Why does $\mathbf{Var}(X) = \mathbf{Var}(-X)$ for random variable $X$? Question from UCLA Math GRE study packet, Problem Set 2, Number 4:
http://www.math.ucla.edu/~cmarshak/GREProb.pdf
Let $X$ and $Y$ be random variables. Which of the following is always true?
\begin{align}
...\\
(II) \ \mathbf{Var}(X) = \mathbf{Var}(-X)\\
...\\
...\\
\end{align}
Answer says (II) is True. Why?

Let $X$ be a random variable and $\alpha \in \mathbb R$. We have
\begin{align*}
  {\rm Var}(\alpha X) &= \def\E{\mathbb E}\E[(\alpha X)^2] - \E[\alpha X]^2\\
    &= \E[\alpha^2 X^2] - \bigl(\alpha \E[X]\bigr)^2\\
    &= \alpha^2 \bigl(\E[X^2] - \E[X]^2\bigr)\\
    &= \alpha^2 {\rm Var}(X)
\end{align*}
In your case $\alpha = -1$, and 
$$ {\rm Var}(-X) = (-1)^2 {\rm Var}(X) = {\rm Var}(X). $$
