Show that $n-n\max\{X_i\}$ equals exponential distribution for $X_i$ i.i.d. and uniformly distributed on $[0,1]$ Say, we have a sequence of random variables $(X_i)_{1\leq i \leq n}$ that are i.i.d. and distributed according to a uniform distribution defined on interval $[0,1]$.
Why is $n-n\cdot\max\{X_1,\cdots,X_n\}$ distributed like an exponential distribution for real and positive $n$?
I know that $\lim_{n\rightarrow\infty}e(x)=(1+\frac{x}{n})^n$...
I also know that $\max\{X_1,\cdots,X_n\}=P(X_1)\cdot\cdots\cdot P(X_n)$ for independent variables...
I have:
\begin{align}
F(x) & =P(n-n\cdot\max\{X_1,\cdots,X_n\}<x)=P(\max\{\ldots\} \\[10pt]
& \geq1-x/n)=1-P(\max\{\ldots\}<1-x/n).
\end{align}
Now, due to the cdf of $U(0,1)$ which is just $x$, this is equivalent to $1-(1-x/n)^n$.
But how do I proceed now? Or are there any mistakes so far?
 A: Let $M_n = \max_{1\leqslant i\leqslant n}X_i$ for each positive integer $n$. For $0<t<1$ we have $$\{M_n \leqslant t\} = \bigcap_{i=1}^n \{X_i\leqslant t\},$$
so by independence it follows that
$$
\mathbb P(M_n\leqslant t) = t^n.
$$
Let $Y_n = n(1-M_n)$. Let $t>0$ and choose $N>1/t$. Then for $n\geqslant N$ we have
\begin{align}
\mathbb P(Y_n\leqslant t) &= \mathbb P(n(1-M_n)\leqslant t)\\
&=\mathbb P(1-M_n\leqslant t/n)\\
&=\mathbb P(M_n\geqslant 1-t/n)\\
&=1-\mathbb P(M_n\leqslant 1-t/n)\\
&=1 - (1-t/n)^n.
\end{align}
Since $$\lim_{n\to\infty}1 - (1-t/n)^n  = 1 - e^{-t},$$
it follows that $Y_n$ converges in distribution to the exponential distribution with mean $1$.
A: You are right in your calculations. There are two points and you are done:
$1.$ From uniform to exponential you have the condition $0 \leq 1-x/n \leq 1$
$2.$ You have to consider $\lim n\rightarrow\infty$. Then you get $\exp(-x)$. Your formula about this is also correct.
So in conclusion you get an exponential distribution with parameter $+1$ and it is $exp(-1)$ safely enough it doesnt blow up.
