امید ریاضی

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

${\displaystyle \mathbb{E}[X]=\sum _{i=1}^n{x}_i{p}_X(x_i)}$

${\displaystyle \mathrm{E}(c)=c}$

${\displaystyle \mathrm{E}(X)\le \mathrm{E}(Y)}$

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

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

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

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

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

${\displaystyle |\mathrm{E}(X)|\le \mathrm{E}(|X|)}$

${\displaystyle \mathrm{E}[X]=x_1{p}_1+x_2{p}_2+\dots +x_k{p}_k.}$

${\displaystyle \mathrm{E}[X]=\frac{x_1{p}_1+x_2{p}_2+\dots +x_k{p}_k}{p_1+p_2+\dots +p_k}.}$

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

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

${\displaystyle \mathrm{E}[X]=\sum _{i=1}^{\mathrm{\infty }}{x}_i{p}_i,}$

${\displaystyle \sum _{i=1}^{\mathrm{\infty }}{x}_i{p}_i=c(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots )}$

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

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

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

• ${\displaystyle \mathrm{E}(g(X))={\int }_a^{\mathrm{\infty }}g(x)\mathrm{d}\mathrm{P}(X\le x)=g(a)+{\int }_a^{\mathrm{\infty }}{g}^{\prime }(x)\mathrm{P}(X>x)\mathrm{d}x}$
  if ${\displaystyle \mathrm{P}(g(X)\ge g(a))=1}$
  ,
• ${\displaystyle \mathrm{E}(g(X))={\int }_{-\mathrm{\infty }}^{a}g(x)\mathrm{d}\mathrm{P}(X\le x)=g(a)-{\int }_{-\mathrm{\infty }}^a{g}^{\prime }(x)\mathrm{P}(X\le x)\mathrm{d}x}$
  if ${\displaystyle \mathrm{P}(g(X)\le g(a))=1}$
  .

${\displaystyle \mathrm{E}({\left|X\right|}^{\alpha })=\alpha {\int }_0^{\mathrm{\infty }}{t}^{\alpha -1}\mathrm{P}(\left|X\right|>t)\mathrm{d}t.}$

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

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

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

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

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

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

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

${\displaystyle =\mathrm{E}\left(\mathrm{E}(X|Y)\right)=\sum _y\mathrm{E}(X|Y=y)\cdot \mathrm{P}(Y=y)}$

${\displaystyle =\sum _y\left(\sum _{x}x\cdot \mathrm{P}(X=x|Y=y)\right)\cdot \mathrm{P}(Y=y)}$

${\displaystyle =\sum _y\sum _{x}x\cdot \mathrm{P}(X=x|Y=y)\cdot \mathrm{P}(Y=y)}$

${\displaystyle =\sum _y\sum _{x}x\cdot \mathrm{P}(Y=y|X=x)\cdot \mathrm{P}(X=x)}$

${\displaystyle =\sum _{x}x\cdot \mathrm{P}(X=x)\cdot \left(\sum _y\mathrm{P}(Y=y|X=x)\right)}$

${\displaystyle =\sum _{x}x\cdot \mathrm{P}(X=x)}$

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

${\displaystyle ={\mathrm{E}}_X(x).}$

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

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

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

${\displaystyle |\mathrm{E}(X)|\le \mathrm{E}(|X|)}$

${\displaystyle \mathrm{E}(XY)=\int \int xyj(x,y)dxdy.}$

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

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

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

${\displaystyle \mathrm{E}(X)=\sum _{i=1}^{\mathrm{\infty }}P(X\ge i).}$

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

${\displaystyle \mathrm{E}(X)={\int }_0^{\mathrm{\infty }}(-x)(-f_X(x))dx={\left[-x(1-F(x))\right]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}(1-F(x))dx}$

${\displaystyle {\int }_0^{\mathrm{\infty }}P(X\ge x)dx={\int }_0^{\mathrm{\infty }}{\int }_x^{\mathrm{\infty }}{f}_X(t)dtdx={\int }_0^{\mathrm{\infty }}{\int }_0^t{f}_X(t)dxdt={\int }_0^{\mathrm{\infty }}t{f}_X(t)dt=\mathrm{E}(X)}$

${\displaystyle \begin{array}{r}\mathrm{E}(X)={\int }_0^{\mathrm{\infty }}{\int }_0^{x}dtdF(x)={\int }_0^{\mathrm{\infty }}{\int }_t^{\mathrm{\infty }}dF(x)dt={\int }_0^{\mathrm{\infty }}(1-F(t))dt.\end{array}}$