I understand how to calculate the expectation of the minimum and the maximum of $n$ random variables, but I'm not sure whether I'm getting the following right.
I'm interested in the expected value of the 2nd-smallest out of 3 I.I.D. random variables with distribution $F$.
Denoting the random variables as $A$, $B$ and $C$, my approach would be to first calculate the distribution of the 2nd-smallest out of 3 as follows:
$$ F_{2/3}(x) = Pr(A \land B < x < C ) + Pr(A \land C < x < B) + Pr (B \land C < x< A) + Pr (A \land B \land C < x) = 3 F(x)^2(1-F(x)) + F(x)^3 = 3 F(x)^2 - 2 F(x)^3 $$
and then the expectation using
$$ \int x f_{2/3} (x) dx. $$
Is this correct?
Yes, for iid continuous random variables, it is so that: $${F_{(2\setminus 3)}(x)} ~{=~{ {\mathsf P\big((A\vee B)\leqslant x\lt C\big)}+{\mathsf P\big((A\vee C)\leqslant x\lt B\big)}\\+{\mathsf P\big((B\vee C)\leqslant x\lt A)}+{\mathsf P\big((A\vee B\vee C)\leqslant x\big)}}\\=~ 3\, F(x)^2\,\big(1-F(x)\big)~+~ F(x)^3 \\=~ 3\, F(x)^2-2\,F(x)^3}$$
Thus by differentiation: $$f_{(2\setminus 3)}(x) ~{=~ 6 f(x)\,F(x)-6f(x)\,F(x)^2\\ =~ 3!\, f(x)\,F(x)\,\big(1-F(x)\big)}$$
Note: In this context, $\wedge$ and $\vee$ are the binary operators for $\min$ and $\max$ respectively. You want the $\max$.