E ⁡ [ ( X − E ⁡ [ X ] ) 2 ]E ⁡ [ X 2 − 2 X E ⁡ [ X ] + E ⁡ [ X ] 2 ]E ⁡ [ X 2 ] − 2 E ⁡ [ X ] E ⁡ [ X ] + E ⁡ [ X ] 2E ⁡ [ X 2 ] − 2 E ⁡ [ X ] 2 + E ⁡ [ X ] 2E ⁡ [ X 2 ] − E ⁡ [ X ] 2 \operatorname {Var} (X)&=\operatorname {E} \left[{\left(X-\operatorname {E} [X]\right)}^{2}\right]\\[4pt]&=\operatorname {E} \left[X^{2}-2X\operatorname {E} [X]+\operatorname {E} [X]^{2}\right]\\[4pt]&=\operatorname {E} \left[X^{2}\right]-2\operatorname {E} [X]\operatorname {E} [X]+\operatorname {E} [X]^{2}\\[4pt]&=\operatorname {E} \left[X^{2}\right]-2\operatorname {E} [X]^{2}+\operatorname {E} [X]^{2}\\[4pt]&=\operatorname {E} \left[X^{2}\right]-\operatorname {E} [X]^{2}\end{aligned}}} In other words, the variance of ⁠ X ⁠ is equal to the mean of the square of ⁠ X ⁠
E ⁡ [ ( X − μ ) 2 ]E ⁡ [ X 2 − 2 μ X + μ 2 ]E ⁡ [ X 2 ] − 2 μ 2 + μ 2E ⁡ [ X 2 ] − μ 2 \operatorname {Var} (X)&=\operatorname {E} \left[(X-\mu )^{2}\right]\\&=\operatorname {E} \left[X^{2}-2\mu X+\mu ^{2}\right]\\&=\operatorname {E} \left[X^{2}\right]-2\mu ^{2}+\mu ^{2}\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\end{aligned}}} Dalam kata lain, variansi dari peubah acak X adalah selisih dari ekspektasi dari X 2 } dengan rerata kuadrat.
That is, μ∑ i1 n p i x i .
∑ x1 n ( x − μ ) 2 p X ( x )∑ x1 n x 2 p X ( x )
1 n 2 ∑ i1 n ∑ j1 n 1 2 ( x i − x j ) 21 n 2 ∑ i ∑ j > i ( x i − x j ) 2 .
− μ 21 n ∑ x1 n x 2 − ( 1 n ∑ x1 n x ) 21 16 ( n + 1 ) ( 2 n + 1 ) − 1 14 ( n + 1 )
(X)={\frac {1}{n^{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n}{\frac {1}{2}}{\left(x_{i}-x_{j}\right)}^{2}={\frac {1}{n^{2}}}\sum _{i}\sum _{j>i}{\left(x_{i}-x_{j}\right)}^{2}.}
\operatorname {Var} (X)&=\sum _{x=1}^{n}(x-\mu )^{2}p_{X}(x)\\&=\sum _{x=1}^{n}x^{2}p_{X}(x)-\mu ^{2}\\&={\frac {1}{n}}\sum _{x=1}^{n}x^{2}-\left({\frac {1}{n}}\sum _{x=1}^{n}x\right)^{2}\\&={\frac {1}{16}}(n+1)(2n+1)-{\frac {1}{14}}(n+1)^{2}\\&={\frac {n^{2}-1}{12}}.\end{aligned}}}
If the random variable X has a probability density function ⁠ f ( x ) ⁠, and F ( x ) is the corresponding cumulative distribution function, then Var ⁡ ( X )
Di sisi lain, jika X merupakan peubah acak kontinu dengan fungsi kepadatan peluang f X ( x ) (x)} untuk x ∈ R } , variansinya adalah Var ⁡ ( X )∫ − ∞ ∞ ( x − μ ) 2 f X ( x )
d x∫ R x 2 f ( x ) d x − 2 μ ∫ R x f ( x )
∑ x1 n ( x − μ ) 2 p X ( x )∑ x1 n x 2 p X ( x )
− 2 μ ∫ R x d F ( x ) + μ 2 ∫ R d F ( x )∫ R x 2 d F ( x )
∑ x1 n ( x − μ ) 2 p X ( x )∑ x1 n x 2 p X ( x )
− 2 μ ⋅ μ + μ 2 ⋅ 1∫ R x 2 d F ( x )
∑ x1 n ( x − μ ) 2 p X ( x )∑ x1 n x 2 p X ( x )
− μ 2 , \operatorname {Var} (X)=\sigma ^{2}&=\int _{\mathbb {R} }{\left(x-\mu \right)}^{2}f(x)\,dx\\[4pt]&=\int _{\mathbb {R} }x^{2}f(x)\,dx-2\mu \int _{\mathbb {R} }xf(x)\,dx+\mu ^{2}\int _{\mathbb {R} }f(x)\,dx\\[4pt]&=\int _{\mathbb {R} }x^{2}\,dF(x)-2\mu \int _{\mathbb {R} }x\,dF(x)+\mu ^{2}\int _{\mathbb {R} }\,dF(x)\\[4pt]&=\int _{\mathbb {R} }x^{2}\,dF(x)-2\mu \cdot \mu +\mu ^{2}\cdot 1\\[4pt]&=\int _{\mathbb {R} }x^{2}\,dF(x)-\mu ^{2},\end{aligned}}} or equivalently, Var ⁡ ( X )
