امید ریاضی - ویکی‌پدیا، دانشنامهٔ آزاد

${\displaystyle \mathbb {E} [X]={\frac {1}{6}}\times 1+{\frac {1}{6}}\times 2+{\frac {1}{6}}\times 3+{\frac {1}{6}}\times 4+{\frac {1}{6}}\times 5+{\frac {1}{6}}\times 6=3.5}$

${\displaystyle \mathbb {E} [X]=\int {xf_{X}(x)dx}}$

${\displaystyle \mathbb {E} [X]=\sum _{i=1}^{n}{x_{i}p_{X}(x_{i})}}$

${\displaystyle \operatorname {E} (c)=c}$.

${\displaystyle \operatorname {E} (X)\leq \operatorname {E} (Y)}$.

${\displaystyle \operatorname {E} (X+c)=\operatorname {E} (X)+c\,}$

${\displaystyle \operatorname {E} (X+Y)=\operatorname {E} (X)+\operatorname {E} (Y)\,}$

${\displaystyle \operatorname {E} (aX)=a\operatorname {E} (X)\,}$

${\displaystyle \operatorname {E} (aX+b)=a\operatorname {E} (X)+b\,}$

${\displaystyle \operatorname {E} (aX+bY)=a\operatorname {E} (X)+b\operatorname {E} (Y)\,}$

${\displaystyle |\operatorname {E} (X)|\leq \operatorname {E} (|X|)}$

${\displaystyle \operatorname {E} [X]=x_{1}p_{1}+x_{2}p_{2}+\ldots +x_{k}p_{k}\;.}$

${\displaystyle \operatorname {E} [X]={\frac {x_{1}p_{1}+x_{2}p_{2}+\ldots +x_{k}p_{k}}{p_{1}+p_{2}+\ldots +p_{k}}}\;.}$

${\displaystyle \operatorname {E} [X]=1\cdot {\frac {1}{6}}+2\cdot {\frac {1}{6}}+3\cdot {\frac {1}{6}}+4\cdot {\frac {1}{6}}+5\cdot {\frac {1}{6}}+6\cdot {\frac {1}{6}}=3.5.}$

${\displaystyle \operatorname {E} [\,{\text{gain from }}\$1{\text{ bet}}\,]=-\$1\cdot {\frac {37}{38}}\ +\ \$35\cdot {\frac {1}{38}}=-\$0.0526.}$

${\displaystyle \operatorname {E} [X]=\sum _{i=1}^{\infty }x_{i}\,p_{i},}$

${\displaystyle \sum _{i=1}^{\infty }x_{i}\,p_{i}=c\,{\bigg (}1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\ldots {\bigg )}}$

${\displaystyle \operatorname {E} [X]=\int _{-\infty }^{\infty }xf(x)\,\operatorname {d} x.}$

${\displaystyle \operatorname {E} [X]=\int _{\Omega }X\,\operatorname {d} P=\int _{\Omega }X(\omega )\,P(\operatorname {d} \omega )}$

${\displaystyle \operatorname {E} (g(X))=\int _{-\infty }^{\infty }g(x)f(x)\,\operatorname {d} x.}$

• ${\displaystyle \operatorname {E} (g(X))=\int _{a}^{\infty }g(x)\,\mathrm {d} \operatorname {P} (X\leq x)=g(a)+\int _{a}^{\infty }g'(x)\operatorname {P} (X>x)\,\mathrm {d} x}$ if ${\displaystyle \operatorname {P} (g(X)\geq g(a))=1}$,
• ${\displaystyle \operatorname {E} (g(X))=\int _{-\infty }^{a}g(x)\,\mathrm {d} \operatorname {P} (X\leq x)=g(a)-\int _{-\infty }^{a}g'(x)\operatorname {P} (X\leq x)\,\mathrm {d} x}$ if ${\displaystyle \operatorname {P} (g(X)\leq g(a))=1}$.

${\displaystyle \operatorname {E} (\left|X\right|^{\alpha })=\alpha \int _{0}^{\infty }t^{\alpha -1}\operatorname {P} (\left|X\right|>t)\,\operatorname {d} t.}$

${\displaystyle \operatorname {E} (X+c)=\operatorname {E} (X)+c\,}$

${\displaystyle \operatorname {E} (X+Y)=\operatorname {E} (X)+\operatorname {E} (Y)\,}$

${\displaystyle \operatorname {E} (aX)=a\operatorname {E} (X)\,}$

${\displaystyle \operatorname {E} (aX+b)=a\operatorname {E} (X)+b\,}$

${\displaystyle \operatorname {E} (aX+bY)=a\operatorname {E} (X)+b\operatorname {E} (Y)\,}$

${\displaystyle \operatorname {E} (X|Y)(y)=\operatorname {E} (X|Y=y)=\sum \limits _{x}x\cdot \operatorname {P} (X=x|Y=y).}$

${\displaystyle \operatorname {E} _{Y}\left[\operatorname {E} _{X|Y=y}(x)\right]=}$

${\displaystyle =\operatorname {E} \left(\operatorname {E} (X|Y)\right)=\sum \limits _{y}\operatorname {E} (X|Y=y)\cdot \operatorname {P} (Y=y)\,}$

