Comme un jeu

The random variable \(X\) with values on \([-1, 1]\) has density \( f(x) = \frac{1}{2} \).
Find \(V(X)\).\FieldText{1}{

\(\begin{aligned}[t]E(X) &= \int_{-1}^{1} x \cdot \frac{1}{2} \, dx \\&= \frac{1}{2} \left[ \frac{x^2}{2} \right]_{-1}^{1} \\&= \frac{1}{2} \left( \frac{1^2}{2} - \frac{(-1)^2}{2} \right) \\&= \frac{1}{2} \left( \frac{1}{2} - \frac{1}{2} \right) \\&= 0 \end{aligned}\)
\(\begin{aligned}[t]\int_{-1}^{1} x^2 \cdot \frac{1}{2} \, dx &= \frac{1}{2} \int_{-1}^{1} x^2 \, dx \\&= \frac{1}{2} \left[ \frac{x^3}{3} \right]_{-1}^{1} \\&= \frac{1}{2} \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) \\&= \frac{1}{2} \left( \frac{1}{3} - \left( -\frac{1}{3} \right) \right) \\&= \frac{1}{2} \cdot \frac{2}{3} \\&= \frac{1}{3} \end{aligned}\)
\(\begin{aligned}[t]V(X) &= \int_{-1}^{1} x^2 \cdot f(x) \, dx - [E(X)]^2 \\&= \frac{1}{3} - (0)^2 \\&= \frac{1}{3} \end{aligned}\)

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