Let $Y_1,Y_2,...,Y_n$ be independent and identically distributed with the following probability density function $f(y) =4(1−y)^3$ for $y$ between 0 and 1
(a) Find the probability density function of $Y_{(1)}=\min(Y_1,...,Y_n)$
(c) Find the expected value of $Y_{(1)}$
I did part a and got that the pdf of the first order statistic is $$f_{Y_{(1)}}(y)=4n(1-y)^3 \cdot [1+(1-y)^4]^{n-1}$$
I tried to find the expected value of this by taking the integral from 0 to 1 of the pdf of the first statistic times y. This integral got way to complicated so I am wondering if there is a way to find the expected value of this first order statistic differently.
For $0\leq y<1$ you should have : $$\begin{split}f_{Y_{(1)}}(y) &= n~f(y)~\mathsf P(Y > y)^{n-1}\\[1ex] &=4n~(1-y)^3\left(\int_y^1 4(1-t)^3\mathsf d t\right)^{n-1}\\[1ex]&= 4n~(1-y)^3(1-y)^{4n-4}\\[1ex]&= 4n~(1-y)^{4n-1}\\[3ex]\therefore\quad\mathsf E(Y_{(1)})&=4n\int_0^1 y~(1-y)^{4n-1}~\mathsf d y\end{split}$$
Which is not too difficult.