# properties of a expectation for a non-negative random variable

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

$$X^\alpha \leq X$$ 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 $$f(\alpha) < 0 \text{ for }\alpha > 1$$
• +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 $\{2,3,\ldots\}.$
– whuber
Mar 13 at 11:52
• @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 $f(\alpha) < 0$? Mar 13 at 11:54
• 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.
– whuber
Mar 13 at 11:58
• 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. Mar 14 at 16:57
• How is $g(X)$ continuous if $X$ is discrete? Or is it just the function $g$ that is continuous? Mar 14 at 20:08