Let $U_1,U_2,...$ denote an i.i.d. sequence of random variables with the uniform distribution on $[0,1]$. For every integer $n\geq1$, we set $M_n = \max\{1/\sqrt{U_1},...,1/\sqrt{U_n}\}$.
a) Compute the distribution function of $M_n:$
$F_n(x)=\mathbb{P}(M_n\leq x),\qquad x\geq 0.$
Let $p>0$. Specify the values of $p$ for which $\mathbb{E}(M_n^p)<\infty$.
I was able to compute the distribution function, which is
$F_n(x) = (1-\frac{1}{x^2})^n$ for $x\geq1$ and zero otherwise.
which is correct regarding the solution. However, I am having trouble understanding the solution of the second part.
The solution says: $\mathbb{E}(M_n^p)\stackrel{\text{1}}{=}\int_0^\infty x^p dF_n(x) = \int_1^\infty x^pn(1-x^{-2})^{n-1}2x^{-3}dx = 2\int_1^\infty x^{p-3}n(1-x^{-2})^{n-1}dx$ and this integral converges if and only if $p-3<-1$ i.e. $p<2$
I cannot see how this is consistent with the definition of the expected value. In my opinion 1) should be $\mathbb{E}(M_n^p)=\int_0^\infty M_n^p dF_n(x)$ I know that does not make much sense, but can somebody explain the equality 1?
The second thing I do not understand is why this integral converges. How do you see that the value of $\mathbb{E}(M_n^p)$ is finite?? Obviously there is not a chance to compute it.