This is for a homework question in which I am trying to find the $E(Y_n = min\{X_1,...,X_n\}$).
So far I have found that the minimum cdf, as below.
The minimum of $X_i$ is when all of $X_i > x$.
$Pr(Y_n \leq y) = Pr(X_i > x) = \prod_{i=1}^{n} Pr(X_i \leq 1-x) = F_X(1- x)^n $
Then I found the pdf like so:
$$ F_Y(y) = Pr(X_i > x) = \left\{ \begin{eqnarray} 0 \textrm{ for } y \leq 0 \\ (1-x)^n \textrm{ for } y \in (0,1) \\ 1 \textrm{ for } y \geq 1 \end{eqnarray} \right. $$
$$\frac{d}{dy}F_Y(y) = \frac{d}{dy}(1-y)^n = n(1-y)^{n-1} = f_Y(y)$$
... and then I tried to integrate for $ y \in (0,1)$ given the common definition of $E(Y) = \int_0^1 y f_Y(y) dy$
$$ \begin{aligned} E(Y_n) &= \int_0^1 yn(1-y)^{n-1}dy\\ &= n\int_0^1(1-u)(u)^{n-1} -du && \textrm{where } &&&u=1-y\\ &&& \textrm{where }&&&du=-dy\\ &= n\int_0^1(u-1)(u)^{n-1}du\\ &= n\int_0^1u^{n}-u^{n-1}du\\ &= n\left(\left[\frac{u^{n+1}}{n+1}\right]_0^1-\left[\frac{u^{n}}{n}\right]_0^1\right)\\ &= n\left(\left[\frac{(1-y)^{n+1}}{n+1}\right]_0^1-\left[\frac{(1-y)^{n}}{n}\right]_0^1\right)\\ &= n\left(\left[ 0 - \frac{1^{n+1}}{n+1}\right]-\left[0 - \frac{1^{n}}{n}\right]\right)\\ &= n\left(-\left[\frac{1^{n+1}}{n+1}\right]+\left[\frac{1^{n}}{n}\right]\right)\\ E(Y_n) &= 1^{n} - \frac{1^{n+1}}{n} \end{aligned} $$
However, this answer is apparently wrong. One source says that the correct answer is $$E(Y_n) = \frac{1}{n+1}$$
Is anyone able to steer me in the right direction in my integration? I haven't done calculus, let alone integration by u substitution, in 7 years or so! Thank you.
