واریانس - ویکی‌پدیا، دانشنامهٔ آزاد

${\displaystyle Var(X)=\sigma ^{2}\equiv \left\langle (X-\mu )^{2}\right\rangle }$

${\displaystyle \sigma ^{2}=\sum _{i=1}^{N}p(x_{i})(x_{i}-\mu )^{2}}$

${\displaystyle S_{N}^{2}\equiv {\frac {1}{N}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}$

${\displaystyle {\overline {x}}={\frac {1}{N}}\sum _{i=1}^{N}x_{i}={\frac {x_{1}+x_{2}+\cdots +x_{N}}{N}}}$

${\displaystyle S_{N-1}^{2}\equiv {\frac {1}{N-1}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}$

${\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\operatorname {E} [(X-\mu )^{2}]\\&=\operatorname {E} [X^{2}-2\mu X+\mu ^{2}]\\&=\operatorname {E} [X^{2}]-2\mu \,\operatorname {E} [X]+\mu ^{2}\\&=\operatorname {E} [X^{2}]-2\mu ^{2}+\mu ^{2}\\&=\operatorname {E} [X^{2}]-\mu ^{2}\\&=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}.\end{aligned}}}$

${\displaystyle \operatorname {Var} (X)=\sum _{i=1}^{n}p_{i}\cdot (x_{i}-\mu )^{2},}$

${\displaystyle \operatorname {Var} (X)=\left(\sum _{i=1}^{n}p_{i}x_{i}^{2}\right)-\mu ^{2},}$

${\displaystyle \mu =\sum _{i=1}^{n}p_{i}x_{i}.}$

${\displaystyle \operatorname {Var} (X)={\frac {1}{n}}\sum _{i=1}^{n}(x_{i}-\mu )^{2},}$

${\displaystyle \mu ={\frac {1}{n}}\sum _{i=1}^{n}x_{i}.}$

${\displaystyle \operatorname {Var} (X)={\frac {1}{n^{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n}{\frac {1}{2}}(x_{i}-x_{j})^{2}={\frac {1}{n^{2}}}\sum _{i}\sum _{j>i}(x_{i}-x_{j})^{2}.}$

${\displaystyle {\begin{aligned}\operatorname {Var} (X)=\sigma ^{2}&=\int (x-\mu )^{2}f(x)\,dx\\[4pt]&=\int x^{2}f(x)\,dx-2\mu \int xf(x)\,dx+\int \mu ^{2}f(x)\,dx\\[4pt]&=\int x^{2}\,dF(x)-2\mu \int x\,dF(x)+\mu ^{2}\int \,dF(x)\\[4pt]&=\int x^{2}\,dF(x)-2\mu \cdot \mu +\mu ^{2}\cdot 1\\[4pt]&=\int x^{2}\,dF(x)-\mu ^{2},\end{aligned}}}$

${\displaystyle \mu =\int x\,f(x)\,dx\,,}$

${\displaystyle \operatorname {Var} (X)\geq 0.}$

${\displaystyle P(X=a)=1\iff \operatorname {Var} (X)=0.}$

${\displaystyle \operatorname {Var} (X+a)=\operatorname {Var} (X).}$

${\displaystyle \operatorname {Var} (aX)=a^{2}\operatorname {Var} (X).}$

${\displaystyle \operatorname {Var} (aX+bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)+2ab\,\operatorname {Cov} (X,Y),}$

${\displaystyle \operatorname {Var} (aX-bY)=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)-2ab\,\operatorname {Cov} (X,Y),}$

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{N}X_{i}\right)=\sum _{i,j=1}^{N}\operatorname {Cov} (X_{i},X_{j})=\sum _{i=1}^{N}\operatorname {Var} (X_{i})+\sum _{i\neq j}\operatorname {Cov} (X_{i},X_{j}).}$

${\displaystyle {\begin{aligned}\operatorname {Var} \left(\sum _{i=1}^{N}a_{i}X_{i}\right)&=\sum _{i,j=1}^{N}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j})\\&=\sum _{i=1}^{N}a_{i}^{2}\operatorname {Var} (X_{i})+\sum _{i\not =j}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j})\\&=\sum _{i=1}^{N}a_{i}^{2}\operatorname {Var} (X_{i})+2\sum _{1\leq i<j\leq N}a_{i}a_{j}\operatorname {Cov} (X_{i},X_{j}).\end{aligned}}}$

${\displaystyle \operatorname {Var} \left(\sum _{i=1}^{N}X_{i}\right)=\sum _{i=1}^{N}\operatorname {Var} (X_{i}).}$

${\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\sum _{i=1}^{6}{\frac {1}{6}}\left(i-{\frac {7}{2}}\right)^{2}\\[5pt]&={\frac {1}{6}}\left((-5/2)^{2}+(-3/2)^{2}+(-1/2)^{2}+(1/2)^{2}+(3/2)^{2}+(5/2)^{2}\right)\\[5pt]&={\frac {35}{12}}\approx 2.92.\end{aligned}}}$

${\displaystyle {\begin{aligned}\operatorname {Var} (X)&=\operatorname {E} (X^{2})-(\operatorname {E} (X))^{2}\\[5pt]&={\frac {1}{n}}\sum _{i=1}^{n}i^{2}-\left({\frac {1}{n}}\sum _{i=1}^{n}i\right)^{2}\\[5pt]&={\frac {(n+1)(2n+1)}{6}}-\left({\frac {n+1}{2}}\right)^{2}\\[4pt]&={\frac {n^{2}-1}{12}}.\end{aligned}}}$

${\displaystyle \operatorname {Var} (X)=\int _{-\infty }^{\infty }{\frac {x^{2}}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}\,dx-\mu ^{2}=\sigma ^{2}}$

${\displaystyle \operatorname {Var} (X)=\int _{0}^{\infty }x^{2}\lambda e^{-\lambda x}\,dx-\mu ^{2}=\lambda ^{-2}}$

${\displaystyle \operatorname {Var} (X)=\left(\sum _{k=0}^{\infty }k^{2}{\frac {\lambda ^{k}}{k!}}e^{-\lambda }\right)-\mu ^{2}=\lambda }$

${\displaystyle \operatorname {Var} (X)=\left(\sum _{k=0}^{n}k^{2}{n \choose k}p^{k}(1-p)^{n-k}\right)-\mu ^{2}=np(1-p)}$

${\displaystyle \operatorname {Var} \left[f(X)\right]\approx \left(f'(\operatorname {E} \left[X\right])\right)^{2}\operatorname {Var} \left[X\right]}$