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均值为 a + b 2 \frac{a + b}{2} 2a+b, 总数n为 ( b − a ) (b-a) (b−a)
方差= ( x − 均 值 ) 2 n \frac{(x-均值)^2}{n} n(x−均值)2
所以[a, b]均匀分布的方差为:
v a r i a n c e = ∫ a b ( x − a + b 2 ) 2 d x ( b − a ) = 1 ( b − a ) ⋅ 1 3 ( x − a + b 2 ) 3 ∣ a b = 1 ( b − a ) ⋅ 1 3 ⋅ [ ( b − a + b 2 ) 3 − ( a − a + b 2 ) 3 ] = 1 ( b − a ) ⋅ 1 3 ⋅ [ ( b − a 2 ) 3 − ( a − b 2 ) 3 ] = 1 ( b − a ) ⋅ 2 3 ⋅ ( b − a 2 ) 3 = ( b − a ) 2 12 \begin{aligned} variance &= \frac{\int_a^b (x – \frac{a + b}{2})^2 dx }{(b – a)}\\ &= \frac{1}{(b – a)} \cdot { \frac{1}{3}(x – \frac{a + b}{2})^3\bigg\rvert_{a}^{b} } \\ &=\frac{1}{(b – a)} \cdot \frac{1}{3} \cdot [ (b – \frac{a + b}{2})^3- (a – \frac{a + b}{2})^3] \\ &=\frac{1}{(b – a)} \cdot \frac{1}{3} \cdot [ (\frac{b – a}{2})^3- (\frac{a – b}{2})^3] \\ &=\frac{1}{(b – a)} \cdot \frac{2}{3} \cdot (\frac{b – a}{2})^3 \\ &=\frac{(b – a)^2}{12} \end{aligned} variance=(b−a)∫ab(x−2a+b)2dx=(b−a)1⋅31(x−2a+b)3∣∣∣∣ab=(b−a)1⋅31⋅[(b−2a+b)3−(a−2a+b)3]=(b−a)1⋅31⋅[(2b−a)3−(2a−b)3]=(b−a)1⋅32⋅(2b−a)3=12(b−a)2
不过也可以用 D ( x ) = E ( x 2 ) − E ( x ) 2 D(x)=E(x^2)-E(x)^2 D(x)=E(x2)−E(x)2来算
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