${\displaystyle =\sum \limits _{y}\left(\sum \limits _{x}x\cdot \operatorname {P} (X=x|Y=y)\right)\cdot \operatorname {P} (Y=y)\,}$

${\displaystyle =\sum \limits _{y}\sum \limits _{x}x\cdot \operatorname {P} (X=x|Y=y)\cdot \operatorname {P} (Y=y)\,}$

${\displaystyle =\sum \limits _{y}\sum \limits _{x}x\cdot \operatorname {P} (Y=y|X=x)\cdot \operatorname {P} (X=x)\,}$

${\displaystyle =\sum \limits _{x}x\cdot \operatorname {P} (X=x)\cdot \left(\sum \limits _{y}\operatorname {P} (Y=y|X=x)\right)\,}$

${\displaystyle =\sum \limits _{x}x\cdot \operatorname {P} (X=x)\,}$

${\displaystyle =\operatorname {E} (X)}$

${\displaystyle =\operatorname {E} _{X}(x).}$

${\displaystyle \operatorname {E} (X)=\operatorname {E} \left(\operatorname {E} (X|Y)\right).}$

${\displaystyle \operatorname {E} _{X}(x)=\operatorname {E} _{Y}\left[\operatorname {E} _{X|Y=y}(x)\right].}$

${\displaystyle \operatorname {E} (X)=\operatorname {E} \left(\operatorname {E} (X|Y)\right).}$

${\displaystyle |\operatorname {E} (X)|\leq \operatorname {E} (|X|)}$

${\displaystyle \operatorname {E} (XY)=\int \int xy\,j(x,y)\,dx\,dy.}$

${\displaystyle \operatorname {Cov} (X,Y)=\operatorname {E} (XY)-\operatorname {E} (X)\operatorname {E} (Y).}$

${\displaystyle {\begin{aligned}\operatorname {E} (XY)&=\int \int xy\,j(x,y)\,dx\,dy=\int \int xyf(x)g(y)\,dy\,dx\\&=\left[\int xf(x)\,dx\right]\left[\int yg(y)\,dy\right]=\operatorname {E} (X)\operatorname {E} (Y)\end{aligned}}}$

${\displaystyle \operatorname {E} (g(X))=\int _{\Omega }g(X)\,\operatorname {d} P\neq g(\operatorname {E} (X)),}$

${\displaystyle \operatorname {Var} (X)=\operatorname {E} (X^{2})-(\operatorname {E} (X))^{2}.}$

${\displaystyle \operatorname {E} (X)=\operatorname {E} {\begin{pmatrix}x_{1,1}&x_{1,2}&\cdots &x_{1,n}\\x_{2,1}&x_{2,2}&\cdots &x_{2,n}\\\vdots &\vdots &\ddots &\vdots \\x_{m,1}&x_{m,2}&\cdots &x_{m,n}\end{pmatrix}}={\begin{pmatrix}\operatorname {E} (x_{1,1})&\operatorname {E} (x_{1,2})&\cdots &\operatorname {E} (x_{1,n})\\\operatorname {E} (x_{2,1})&\operatorname {E} (x_{2,2})&\cdots &\operatorname {E} (x_{2,n})\\\vdots &\vdots &\ddots &\vdots \\\operatorname {E} (x_{m,1})&\operatorname {E} (x_{m,2})&\cdots &\operatorname {E} (x_{m,n})\end{pmatrix}}.}$

${\displaystyle \operatorname {E} (X)=\sum \limits _{i=1}^{\infty }P(X\geq i).}$

${\displaystyle {\begin{aligned}\sum \limits _{i=1}^{\infty }P(X\geq i)&=\sum \limits _{i=1}^{\infty }\sum \limits _{j=i}^{\infty }P(X=j)\end{aligned}}}$

${\displaystyle {\begin{aligned}\sum \limits _{i=1}^{\infty }P(X\geq i)&=\sum \limits _{j=1}^{\infty }\sum \limits _{i=1}^{j}P(X=j)\\&=\sum \limits _{j=1}^{\infty }j\,P(X=j)\\&=\operatorname {E} (X)\end{aligned}}}$

${\displaystyle \sum _{k=1}^{\infty }r^{k}={\frac {r}{1-r}}.}$

${\displaystyle \operatorname {E} (X)=\int _{0}^{\infty }(-x)(-f_{X}(x))\;dx=\left[-x(1-F(x))\right]_{0}^{\infty }+\int _{0}^{\infty }(1-F(x))\;dx}$

${\displaystyle \int _{0}^{\infty }P(X\geq x)\;dx=\int _{0}^{\infty }\int _{x}^{\infty }f_{X}(t)\;dt\;dx=\int _{0}^{\infty }\int _{0}^{t}f_{X}(t)\;dx\;dt=\int _{0}^{\infty }tf_{X}(t)\;dt=\operatorname {E} (X)}$

${\displaystyle {\begin{aligned}\operatorname {E} (X)=\int _{0}^{\infty }\int _{0}^{x}\;dt\;dF(x)=\int _{0}^{\infty }\int _{t}^{\infty }dF(x)\;dt=\int _{0}^{\infty }(1-F(t))\;dt.\end{aligned}}}$