Let $Z_1, Z_2, Z_3$ be independent standard Normal R.V.'s. Which of the following has a Chi-Square distribution with 1 degree of freedom.
$$ \begin{align} A) & & & \frac{Z_1^2, Z_2^2}{2} \\ B) & & & (Z_1+Z_2)^2-Z_1^2 \\ C) & & & Z_1^2+Z_2^2-Z_3^2 \\ D) & & & \frac{(Z_1+Z_2-Z_3)^2}{3} \\ E) & & & (Z_1+Z_2-Z_3)^2 \end{align} $$
I understand why A and B & E aren't Chi-Squared, as well as why D is chi-squared with one degree of freedom, but cannot figure out why C isn't a chi-squared with one degree of freedom.
If $Z_3 \sim N(0,1)$ then $-Z_3 \sim N(0,1)$ which should work... Can someone explain why this is wrong?
If $X\sim \chi^2_n$ then $\Pr(X\ge 0) =1$. But $\Pr(X_1^2+X_2^2 - X_3^2 \ge 0 ) <1$, because the probability that $X_1^2+X_2^2$ is small and $X_3^2$ is large is not $0$.