Convergence in Probability where Yn= the max of n independent exponentially distributed rv's

Question

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.

Answer 1

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 $1$.