$A\geq 0$ a.s., $E[A^a]<1,E[A^{a+d}]<\infty$. Show: $\exists e>0,g\in(0,1): \forall z\in [-e,e]:\mathbb{E}[(A/g)^{a+z}]<1$.

Question

Assume a random variable $A\geq 0$ a.s., $\exists \: \alpha,\delta>0: \mathbb{E}[A^\alpha]<1, \: \mathbb{E}[A^{\alpha+\delta}]<\infty$. Show: $\exists \: \epsilon>0: \: \exists \: \gamma\in(0,1): \: \forall z\in [-\epsilon,\epsilon]: \: \mathbb{E}[(A/ \gamma)^{\alpha+z}]<1$.

This problem is quite intuitive since $\alpha 's \text{ and } (\alpha+\delta) 's$ moment of $A$ and all the moments in between exist and also $\mathbb{E}[A^\alpha]$ is strictly less than 1, so there is still some gap before reaching 1, but how do I show it mathematically rigorous, analytically?

Thank you very much for your help!

Answer 1

I will assume that $a$ and $d$ are positive.

Hints:

Step 1: There exists $g \in (0,1)$ such that $E(\frac A g)^{a} <1$. [You have to to take $g$ close to $1$].

Step 2. $(\frac A g)^{a+t}\leq (\frac A g)^{a}+(\frac A g)^{a+d}$ if $|t| \leq d$. [Consider the cases $\frac A g \leq 1$ and $\frac A g > 1$ separately.

Step 3. $E(\frac A g)^{a+t} \to E(\frac A g)^{a}<1$ as $t \to 0$ by DCT.

Now you can finish the proof.