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 $$ {\rm Var}(-X) = (-1)^2 {\rm Var}(X) = {\rm Var}(X). $$
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 $(EX)^{2}=(E(-X))^{2}=(-EX)^{2}$
Another way to realize this is to $$ \begin{align} 0&=\mathrm{Var}(0)=\mathrm{Var}(X+(-X))=\mathrm{Var}(X)+\mathrm{Var}(-X)-2\mathrm{Cov}(X,X)\\ &=\mathrm{Var}(X)+\mathrm{Var}(-X)-2\mathrm{Var}(X)=\mathrm{Var}(-X)-\mathrm{Var}(X). \end{align} $$
