let $g(x)=E[Y\mid Y<x],$ where $Y=\max(Z_1,\ldots, Z_n)$ and each $Z_i$ is i.i.d. with density function $f(z)>0$ for any $z$ in some interval $(0,a)$, $a>0.$ Is $g(x)$ increasing in $x$, for $x\in (0,a)$? By rewriting the expectation, $$g(x)=\frac{n}{F^n(x)}\cdot \int_0^{x}yF^{n-1}(y)f(y)dy.$$
The problem is that while integral in the numerator increases, distribution function in the denominator of the conditional expectation increases as well and I do not know which way will it move. I tried taking the derivative but resulting expression doesn't help me much. I expect that $g$ should be strictly increasing.
Thanks for your help! :)
First assume $Y \ge 0$.
We have: $$ E(Y|Y<x) = \int_0^\infty \Pr(Y \ge z | Y < x) \, dz = \int_{0}^\infty \frac{(\Pr(Y\ge z) - \Pr(Y \ge x))^+}{1 - \Pr(Y \ge x)} \, dz .$$ Now suppose $x_1 < x_2$. Let $$s_1 = \Pr(Y \ge x_1) \\ s_2 = \Pr(Y \ge x_2) \\ \alpha = \Pr(Y \ge z). $$ Then $0 \le s_2 \le s_1 \le 1$ and $0 \le \alpha \le 1$, and hence simple algebra shows $$ \frac{(\alpha - s_1)^+}{1-s_1} \le \frac{(\alpha - s_2)^+}{1-s_2} .$$ Hence $$ E(Y|Y<x_1) \le E(Y|Y<x_2) .$$
To remove the assumption that $Y \ge 0$, first use $$ E(Y | Y<x) = E(Y+y | (Y+y < x+y) - y $$ so that without loss of generality we can assume $x_1, x_2 > 0$, and then see that for $x > 0$ $$ E(Y | Y<x) = E(Y^+ | Y^+ < x) - \frac{E(Y^-)}{\Pr(Y<x)} .$$