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$X$ is a random variable with mean $\mu$, variance $\sigma^2$, skewness $\gamma$ and kurtosis $\kappa$, and $ \mathtt{Var}[X^2]=0.$

Under what conditions can $\mathtt{Var}[X]=\sigma^2 >0$?

Is my reasoning below correct or can it be done in a simpler way?

We have that $$ \mathtt{Var}[X^2] = \sigma^2 \left( (\kappa-1)\sigma^2 + 4\mu \gamma \sigma + 4\mu^2 \right ) =0. $$ So $\sigma=0$ or $$ \sigma = \frac{-2\mu \left [ \gamma \pm \sqrt{\gamma^2-(\kappa-1)} \right ] }{\kappa-1}.$$

As $\sigma$ has to be real and non-negative, $\gamma^2 \ge \kappa-1$ (by definiton $\kappa-1>0$).

For example, if $\mu>0$ then we require negative skew, $\gamma<0$, to get a positve real solution for $\sigma$.

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If $Var(X^2) = 0$ then $X^2 = a^2 = \text{constant}$. Now if $Var(X)>0$ then $X$ cannot be constant, but does take values given by $\pm a$ (with $a>0$).

This means that the only class of suitable distributions for $X$ are discrete and characterised by \begin{equation} X = \begin{cases} a, \text{with probability } p\\ -a, \text{with probability} 1-p \end{cases} \end{equation} where $p \in (0,1)$. You can find the mean, variance and other moments from this form.

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