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I am looking for a sequence of probability distributions (parameterized by $h \in \{1,2,3,4,..\}$) supported on the compact interval $x \sim [a(h),b(h)]$ such that,

  • $a(h) > \frac{b(h)}{h^{\nu^2}} >0$ for some $\nu \in \mathbb{R}$ (Hence $x \geq 0$)
  • $\frac {\mathbb{E}[x^2]}{\mathbb{E}^2[x]}$ is an (preferably polynomially) increasing function of $h$.

It would be great if someone could give an example of such a sequence or let me know if there is any argument which shows that such a thing is impossible.

EDIT : Its okay if the properties are satisfied only in the large $h$ asymptotics.

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    $\begingroup$ How about $X_n$ takes a uniformly chosen element of $\{2,4,8,\ldors,2^n\}$? $\mathbb E X_n^2$ is essentially $2^{2n}/n$, while $(\mathbb E X_n)^2$ is essentially $2^{2n}/n^2$. $\endgroup$
    – Anthony Quas
    Commented Sep 19, 2017 at 6:03
  • $\begingroup$ Thanks! Thats interesting! I was actually thinking of continuous distributions when I framed the question. $\endgroup$
    – gradstudent
    Commented Sep 19, 2017 at 6:33
  • $\begingroup$ Also see the new EDIT. $\endgroup$
    – gradstudent
    Commented Sep 19, 2017 at 6:34
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    $\begingroup$ So: same example $X_n\sim \exp(\text{Unif}[0,n])$. Again, $\mathbb EX_n^2\sim e^{2n}/n$ and $(\mathbb EX_n)^2\sim e^{2n}/n^2$. $\endgroup$
    – Anthony Quas
    Commented Sep 19, 2017 at 8:14

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This is more a comment than an answer (which was given in one of the comments above), but it is too long for a MO comment.

By scaling the distribution appropriately, you may assume that your random variables have mean one. Among mean one distributions supported in a given interval, the second moment is maximised by a two-point distribution, concentrated on the endpoints. Thus, an optimal choice for an interval $[a,b]$ is $X = b$ with probability $(1-a)/(b-a)$ and $X = a$ with probability $(b-1)/(b-a)$. The second moment is then $(b-a)(a+b-ab)$. Optimising this quantity with a fixed ratio $h=b/a>2$ gives $a=\tfrac23(1+1/h)$ and $b=2/3(h+1)$, with second moment $\tfrac4{27}(h+1)^3(h-1)/h^2 \approx \tfrac4{27}h^2$.

This means that if $h$ is large, $b(h) = h a(h) > 0$ and $X$ is supported in $[a(h),b(h)]$ then the best one can have is $\mathbb{E}X^2 / (\mathbb{E} X)^2 \approx \tfrac4{27} h^2$.

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