# If the variance of $X^2$ is zero, when is the variance of $X$ not zero

$$X$$ 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 $$\sigma$$.

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 $$$$X = \begin{cases} a, \text{with probability } p\\ -a, \text{with probability} 1-p \end{cases}$$$$ where $$p \in (0,1)$$. You can find the mean, variance and other moments from this form.