Consider a random variable X with a uniform distribution on $[0, a]$, with $(a > 0)$, where the value of a is unknown. To estimate a, we draw n independent random samples of $X$, denoted by $X_1,\ldots,X_n$, and we consider using :
$A_n$ = $\max(X_1,\ldots,X_n)$ to estimate the value of a.
1) What is the cdf and pdf of $An$?
2) Find $E[A_n]$ and $Var(A_n)$?
First I point out how $X_i$ is distributed. In general the pdf of a uniform distributed variable has the pdf
$f_X(x)=\begin{cases} \frac1{u-l}, l<x<u \\0, \ \ \text{elsewhere}\end{cases}$
$u$ ist upper bound and $l$ ist the lower bound. In your cases $l=0$ and $u=a$
$f_X(x)=\begin{cases} \frac1{a}, 0<x<a \\0, \ \ \text{elsewhere}\end{cases}$
To get the cdf you calculate $\int_0^x f_X(t) \, dt=\int_0^x \frac1{a} \, dt= \ \frac{t}{a} \ \ \large|_0^x=\frac{x}{a}-\frac{0}{a}=\frac{x}{a}$
$F_X(x)=\begin{cases} 0, x\leq 0\\ \frac{x}{a}, 0<x<a \\1, \ \ x\geq a\end{cases}$
If $max(X_1, X_2,\dots, X_n)=x$, then each of the $n$ random variables, $X_i$, must be smaller or equal to $x$.
Since the samples are independent every sample has the same probability.
The probability that $X_1$ is smaller or equal than x is $\frac{x}a$, if $0<x<a$.
The probability that $X_2$ is smaller or equal than x is $\frac{x}a$, if $0<x<a$ as well.
Now we can conclude what the probability is that every of the $n$ random variables are smaller or equal than $x$, if $0<x<a$.
I post it to make clear how the expected value has to be calculated.
$$P(A_n\leq x)=\left( \frac{x}a \right)^n$$
Differentiating w.r.t x gives $\left(\frac{1}a\right)^n\cdot n\cdot x ^{n-1}$
Thus $$E(A_n)=\int_0^a x\cdot \left(\frac{1}a\right)^n\cdot n\cdot x ^{n-1}\, dx=a\cdot \frac{n}{n+1}$$