Knowing that $X$ is a positive random variable, and that for some $m>0$ and $M>0$ we have $\mathbb{P}\left( X \leq m \right) \leq M$, how can one show the following result ?

$$\forall \alpha > 0 \text{ s.t } \mathbb{E}\left[X^{2\alpha}\right] < +\infty \text{ and } \frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]} < 1, \quad \mathbb{E}\left[X^{\alpha}\right] \leq \frac{m^{\alpha}}{1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}}$$

Applying the Paley-Zygmund bound on $X^{\alpha}$ and choosing $\theta = 1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}$ yields

$$\mathbb{P} \left\{ X^{\alpha} > \left(1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}\right) \mathbb{E}\left[ X^{\alpha} \right] \right\} \geq M$$

Then applying the Markov inequality to get rid of the probability gives :

$$\frac{1}{1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}} \geq M$$

It would help if $\mathbb{E}\left[ X^{\alpha} \right] \leq M m^{\alpha}$, but sadly I can't prove it, nor do I think it is even true.

One can however show a weaker result :

$$\frac{m^{\alpha}}{1-\frac{\sqrt{M \mathbb{E}\left[X^{2 \alpha}\right]}}{\mathbb{E}\left[X^{\alpha}\right]}} \geq \mathbb{E}\left[ X^{\alpha} \left| X \leq m \right. \right] \mathbb{P}\left( X \leq m \right)$$


Okay found it, using a Markov inequality where we condition by $X \leq m$ instead of the "vanilla" version of the Markov inequality in my previous attempt yields the result.

EDIT : false flag sorry, I made a mistake and a direct application of Markov doesn't yield the result. I will update this answer if I find something.


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