Answer to Question #129604 in Statistics and Probability for Zuga

Question

If you receive a negative value for a variance and standard, you must check your computations. Is this statement true or false?

Answer 1

This statement is true because a variance (and a standard deviation) cannot be negative.

Let we have a r. v. $\xi$. Then $D\xi=M(\xi-M\xi)^2$. So we consider a r. v. $(\xi-M\xi)^2$ (which takes on only non-negative values) and find its expectation. We get $D\xi\geq 0$.

$D\xi=0 \Leftrightarrow \xi-M\xi=0\ (\text{when }\xi \text{ is constant}).$

$\sigma_\xi=\sqrt{D\xi}$. So $\sigma_\xi\geq 0$.