Uniform(0,1)$. Let $Y_{(n)}=MAX{(X_1,...,X_n)}$. Define $W_n=n(1-Y_{(n)})$. Find the limiting distribution of $W_n$ as $n$ increases without bound. Can you identify this limiting distribution? Give an interpretation of the result obtained above.

So, I managed to find that the limiting distribution follows an exponential distribution with mean 1. However, I'm not quite sure how to "interpret" this. What's so special about $W_n$ that makes this result significant when $n$ grows large? Is there some intuitive understanding to this that I'm just not seeing?

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  • $\begingroup$ I would like to humbly request a notation change from $X_{(n)}=MAX(X_1, ..., X_n)$ to $Y_n = MAX(X_1, ..., X_n)$. So then $W_n = n(1-Y_n)$. $\endgroup$ – Michael Dec 13 '17 at 22:20
  • $\begingroup$ My own 2 cents on interpretation (others can likely word it better): Since $Y_n$ is getting very close to 1, $1-Y_n$ is getting very close to $0$, so without scaling by $n$, the value $1-Y_n$ just goes to zero. The significance of $n(1-Y_n)$ is that it shows, when these very small values are scaled, the (scaled) "deviation from zero" looks exponential, which is interesting. $\endgroup$ – Michael Dec 13 '17 at 22:23
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Here is one possible explanation. Mathematically, it's pretty meaningless, but may shed a light on the emergence of exponential distribution.

Instead of $W_n$, let us look at $V_n=n\min_{1\le k \le n} X_k$, which has the same distribution. It is a minimum of $n$ iid variables, uniform on $[0,n]$. Then its limit $W_\infty$ is a "minimum of infinitely many random variables, uniformly distributed on $[0,\infty)$". What is the distribution of $W_\infty - t$ given $W_\infty > t$? It is the distribution of minimum of infinitely many random variables, uniformly distributed on $[t,\infty)$, minus $t$; this is the distribution of $W_\infty$. So $W_\infty$ has no memory, hence the exponential distribution.

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