Good bound on $ \max_{|z| \le A} \left | \frac{\mathbb{E} [ X^2 \exp( -z X) ]}{ \mathbb{E} [ X \exp( -z X) ]} \right| $ where $z \in \mathbb{C}$

Question

Consider a random variable $X \in (0,1]$ and consider the following quantity \begin{align} \max_{|z| \le A} \left | \frac{\mathbb{E} [ X^2 \exp( -z X) ]}{ \mathbb{E} [ X \exp( -z X) ]} \right| \end{align} in the above $z \in \mathbb{C}$.

The question is the following: Can we find a good upper bound on this quantity?

Here is what I did: \begin{align} \max_{|z| \le A} \left | \frac{\mathbb{E} [ X^2 \exp( -z X) ]}{ \mathbb{E} [ X \exp( -z X) ]} \right| & \le \max_{|z| \le A} \frac{\mathbb{E} [ \left | X^2 \exp( -z X) \right| ]}{| \mathbb{E} [ X \exp( -z X) ] | } \\ &\le \exp(A) \max_{|z| \le A} \frac{\mathbb{E} [ X^2 ]}{ | \mathbb{E} [ X \exp( -z X) ]| }\\ &\le \exp(A) \frac{\mathbb{E} [ X^2 ]}{ \exp(-A) \mathbb{E} [ X ] } \text{ **Edit:** this step is actually not true see comment below }\\ &\le \exp(2 A) \frac{\mathbb{E} [ X^2 ]}{ \mathbb{E} [ X ] }\\ & \le \exp(2 A) \end{align}

My question is, can we do better? Can some kind of joint optimization be done?

Edit: It might be useful to add some examples.

If $X$ uniform, then we have that \begin{align} \frac{\mathbb{E} [ X^2 \exp( -z X) ]}{ \mathbb{E} [ X \exp( -z X) ]} = \frac{z}{z-e^z+1}+\frac{2}{z} \end{align}

Edit2:

Note that $ L(z) = \mathbb{E} [ \exp( -z X) ]$ is the Laplace transform of a random variable $X$. The question can be equivalently rested as \begin{align} \max_{|z| \le A} \left| \frac{L^{(2)}(z)}{L^{(1)}(z)} \right| \end{align}

Answer 1

Partial result: It can be shown that the ratio in question can never exceed unity for any complex $z$ in the strip $|\text{Im}(z)|<\pi/2.$

To simplify matters a bit, we assume that the pdf of $X$ exists. One first needs to prove the following:

Lemma: Suppose for $x\in S$, $G(x)>0$, $F/G$ is bounded above and the $\int_S G(x)dx$ exists. Then

$$\frac{\int_S F(x)dx}{\int_S G(x)dx}\leq \sup_{x\in S}\frac{F(x)}{G(x)}$$

The proof is simple and I will omit it. Now to prove the statement, first note that the square of the ratio in question can be rewritten as $(z=\alpha+\beta i$, pdf of $X$ is denoted by $f)$:

$$\Bigg|\frac{\mathbb{E}(X^2e^{-zX})}{\mathbb{E}(Xe^{-zX})}\Bigg|^2=\frac{\int _{0}^1dx\int_0^x dy~(xy)^2e^{-\alpha (x+y)}\cos(|\beta|(x-y))f(x)f(y)}{\int _{0}^1dx\int_0^x dy~~xy ~e^{-\alpha (x+y)}\cos(|\beta|(x-y))f(x)f(y)}$$

Apply the lemma with $F$ in the numerator and $G$ in the denominator with $|\beta|<\pi/2$ (to ensure the cosine is positive) and noticing that $\sup_{x\in[0,1]\times[0,1]} xy=1$ we conclude that

$$\Bigg|\frac{\mathbb{E}(X^2e^{-zX})}{\mathbb{E}(Xe^{-zX})}\Bigg|^2\leq 1~~, ~~|\text{Im(z)}|\leq \pi/2$$

We note that this bound is optimal in the strip, since it is saturated by the distribution that has $X=1$ with probability $1$.

I have not been able to extend the proof past this strip since that would require a serious refinement of the technique used above, but I believe the conjecture could hold in an even larger domain, since asymptotically as $\beta\to \infty$ the integral favors the values of the integrand close to $x-y=n\pi/\beta$, and for those, $n=0$ is the largest, if $\alpha$ has a large enough value.