Question 705030

The mean is $\langle x \rangle = \int_{-\infty}^{+\infty} x P(x) dx = \int x \frac{1}{2}(\delta(x-a) + \delta(x+a)) dx = \frac{1}{2}(a - a) = 0$. The variance is $\langle x^2 \rangle - \langle x \rangle^2 = \langle x^2 \rangle$. $\langle x^2 \rangle = \int_{-\infty}^{+\infty} x^2 \frac{1}{2}(\delta(x-a) + \delta(x+a)) dx = \frac{1}{2}(a^2 + (-a)^2) = \frac{1}{2}(2a^2) = a^2$.