Let $M_n = \max(U_1,\ldots,U_n)$ , the maximum of a sample of size n from $U(0,1)$ distribution. We want to see what happens with the distribution of $M_n$ (properly standardised or normalised) as $n \rightarrow \infty$. Can someone please show where my misunderstanding is. I'll show what I did vs what the correct solution is.
What I did:
The df of $M_n$ is $F_n(u) = u^n$, $0\leq u \leq 1 $
I found the expected value and the variance of $M_n$ then I tried to normalise by $$ \frac{M_n - \operatorname E(M_n)}{\sqrt{\operatorname{Var}(M_n)}} $$
But this wasn't the same as what the book did. This is the correct solution:
$$ P(n(1-M_n) \leq u) = P(M_n \geq 1 - \frac{u}{n}) $$ $$ =1 - (1-\frac{u}{n})^n $$ $$ \rightarrow 1-e^{-u}, n \rightarrow \infty $$ How did they go about getting $n(1-M_n)$?