 Solution

The minimum exceeds $t$ exactly when both times exceed $t$. By independence,Differentiating its survivor function givesThe tail integral formula for moments then givesEquivalently, integration by parts in ${\int }_0^{R}t{f}_X(t)dt$ gives $-R{F}_A(R)F_B(R)+{\int }_0^R{F}_A(t)F_B(t)dt$. The assumed $t{F}_B(t)\to 0$ makes the boundary term vanish, since $0\le F_A\le 1$.

The stated tail condition does not itself guarantee a finite mean. For example, the proper survivor functionssatisfy $t{F}_B(t)\to 0$, but their product behaves as $1/[t\log t\log \log t]$, whose integral diverges. Thus the displayed expectation is valid as an extended nonnegative value; finiteness needs integrability of the product. The exponential distribution in the next part guarantees that finiteness.