How to show $\max\{Y_{1},Y_{2},\cdots,Y_{n}\}$ converges in probability to $\theta$ as $n \to \infty$. Let $Y_{1},Y_{2},\ldots,Y_{n} $ be independent random variables , each uniformly distributed over the interval $(0,\theta)$. 
Show that $\max\{Y_{1},Y_{2},\ldots,Y_{n}\}$ converges in probability to $\theta$ as $n \to \infty$.
 A: Let $M_n=\max \{ X_1, \dots, X_n\}$ and $\varepsilon>0$.
$$\mathrm{P}\left(\theta - M_n \geq \varepsilon \right) = \mathrm{P}\left((X_1\leq\theta-\varepsilon) \wedge \dots \wedge (X_n \leq\theta-\varepsilon)\right)$$
Now use independance to develop your probability in a product, and use the CDF of uniform distribution.
$$\mathrm{P}\left(\theta - M_n \geq \varepsilon \right) = \mathrm{P}\left(X_1\leq\theta-\varepsilon\right)^n = \left(\frac{\theta-\varepsilon}{\theta}\right)^n \underset{n\to\infty}{\longrightarrow} 0$$
A: I'll give you a rough sketch of the proof.
Let $\max\{Y_1,Y_2,\ldots,Y_k\} =\alpha$. Then, $Y_i \le \alpha \quad \forall 1\le i\le k$.
$\Pr(Y_i\le \alpha) = \frac{\alpha}{\theta} $
As all the random variables are independent, $$\Pr(\max_{1\le i\le k}{Y_i} \le \alpha)=\Pr(Y_i \le \alpha, \forall 1\le i\le k) = \left(\frac{\alpha}{\theta}\right)^k$$
As $k \to \infty$, $\Pr(\max_{1\le i\le k}{Y_i} \le \theta) \to 1$.
However, for any $\alpha < \theta$, the probability goes to $0$. So, the sequence converges to $\theta$.
