Say I have a non-negative discrete random variable $X$ (values of $X$ can be mapped to integers $(0, 2^n -1)$ for $n \in \mathbb{Z}$) and an associated distribution $P(X)$. Given a non-negative scalar $\alpha$, let us define the following

$f(\alpha)$= $\langle X^2 \rangle - \langle X \rangle \langle X^\alpha \rangle$.

where $\langle...\rangle$ is over $P(X)$ which is arbitrary discrete random distribution. Note that $\alpha$ is non-negative and so is $X$. My question is when can I say $f(\alpha) \ge 0$ ? I know for specific values like for $\alpha = 1$ it is true. Similarly for $\alpha = 0$ it is true if $X \ge 1$ and false otherwise and $\alpha=2$ it is false if $\langle X\rangle \ge 1$. Can I say something more for certain values of $\alpha$ in general (say $ 0.5 \le \alpha \le 1.5$)? Any help will be greatly appreciated


1 Answer 1


$X^\alpha \leq X$ for $\alpha <1$ and so you can sharpen the sufficient condition $\alpha = 1$ to $\alpha \leq 1$.

This is the sharpest sufficient condition of the simple form $\alpha \leq x$. For $\alpha > 1$ it will depend on additional condition. For example

  • If $P(X>1) = 0$ then $$f(\alpha) \geq 0 \text{ for any } \alpha$$
  • If $X$ is a degenerate distribution with a value above $1$ then $$f(\alpha) < 0 \text{ for }\alpha > 1$$
  • $\begingroup$ +1 -- but with one more line you can go much further than that by pointing out that when $\alpha\gt 1$ (whence $1+\alpha \gt 2$) there exist $X$ for which $f(\alpha)\lt 0$ and that all such $X$ have some support on the subset $\{2,3,\ldots\}.$ $\endgroup$
    – whuber
    Mar 13 at 11:52
  • $\begingroup$ @whuber you mean to state that the sufficient condition $\alpha \leq 1$ is the sharpest we can get, as for $\alpha > 1$ it is possible to get $f(\alpha) < 0$? $\endgroup$ Mar 13 at 11:54
  • $\begingroup$ Consider, as an extreme (but illustrative) example, the variable $X=2$ a.s. and notice that for any $\alpha\gt 1,$ $f(\alpha)=2^2 - 2^{1+\alpha}\lt 0.$ The general result follows immediately. $\endgroup$
    – whuber
    Mar 13 at 11:58
  • $\begingroup$ Thanks a lot for the comments. I am now able to sidestep the problem by doing some tricks from literature.....however we now have a new polynomial $\tilde{f(\alpha)} = \langle g(X)^2 \rangle - \langle g(X)^{2- \alpha} \rangle \langle g(X)^{\alpha}\rangle$ ...where $g(X) \ge 0$ is a continuous r.v. even though $X$ is discrete. Statement about $\alpha \ge 0$ as above still holds. $\langle .....\rangle$ is over the distribution of $P(X)$ which is discrete probability distribution. Under what conditions this new $\tilde{f(\alpha)}\ge 0$? I can ask in a separate post too if that helps. $\endgroup$ Mar 14 at 16:57
  • $\begingroup$ How is $g(X)$ continuous if $X$ is discrete? Or is it just the function $g$ that is continuous? $\endgroup$ Mar 14 at 20:08

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