I'm a beginner, I'm so sorry and I may apologize in advance if my question is too basic to be asked.
Let $X_1, \dotsc, X_n$ be i.i.d. random variables with cumulative distribution function $F$.
a) Write down a formula for the cumulative distribution function $F_{Y_n}$ of $$Y_n = \max (X_1, \dotsc, X_n)$$ in terms of $F$. (Hint: $\{\omega:Y_n (\omega) \leqslant y\}=\bigcap_{i=1}^n \{\omega:X_i(\omega)\leqslant y\}$).b) Let $X_i$ be uniformly distributed on $\{1,...,5\}$. That is, $\mathbb P(X_i=k)=1/5$ for $k =1, \dotsc, 5$. Use part a. to compute $$q_n = \mathbb P(\max (X_1, \dotsc, X_n) = 5).$$ Does $q_n\to 1$ as $n\to\infty$? Should it?
Any help would be really appreciated, thanks!
Update:
(a)
\begin{align}Pr(\max(X_1, \ldots, X_n) )&= [{\int _{-\infty }^{y}f(y)\,dy}]^n \end{align}
(b) \begin{align}Pr(\max(X_1, \ldots, X_n) = 5)&=Pr(\max(X_1, \ldots, X_n) \leq 5)-Pr(\max(X_1, \ldots, X_n) \leq 4)\\&=1-Pr(\max(X_1, \ldots, X_n) \leq 4)= 1- {\int _{-\infty }^{y}f(y)\,dy} = 1- {\int _{-\infty }^{4}f(4)\,dy}\end{align}
Is my answer correct, thanks!
Hint for part (a):
$$Pr(\max(X_1, X_2) \leq y)=Pr(X_1 \leq y \text{ and } X_2 \leq y)=Pr(X_1 \leq y)Pr(X_2 \leq y)$$
Reasoning for first equality: if the maximum of two numbers are smaller or equal to $y$, each items must be smaller of equal to $y$. If each item is smaller than equal to $y$, the maximum must be smaller than equal to $y$ as well.
Reasoning for second equality: Independence.
Also note that $Pr(X_1 \leq y) =F(y)$.
Try to figure out write out what is $Pr(X_2 \leq y)$.
After you can do it for $2$ random variable, do it for $n$ random variables.
Hint for part $(b)$:
\begin{align}Pr(\max(X_1, \ldots, X_n) = 5)&=Pr(\max(X_1, \ldots, X_n) \leq 5)-Pr(\max(X_1, \ldots, X_n) \leq 4)\\&=1-Pr(\max(X_1, \ldots, X_n) \leq 4)\end{align}