Could anyone give me an example of a sequence of cumulative distribution functions $F_i$ such that $F_i(\mu_i)\to 0$ as $i\to \infty$, where $\mu_i$ is the expected value of $F_i$?

It would be better if the support of $F_i$ is $\mathbb{R}_{++}$.

I would like to use this result for my research on OR. Thanks!

  • $\begingroup$ A quick-and-dirty approach, violating your $\mathbb{R}_{++}$ assumption, would be $X_i=-i$ with probability $1/i$ and $\frac{1}{1-1/i}$ with probability $1-1/i$, for $i=2,3,\dots$. $\endgroup$
    – Ian
    Jan 27, 2016 at 3:14
  • $\begingroup$ Should the means all be the same? (This question is of no consequence without the $\mathbb{R}_{++}$ assumption, because you can just subtract off each mean to make all the means the same. But with the $\mathbb{R}_{++}$ assumption, this makes a difference.) $\endgroup$
    – Ian
    Jan 28, 2016 at 0:06
  • $\begingroup$ Thank you for a nice answer! There is no problem to take different means, even if it diverges. I greatly appreciate your help! $\endgroup$
    – tarou2
    Jan 28, 2016 at 4:23
  • 1
    $\begingroup$ Actually, now that I think about it carefully, there is a simpler way to do it: pick two positive numbers $x,y$ with $x<y$, give $x$ probability $p_i$ and $y$ probability $1-p_i$ with $p_i \to 0^+$. Then you get your result for free. $\endgroup$
    – Ian
    Jan 29, 2016 at 16:33
  • 1
    $\begingroup$ @Ian this post just got bumped because it doesn’t have any upvoted answers. You should convert these comments into answers. $\endgroup$ Oct 21, 2017 at 16:43

1 Answer 1


Take $p_i$ such that $p_i \in (0,1)$ and $\lim_{i \to \infty} p_i=0$ and $x_1<x_2$. Then

$$F_i(x)=\begin{cases} 0 & x \in (-\infty,x_1) \\ p_i & x \in [x_1,x_2) \\ 1 & x \in [x_2,\infty) \end{cases}$$

has the desired property, since $\mu_i=p_i x_1 + (1-p_i) x_2$, which is in $[x_1,x_2)$, hence $F_i(\mu_i)=p_i$.


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