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.

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    $\begingroup$ Do you have further information on the support of the distributions, are these positive for instance? $\endgroup$ Jun 1, 2022 at 10:24
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    $\begingroup$ in terms of the cumulative distribution function $F(x)$ of the six $X_i$'s you wish to know $\mathbb{E}[U]=\int_0^\infty \bigl(1-F(x)\bigr)^p\,dx,$ with $p=6$, given only the first two moments as input, so the only input you have is $\mathbb{E}[X^k]=k\int_0^\infty x^{k-1}\bigl(1-F(x)\bigr)\,dx,$ with $k=1,2$. I don't see how any progress can be made in general, without more information on $F(x)$. $\endgroup$ Jun 1, 2022 at 11:21
  • $\begingroup$ @CarloBeenakker : If you mean that the mean and variance don't give enough information to determine the expectation of the minimum, you are probably right, but if one finds bounds on the expectation of the minimum, expressed in terms of the mean and variance, isn't that "progress"? $\endgroup$ Jun 1, 2022 at 14:37

2 Answers 2


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.


$\newcommand\ep\varepsilon\newcommand\E{\operatorname{\mathsf{E}}}\newcommand\P{\operatorname{\mathsf{P}}}\newcommand\Var{\operatorname{\mathsf{Var}}}$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.

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    $\begingroup$ Ignoring the variance condition, the highest one can achieve is $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6$, by a probability distribution supported on $\lfloor \mu \rfloor$ and $\lceil \mu \rceil$. One can than take this distribution and increase its variance as much as desired while keeping its mean fixed and changing this expectation as little as desired by shifting $\epsilon$ mass from $\lfloor \mu \rfloor $ to $\lfloor \mu \rfloor -1$ and then $\epsilon/m$ mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor + m$, which raises variance by $\epsilon (m+1)$. $\endgroup$
    – Will Sawin
    Jun 1, 2022 at 14:34
  • $\begingroup$ I claim that I've done the exact analysis, and the optimum is given by the formula above, regardless of the variance. $\endgroup$
    – Will Sawin
    Jun 1, 2022 at 14:51
  • $\begingroup$ @WillSawin : Yes, you are right. $\endgroup$ Jun 1, 2022 at 14:56
  • $\begingroup$ @WillSawin For posterity's sake, and since you already did the analysis, maybe you can post your result as an answer? $\endgroup$ Jun 1, 2022 at 15:12
  • $\begingroup$ @NawafBou-Rabee Sure, done. $\endgroup$
    – Will Sawin
    Jun 1, 2022 at 15:58

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