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?
 A: 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 
$$ {\rm Var}(-X) = (-1)^2 {\rm Var}(X) = {\rm Var}(X). $$
A: By definition 
$$
Var(X):=E(X-EX)^{2}
$$
Let $Z=-X$ then by linearity of the expectation 
$$
Var(Z)=E(Z-EZ)^{2}=E(-X-(-EX))^{2}=E(-X+EX)^{2}=E(EX-X)^{2}=E(X-EX)^{2}=Var(X)
$$
Note that that I have used that equality $(EX-X)^{2}=(X-EX)^{2}$
.
This can also be seen in a similar manner from the identity 
$$
Var(X)=EX^{2}-(EX)^{2}
$$
Since $X^{2}=(-X)^{2}$ and $(EX)^{2}=(E(-X))^{2}=(-EX)^{2}$ 
A: Another way to realize this is to 
$$
\begin{align}
0&=\mathrm{Var}(0)=\mathrm{Var}(X+(-X))=\mathrm{Var}(X)+\mathrm{Var}(-X)-2\mathrm{Cov}(X,X)\\
&=\mathrm{Var}(X)+\mathrm{Var}(-X)-2\mathrm{Var}(X)=\mathrm{Var}(-X)-\mathrm{Var}(X).
\end{align}
$$
