Let $X_1, X_2, X_3,\ldots$ i.i.d. $\mathrm{Unif}(a,b).$ $a<b$.
1) $Y_n = \max(X_1, X_2, X_3,\ldots,X_n)$, what is the method I should apply to find $Y_n$'s distribution?
2) Show that $Y_n$ converges to $b$ in probability.
3) What is the asymptotic distribution for $T=n(b-Y_n)$?
Could someone let me follow step by step? The proof is itself always highly appreciated; still, If I can point out which part I should revise it will be very fine. Actually, I got stuck with the maximum function and the uniform distribution at first.
Clearly, $Y_n$ is concentrated on $(a,b)$. Now take $x\in(a,b)$. Then
$P(Y_n < x) = P(X_1 \leq x, \ldots, X_n \leq x) = \prod_{i=1}^n P(X_i \leq x) = \left(\frac{x-a}{b-a}\right)^n$,
where we have used the definition of $Y_n$ and that the sequence $(X_i)$ is iid. Note: I don't really care about weak or strong inequalities due to continuity of the random variables.
To show convergence in probability, show that for any small $\epsilon>0$, $\lim_{n\to\infty} P(b-Y_n\geq \epsilon)=0$.
Now $P(Y_n \leq b - \epsilon) = \left(\frac{b - a - \epsilon}{b-a}\right)^n \to 0$ as $n\to\infty$, because $b - a - \epsilon < b - a$. Thus $Y_n$ converges to $b$ in probability.
Consider now $T_n = n (b - Y_n)$. We have for any $t >0$, that for $n$ sufficiently large,
$P(T_n \geq t) = P(Y_n \leq b - t/n) = \left(\frac{b-a - t/n}{b-a}\right)^n = \left(1 + \frac{-t/(b-a)}{n}\right)^n$,
thus $P(T_n \geq t) \to e^{-t/(b-a)}$ point-wise as $n\to\infty$. Hence $T_n$ converges in distribution to a $\exp(1/(b-a))$-distribution, so that is the asymptotic distribution of $T_n$. Note that this especially is a Weibull distribution, so the result is in accordance with general extreme value theory.