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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.