A random variable is said to be symmetric around zero if we have P(X ≥ t) = P(X ≤ −t)

for any t ∈ ℝ. Prove that, if X is symmetric around zero with a distribution function F, then the following hold:

(i) P(|X| ≤ t) = 2F(t) − 1;

(ii) P(X = t) = F(t) + F(−t) − 1.

If, in addition, the distribution function F is continuous at t = 0, show that F(0) = 0.5.

(Hint: For the last result, use the relations P(X > 0) + P(X ≤ 0) = 1 and P(X ≥

0) = 1 − F(0−) = 1 − F(0).)