E ⁡ [ ( X − μ ) 2 ]E ⁡ [ X 2 − 2 μ X + μ 2 ]E ⁡ [ X 2 ] − 2 μ 2 + μ 2E ⁡ [ X 2 ] − μ 2 \operatorname {Var} (X)&=\operatorname {E} \left[(X-\mu )^{2}\right]\\&=\operatorname {E} \left[X^{2}-2\mu X+\mu ^{2}\right]\\&=\operatorname {E} \left[X^{2}\right]-2\mu ^{2}+\mu ^{2}\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\end{aligned}}} Dalam kata lain, variansi dari peubah acak X adalah selisih dari ekspektasi dari X 2 } dengan rerata kuadrat.
d x − μ 2 , (X)=\int _{\mathbb {R} }x^{2}f(x)\,dx-\mu ^{2},} where μ is the expected value of X given by μ∫ R x f ( x )
d x , (X)=\int _{-\infty }^{\infty }(x-\mu )^{2}f_{X}(x)\,dx,} dengan μE ⁡ [ X ] [X]} rerata dari X , yaitu μ∫ − ∞ ∞ x f X ( x )
}xf(x)\,dx=\int _{\mathbb {R} }x\,dF(x).}
d x ^{\infty }xf_{X}(x)\,dx} .
The exponential distribution with parameter ⁠ λ > 0 ⁠ is a continuous distribution whose probability density function is given by f ( x )
Misalkan peubah acak X memiliki distribusi eksponensial dengan parameter laju λ .
\operatorname {E} \left[X^{2}\right]&=\int _{0}^{\infty }x^{2}\lambda e^{-\lambda x}\,dx\\&={\left[-x^{2}e^{-\lambda x}\right]}_{0}^{\infty }+\int _{0}^{\infty }2xe^{-\lambda x}\,dx\\&=0+{\frac {2}{\lambda }}\operatorname {E} [X]\\&={\frac {2}{\lambda ^{2}}}.\end{aligned}}}
\operatorname {Var} (X)&=\int _{0}^{\infty }(x-\mu )^{2}f(x)\,dx\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\\&=2\lambda ^{-2}-\lambda ^{-2}\\&=\lambda ^{-2}.\end{aligned}}} Makalah asli Fisher mengenai varians Diarsipkan 2005-12-13 di Wayback Machine.
E ⁡ [ X 2 ] − E ⁡ [ X ] 22 λ 2 − ( 1 λ ) 21 λ 2 .
∫ 0 ∞ ( x − μ ) 2 f ( x ) d xE ⁡ [ X 2 ] − μ 22 λ − 2 − λ − 2λ − 2 .
(X)=\operatorname {E} \left[X^{2}\right]-\operatorname {E} [X]^{2}={\frac {2}{\lambda ^{2}}}-\left({\frac {1}{\lambda }}\right)^{2}={\frac {1}{\lambda ^{2}}}.}
\operatorname {Var} (X)&=\int _{0}^{\infty }(x-\mu )^{2}f(x)\,dx\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\\&=2\lambda ^{-2}-\lambda ^{-2}\\&=\lambda ^{-2}.\end{aligned}}} Makalah asli Fisher mengenai varians Diarsipkan 2005-12-13 di Wayback Machine.
\operatorname {Var} (X)&=\operatorname {E} \left(X^{2}\right)-(\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}}} The following table lists the variance for some commonly used probability distributions.
\operatorname {Var} (X)&=\int _{0}^{\infty }(x-\mu )^{2}f(x)\,dx\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\\&=2\lambda ^{-2}-\lambda ^{-2}\\&=\lambda ^{-2}.\end{aligned}}} Makalah asli Fisher mengenai varians Diarsipkan 2005-12-13 di Wayback Machine.
