Let $X_{1},X_{2}, \dots, \sim Exp(1)$ i.i.d. - Calculate the probability of $P[\max{(X_{1},\dots,X_{n},)} < \log(n) - 5] $ for $ n > e^{5}$

Question

Let $X_{1},X_{2}, \dots, \sim Exp(1)$ i.i.d. - Calculate the probability of $P[\max{(X_{1},\dots,X_{n},)} < \log(n)-5] $ for $ n > e^{5}$ as well as $n \rightarrow \infty $

The correct solution is: $(1 - e^{-(\log n-5)})^{n} = (1 - \frac{e^5}{5})^{n} \rightarrow e^{-e^{5}}$

My solution would be only $(1 - e^{-\lambda \ln({n-5})})$ according to the density function $F(x)= \int\limits_{-\infty}^x f_\lambda\left(t\right)\ {\rm d}t = \begin{cases} 1-\mathrm{e}^{-\lambda x}& x\geq 0, \\ 0 & x < 0. \end{cases}$

How do I get the stated term?

Answer 1

Assuming that $X_1,\ldots,X_n$ are independent, the probability that $\max(X_1,\ldots,X_n)$ is less than $k$ is just the probability that $X_i\leq k$ holds for any $i\in[1,n]$, hence: $$\begin{eqnarray*}\mathbb{P}[\max_i X_i\leq\log(n)-5]&=&\left(\mathbb{P}[X_1\leq\log(n)-5]\right)^n=\left(\int_{0}^{\log(n)-5}e^{-x}\,dx\right)^n\\&=&\left(1-\frac{e^5}{n}\right)^n\approx e^{-e^5}.\end{eqnarray*}$$

Answer 2

The event $E = \{\max(X_1, .., X_n) < \ln(-5)\}$ is equal to the intersection of the events $E_i = \{ X_i < \ln(-5)\}$; i.e., $$ E = \bigcap\limits_{i=1}^n E_i.$$ By independence then $$ P(E) = \prod\limits_{i=1}^n P(E_i)$$ and $P(E_i)$ seems to be what you're calculating...