Given a sequence of uniform random variables $(0, x)$, how do you find the PDF and expectation of the maximum?
I know the PDF of any one variable is $\frac{1}{x}$ and the CDF is $x$, and since they're independent, you can get a JCDF by multiplying them together.
Working from the JCDF:
$P(X_n \leqslant x)=P(X_1 \leqslant x)*...*P(X_n \leqslant x)=(F(x))^n$
And if you differentiate this wrt $x$, this you get $n(F(x))^{n-1}f(x)$... is this process correct?
Then E(X) is usually the integral of $xf(x)$, but I'm not sure how to apply that here.
Help appreciated!
Let $x>0$ and $X_n\stackrel{\mathrm{i.i.d.}}\sim U(0,x)$. For each $0<t<x$ we have $$ \left \{\max_{1\leqslant i\leqslant n} U_i \leqslant t\right\} = \bigcap_{i=1}^n \{U_i\leqslant t\}, $$ and by independence, $$ \mathbb P\left(\bigcap_{i=1}^n \{U_i\leqslant t\}\right) = \prod_{i=1}^n \mathbb P(U_i\leqslant t). $$ Let $V:=\max_{1\leqslant i\leqslant n} U_i$. Since the $U_i$ have the same distribution function $F_U(t) = \frac tx\mathsf 1_{(0,x)}(t)$, it follows that $$ F_V(t)= \mathbb P\left(V \leqslant t\right) = \left(\frac tx\right)^n\mathsf 1_{(0,x)}(t). $$ The density of $V$ is obtained by differentiating: $$ f_V(t) = \frac{\mathsf d}{\mathsf dt} F_V(t) = \frac nx\left(\frac tx\right)^{n-1}\mathsf 1_{(0,x)}(t). $$ The expectation of $V$ is computed as follows: $$ \mathbb E[V] = \int_\mathbb R tf_V(t)\ \mathsf dt = \int_0^x n\left(\frac tx\right)^n\ \mathsf dt = \left(\frac n{n+1}\right)x. $$