∑ i p i μ i 2 − ( ∑ i p i μ i ) 2∑ i p i μ i 2 − μ 2 , (\operatorname {E} [X\mid Y])=\sum _{i}p_{i}\mu _{i}^{2}-\left(\sum _{i}p_{i}\mu _{i}\right)^{2}=\sum _{i}p_{i}\mu _{i}^{2}-\mu ^{2},} where ⁠ 1 } ⁠ and ⁠ μ∑ i p i μ i p_{i}\mu _{i}} ⁠.
∑ ∀ x ∈ D ( x − μ ) 2 p X ( x ) , (X)=\sum _{\forall x\in {\mathcal {D}}}(x-\mu )^{2}p_{X}(x),} dengan μE ⁡ [ X ] [X]} rerata dari X , yaitu μ∑ ∀ x ∈ D x p X ( x ) {\textstyle \mu\sum _{\forall x\in {\mathcal {D}}}xp_{X}(x)} .
a 2 Var ⁡ ( X ) + b 2 Var ⁡ ( Y ) − 2 a b Cov ⁡ ( X , Y ) \operatorname {Var} (aX+bY)&=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)+2ab\,\operatorname {Cov} (X,Y)\\[1ex]\operatorname {Var} (aX-bY)&=a^{2}\operatorname {Var} (X)+b^{2}\operatorname {Var} (Y)-2ab\,\operatorname {Cov} (X,Y)\end{aligned}}} where Cov ⁡ ( X , Y ) (X,Y)} is the covariance.
\operatorname {Var} (X)&=\int _{0}^{\infty }(x-\mu )^{2}f(x)\,dx\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\\&=2\lambda ^{-2}-\lambda ^{-2}\\&=\lambda ^{-2}.\end{aligned}}} Makalah asli Fisher mengenai varians Diarsipkan 2005-12-13 di Wayback Machine.
\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\neq 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}}}
\operatorname {Var} (X)&=\sum _{x=1}^{n}(x-\mu )^{2}p_{X}(x)\\&=\sum _{x=1}^{n}x^{2}p_{X}(x)-\mu ^{2}\\&={\frac {1}{n}}\sum _{x=1}^{n}x^{2}-\left({\frac {1}{n}}\sum _{x=1}^{n}x\right)^{2}\\&={\frac {1}{16}}(n+1)(2n+1)-{\frac {1}{14}}(n+1)^{2}\\&={\frac {n^{2}-1}{12}}.\end{aligned}}}
\left({\overline {X}}\right)=\operatorname {Var} \left({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\right)={\frac {1}{n^{2}}}\sum _{i=1}^{n}\operatorname {Var} \left(X_{i}\right)={\frac {1}{n^{2}}}n\sigma ^{2}={\frac {\sigma ^{2}}{n}}.}
\operatorname {Var} (X)&=\sum _{x=1}^{n}(x-\mu )^{2}p_{X}(x)\\&=\sum _{x=1}^{n}x^{2}p_{X}(x)-\mu ^{2}\\&={\frac {1}{n}}\sum _{x=1}^{n}x^{2}-\left({\frac {1}{n}}\sum _{x=1}^{n}x\right)^{2}\\&={\frac {1}{16}}(n+1)(2n+1)-{\frac {1}{14}}(n+1)^{2}\\&={\frac {n^{2}-1}{12}}.\end{aligned}}}
\operatorname {Var} (X+Y)&=\operatorname {E} \left[(X+Y)^{2}\right]-(\operatorname {E} [X+Y])^{2}\\[5pt]&=\operatorname {E} \left[X^{2}+2XY+Y^{2}\right]-(\operatorname {E} [X]+\operatorname {E} [Y])^{2}.\end{aligned}}} Using the linearity of the expectation operator and the assumption of independence (or uncorrelatedness) of X and Y, this further simplifies as follows: Var ⁡ ( X + Y )
\operatorname {Var} (X)&=\int _{0}^{\infty }(x-\mu )^{2}f(x)\,dx\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\\&=2\lambda ^{-2}-\lambda ^{-2}\\&=\lambda ^{-2}.\end{aligned}}} Makalah asli Fisher mengenai varians Diarsipkan 2005-12-13 di Wayback Machine.
