# $\mathbb{E}[X^2]\leq k \mathbb{E}[X]^2$, upper bound second moment from first moment

Let $$X$$ be a non-negative random variable bounded on $$[0,1]$$. Is it true that $$\mathbb{E}[X^2]\leq k \mathbb{E}[X]^2$$ for some constant $$k$$? If not, are there any minimal assumptions on $$X$$ where this holds?

It's easy to show that this holds for Bernoulli random variables, but I am not sure they are the worst examples.

A related question is https://stats.stackexchange.com/questions/617592/expectation-of-first-of-moment-of-symmetric-r-v-in-terms-of-variance, which shows the statement is false for general $$X$$'s (that are not non-negative).

Despite your "It's easy to show ...", Bernoulli random variables with $$\mathbb P(X=1)=p$$ are a counter-example: $$\dfrac{\mathbb{E}[X^2]}{ \mathbb{E}[X]^2} = \dfrac{p}{p^2} = \dfrac1p$$ and that is unbounded as $$p \to 0$$.
If there were some $$\delta>0$$ with $$\mathbb{P}(X \in [\delta,1])=1$$, then there would be a finite upper bound on $$\dfrac{\mathbb{E}[X^2]}{ \mathbb{E}[X]^2}$$, and clearly there is a lower bound of $$1$$.
• With my $\mathbb{P}(X \in [\delta,1])=1$ suggestion, $k=\frac{1}{\delta^2}$ is certainly an upper bound since $\mathbb{E}[X^2]\le 1$ and $\mathbb{E}[X]^2\ge \delta^2$, and I suspect $k =\frac{(1+\delta)^2}{4\delta}$ may also be an upper bound. Apr 12 at 23:58