For random variable $Z=\max_i X_i$, can we bound $\mathbb{E}(Z|Z>\tau)$ with $\mathbb{E}(Z)$ Let $X_1,…,X_n$ be independent, but not necessarily identical, non-negative random variables. Let $Z=\max_i(X_i)$. Fix a real $\tau > 0$. Is there a way to lower bound $$\mathbb{E}(Z|Z>\tau) > c \mathbb{E}(Z)$$ for constant $c>1$ (that can obviously depend on $\tau$). For example if $\tau$ is the median of $Z$ or $\mathbb{E}(Z)/2$

Requested in comments:
Clearly in general  $\mathbb{E}[Z\mid Z\le \tau] \le \tau \lt \mathbb{E}[Z\mid Z>\tau]$,  assuming the expectations exist.
If $\tau$ is the median of $Z$ then:

*

*$\mathbb P(Z>\tau) \le \frac12 \le \mathbb P(Z\le \tau)$

*$\mathbb{E}[Z] =  \mathbb P(Z>\tau)\,\mathbb{E}[Z\mid Z>\tau] + \mathbb P(Z\le\tau)\,\mathbb{E}[Z\mid Z\le\tau]$

*so $\mathbb{E}[Z] \le \frac12  \mathbb{E}[Z\mid Z>\tau] + \frac12  \mathbb{E}[Z\mid Z\le\tau] $

*and thus  $\mathbb{E}[Z] \le \frac12  \mathbb{E}[Z\mid Z>\tau]+ \frac12 \tau $

*implying $\mathbb{E}[Z\mid Z>\tau] \ge 2 \mathbb{E}[Z] -\tau$

*which combined with the earlier result gives $$\mathbb{E}(Z\mid Z>\tau)\ge \max(\tau,2\mathbb{E}(Z) -\tau).$$
If you extended this to the $p$th quantile of $Z$ so $\mathbb P(Z\le \tau) \ge  p$ and  $\mathbb P(Z>\tau)\le 1-p$ then the same argument would give you  $\mathbb{E}[Z\mid Z>\tau] \ge \max\left(\tau,\frac{\mathbb{E}[Z]-p\tau}{1-p}\right). $
