10.3.0.2 解答

1. $$1=\int_{-\infty}^\infty f(x) \Dx =\int_{0}^1 a x(1-x) \Dx=\frac{\;a\;}{6}.$$
   ゆえに $a=6$.

2. $$E(x)=\int_{-\infty}^\infty x f(x) \Dx =\int_{0}^1 6 x^2(1-x) \Dx =6\left(\frac{1}{3}-\frac{1}{4}\right) =6\cdot\frac{1}{12}=\frac{1}{\;2\;}.$$
   $$E(X^2)=\int_{-\infty}^\infty x^2 f(x) \Dx =\int_{0}^1 6 x^3(1-x) \Dx =6\left(\frac{1}{4}-\frac{1}{5}\right) =6\cdot\frac{1}{20}=\frac{3}{10}.$$
   $$V(X)=E(X^2)-E(X)^2=\frac{3}{10}-\left(\frac{1}{2}\right)^2= \frac{3}{10}-\frac{1}{4}=\frac{6-5}{20}=\frac{1}{20}. \quad\qed$$