Expected value of min of variables - what informations do I need?

Question

I encountered a problem where I need to compute:

$$\mathbb{E}(U) = \mathbb{E}(\min(X_1, .. , X_6))$$

The problem is that I have little information on the $X_i$. Basically I know $\mathbb{E}(X_i)$ and $\operatorname{var}(X_i)$ which are identical for all $i$; say $\mu$ and $\nu$ respectively. One may assume for simplicity that the $X_i$ are independant.

I found this question that answers the problem as long as we know the distributions; the thing is that I don't know these.

Question: Can we figure out an upper bound on $\mathbb{E}(U)$ which is not $\mathbb{E}(U) \le \mu$ with little information?

Edit from comments: $X_i$ are non-negative, finite, integer.

Answer 1

One has an upper bound of $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $, and this is best possible - i.e. one can obtain $\mathbb E(U)$ arbitrarily close to this.

To see this, let's first consider the case where we know the $X_i$ are integer-valued with mean $\mu$ but the variance is unrestricted. Then we can obtain $\mathbb E(U) =\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ by a distribution with probability $1- (\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor $ and $(\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor +1$.

This is optimal since we are trying to maximize $\sum_{n=1}^\infty \mathbb P(X \geq n)^6$ given $\sum_{n=1}^\infty \mathbb P(X \geq n)=\mu$, for which increasing the larger values of $\mathbb P(X \geq n)$ and decreasing the smaller values always gives an improvement, so an optimum is obtained when the larger values are all $1$ and can't be increased and the smaller values are all $0$ and can't be decreased, i.e. when the probability distribution is supported on at most two values.

Let's now see that we can achieve $\mathbb E(U)$ arbitrarily close to $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ with a given variance $\nu$. Our previous construction works for $\nu= (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor))$, and it is not possible to have variance smaller than this, so it suffices to handle the case when $\nu$ is larger.

If we shift $\epsilon$ probability mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+1$ and $\epsilon/m$ mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+m$, we have not changed the mean but have raised the variance by $\epsilon (m+1)$. Taking $$\epsilon = \frac{\nu- (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor)) }{ m+1} $$

we see that for $m$ sufficiently large, the distribution is still well-defined after the mass shift, and taking $m$ sufficiently large we may take $\mathbb E(U)$ arbitrarily close to its initial value.

Answer 2

$\newcommand\ep\varepsilon\newcommand\E{\operatorname{\mathsf{E}}}\newcommand\P{\operatorname{\mathsf{P}}}\newcommand\Var{\operatorname{\mathsf{Var}}}$As you noted, $\mu$ is a trivial upper bound on $\E M$, where $$M:=\min_{i=1}^6 X_i.$$

On the other hand, forgetting about the condition that the $X_i$'s are integer-valued, this trivial upper bound cannot be improved.

Indeed, without loss of generality $\mu>0$ and $\nu>0$. Take any $\ep\in(0,\mu)$. Let $X_1,\dots,X_6$ be independent random variables such that $\P(X_i=n)=p=1-\P(X_i=m)$ for all $i$, where $$m:=\mu-\ep,\quad n:=\mu+\frac\nu\ep,\quad p:=\frac{\ep^2}{\nu+\ep^2}.$$ Then $\E X_i=\mu$ and $\Var X_i=\nu$ for all $i$, as required. However, $$\E M\ge m\P(M=m)=m(1-p^6)\to\mu$$ as $\ep\downarrow0$. Thus, if the $X_i$'s are allowed to take any nonnegative real values, the trivial upper bound $\mu$ on $\E M$ cannot be improved.

Added: It has now been shown in the answer by Will Sawin that, under the condition that the $X_i$'s be integer-valued, for any feasible prescribed value $\nu$ of the variance, the exact upper bound on $\E M$ is $\lfloor\mu\rfloor+(\mu-\lfloor\mu\rfloor)^6$, which is relatively close to the trivial bound $\mu$ if $\mu$ is large.