Question about squares of the coefficients of variation Suppose that $X$ is a random variable with any distribution that takes only positive values. Can the following inequality hold for constants  $c_2 >c_1 >0$?
$$\text{Var}(Z^{c_2})/\text{E}[Z^{c_2}]^2  > \text{Var}(Z^{c_1})/\text{E}[Z^{c_1}]^2$$
In other words:
$$\text{E}[Z^{2c_2}]/\text{E}[Z^{c_2}]^2  > \text{E}[Z^{2c_1}]/\text{E}[Z^{c_1}]^2$$
where $Z=X+1$.
With many thanks for your answer.
 A: It appears  that the coefficient of variation $\textrm{CV}(Z^c)$ of $Z^c$  is increasing in $c$ 
provided that $Z > 0 $ with probability one. Indeed with $f(c) := \textrm{CV}^2(Z^c)+1$
for $c > 0$, we  have
$$
   f(c) = \frac{\textrm{E}[Z^{2c}]}{\textrm{E}^2[Z^c]} =  \frac{\textrm{E}[e^{2c Y}]}{\textrm{E}^2[e^{c Y}]} =  
   \frac{m_Y(2c)}{m_Y^2(c)}
$$
where  $Y:= \log Z$  and  $ m_Y(c) := \textrm{E}[e^{c Y}]$ is the moment generating
function of $Y$. A quite well-known fact is  that a moment generating function
is  log-convex on the interval where it exists. 
Thus $r(c) := \log m_Y(c)$ defines a convex function of
$c$. Now $\log f(c) = r(2c) - 2 \,r(c)$, and for $c > 0$
$$ 
 \frac{\textrm{d}}{\textrm{d}c} \log f(c)= 2 \,r'(2c) - 2\,r'(c) \geqslant 0
$$
since the derivative $r'(c)$ is increasing. Thus $\log f(c)$ is
increasing for $c >0$, and so is $\textrm{CV}(Z^c)$.
Note that $\textrm{CV}(Z^c)$ exists when
the moment of order $c$ of $Z^2$ is finite. This condition holds on an
interval with upper end-point $c^\star \geq 0$ depending on the distribution of
$Z$. The value of $c^\star$ can be finite and even be $0$ for some heavy tailed distributions.
The preceding derivation is valid for $0 < c < c^\star$, hence assumes that $c^\star>0$.
