mathematica在概率论与数理统计中的应用
计算期望和方差
- 给定概率密度函数f(x), 用
Integrate[f(x)*x, {x, 0, Infinity}]计算期望， 用二阶矩Integrate[f(x)*x^2, {x, 0, Infinity}]减去期望的平方计算方差. - 给定概率分布dist, 用
Mean[dist]计算分布dist的期望, 用Variance[dist]计算分布dist的方差.
例
$$ X \verb|~| f(x)=\left\{ \begin{matrix} \frac{1}{9}x{{e}^{-\frac{x}{3}}}, & x>0 \\ 0, & 其它 \\ \end{matrix} \right. $$， 求$E(X),D(X)$.
输入如下
EX=Integrate[x^2/9*Exp[-x/3],{x,0,Infinity}]
E2=Integrate[x^3/9*Exp[-x/3],{x,0,Infinity}]
DX=E2-EX^2得到结果$E(X)=6, D(X)=18$.
- $X \sim N\left(\mu, \sigma^2\right)$, 求 $E(X), D(X)$
Mean[NormalDistribution[μ,σ]]
Variance[NormalDistribution[μ,σ]]输出结果为:
$μ, σ^2$
求区间估计
- 计算均值的置信水平为95%的置信区间用
MeanCI[list] - 计算方差的置信水平为95%的置信区间用
VarianceCI[list]
例
- 55.96,56.54,57.58,55.13,57.48,56.06,59.93,58.30,52.57,58.46，已知方差为4, 求μ的置信水平为0.95的置信区间
<<Statistics\ConfidenceIntervals.m
data1={55.96,56.54,57.58,55.13,57.48,56.06,59.93,58.30,52.57,58.46}
MeanCI[data1,ConfidenceLevel->0.95,KnownVariance->4]输出结果为{55.5614,58.0406}
- 13.1,5.1,18.0,8.7,16.5,9.8,6.8,12.0,17.8,25.4,19.2,15.8,23.0，求σ的置信水平为0.95的置信区间
data2={13.1,5.1,18.0,8.7,16.5,9.8,6.8,12.0,17.8,25.4,19.2,15.8,23.0}
V=VarianceCI[data2,ConfidenceLevel->0.95]输出结果为 {4.40589,10.1424}