\left(\sum _{i=1}^{n}X_{i}\right)=\sum _{i=1}^{n}\sum _{j=1}^{n}\operatorname {Cov} \left(X_{i},X_{j}\right)=\sum _{i=1}^{n}\operatorname {Var} \left(X_{i}\right)+2\sum _{1\leq i<j\leq n}\operatorname {Cov} \left(X_{i},X_{j}\right).} (Note: The second equality comes from the fact that Cov(Xi, Xi)Var(Xi).)
\operatorname {Var} (X)&=\sum _{x=1}^{n}(x-\mu )^{2}p_{X}(x)\\&=\sum _{x=1}^{n}x^{2}p_{X}(x)-\mu ^{2}\\&={\frac {1}{n}}\sum _{x=1}^{n}x^{2}-\left({\frac {1}{n}}\sum _{x=1}^{n}x\right)^{2}\\&={\frac {1}{16}}(n+1)(2n+1)-{\frac {1}{14}}(n+1)^{2}\\&={\frac {n^{2}-1}{12}}.\end{aligned}}}
(XY)=\operatorname {E} \left(X^{2}\right)\operatorname {E} \left(Y^{2}\right)-[\operatorname {E} (X)]^{2}[\operatorname {E} (Y)]^{2}.}
\operatorname {Var} (X)&=\int _{0}^{\infty }(x-\mu )^{2}f(x)\,dx\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\\&=2\lambda ^{-2}-\lambda ^{-2}\\&=\lambda ^{-2}.\end{aligned}}} Makalah asli Fisher mengenai varians Diarsipkan 2005-12-13 di Wayback Machine.
− 2 μ ( 1 N ∑ i1 N x i ) + μ 2E ⁡ [ x i 2 ] − μ 2 , \sigma ^{2}&={\frac {1}{N}}\sum _{i=1}^{N}{\left(x_{i}-\mu \right)}^{2}={\frac {1}{N}}\sum _{i=1}^{N}\left(x_{i}^{2}-2\mu x_{i}+\mu ^{2}\right)\\[5pt]&=\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-2\mu \left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)+\mu ^{2}\\[5pt]&=\operatorname {E} [x_{i}^{2}]-\mu ^{2},\end{aligned}}} where the population mean is μE ⁡ [ x i ]1 N ∑ i1 N x i {\textstyle \mu\operatorname {E} [x_{i}]={\frac {1}{N}}\sum _{i=1}^{N}x_{i}} and ⁠ E ⁡ [ x i 2 ]( 1 N ∑ i1 N x i 2 ) [x_{i}^{2}]=\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)} ⁠, where E {\textstyle \operatorname {E} } is the expectation value operator.
E ⁡ [ ( X − μ ) 2 ]E ⁡ [ X 2 − 2 μ X + μ 2 ]E ⁡ [ X 2 ] − 2 μ 2 + μ 2E ⁡ [ X 2 ] − μ 2 \operatorname {Var} (X)&=\operatorname {E} \left[(X-\mu )^{2}\right]\\&=\operatorname {E} \left[X^{2}-2\mu X+\mu ^{2}\right]\\&=\operatorname {E} \left[X^{2}\right]-2\mu ^{2}+\mu ^{2}\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\end{aligned}}} Dalam kata lain, variansi dari peubah acak X adalah selisih dari ekspektasi dari X 2 } dengan rerata kuadrat.
1 2 N ∑ j1 N ( 1 N ∑ i1 N x i 2 ) − ( 1 N ∑ i1 N x i ) ( 1 N ∑ j1 N x j ) + 1 2 N ∑ i1 N ( 1 N ∑ j1 N x j 2 )1 2 ( σ 2 + μ 2 ) − μ 2 + 1 2 ( σ 2 + μ 2 )σ 2 .
− μ 21 n ∑ x1 n x 2 − ( 1 n ∑ x1 n x ) 21 16 ( n + 1 ) ( 2 n + 1 ) − 1 14 ( n + 1 )
\left[S^{2}\right]=\operatorname {Var} \left({\frac {\sigma ^{2}}{n-1}}\chi _{n-1}^{2}\right)={\frac {\sigma ^{4}}{{\left(n-1\right)}^{2}}}\operatorname {Var} \left(\chi _{n-1}^{2}\right)={\frac {2\sigma ^{4}}{n-1}}.}
\operatorname {Var} (X)&=\int _{0}^{\infty }(x-\mu )^{2}f(x)\,dx\\&=\operatorname {E} \left[X^{2}\right]-\mu ^{2}\\&=2\lambda ^{-2}-\lambda ^{-2}\\&=\lambda ^{-2}.\end{aligned}}} Makalah asli Fisher mengenai varians Diarsipkan 2005-12-13 di Wayback Machine.