Let $X_1, X_2, \cdots$ be independent random variables, each with exponential distribution with parameter $\lambda = 1$. For any $n ≥ 1$, let $Y_n := \max(X_1,\cdots, X_n)$. Let $0 < a < 1 < b$. Show that $P(Y_n ≤ a \log n) \to 0$ as $n \to \infty$, and $P(Y_n ≤ b \log n) \to 1$ as $n \to \infty$. Conclude that $Y_n/ \log n$ converges to $1$ in probability as $n\to \infty$.

I have no idea how to approach this problem. I know the formula for convergence in probability, but this seems foreign to me.


Hint: $$P(\max_{i\leq n} X_i \leq u) = \prod_{i=1}^nP(X_i\leq u)$$ This is because the max of a bunch of things less than $u$ is equivalent to all those things less than $u$. The product form happens because the variables are independent.

Can you work out the rest? If you need more help, let me know in comments.

Update: So basically you have to evaluate $(1-e^{-u})^n$ which happens to be the RHS above.

Put $u=a\log n$. You get $$\left(1-\frac{1}{n^a}\right)^n = \left(1-\frac{1}{n^a}\right)^{n^a(n/n^a)}$$

Now we know $(1-\frac{1}{n})^n\to e^{-1}$. Hence as $n^a\to \infty$, we get $(1-\frac{1}{n^a})^{n^a}\to e^{-1}$. Now $e^{-1} < 1$, so we have that for any $\delta>0$, for $n$ large enough $$e^{-1}-\delta\le \left(1-\frac{1}{n^a}\right)^{n^a} \leq e^{-1}+\delta$$ Here pick any $\delta$ such that $e^{-1}+\delta < 1$. Now $$0\leq \left(1-\frac{1}{n^a}\right)^{n^a(n/n^a)} \leq (e^{-1}+\delta)^{n^{1-a}}\to 0 $$

For the $b$ case, use a similar argument except you need the lower bound now. Just pick $\delta$ such that $e^{-1}+\delta>0$ and note that probabilities cannot exceed $1$.

  • $\begingroup$ I was able to show the second part, how Yn/log(n) converges to 0. This is because P(Yn>(e+1)logn) + P(Yn<(1-e)logn) = 1 - 1 + 0 = 0, as e+1 is the case where the coefficient = b (thus ->1) and 1-e is the case where the coefficient = a (thus ->0). I'm still having trouble though proving why probability with a coef converges to 0 and with b coef converges to 1. I'm stuck on finding the limit of P(xi<alog(n))^n. $\endgroup$ – Lebron James Feb 16 '17 at 19:20
  • $\begingroup$ Good. I'll update my answer to assist you with the part you were stuck on. $\endgroup$ – Gautam Shenoy Feb 16 '17 at 19:42
  • $\begingroup$ Thank you, I think I understand it now. $\endgroup$ – Lebron James Feb 16 '17 at 20:20

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