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離散型確率変数(discrete random variable) / 分散(variance)

$$\begin{array}{rcl} 母集団(population)の確率変数&:&X,Y\\ 確率質量凾数(probability\,mass\,function,\,PMF)&:&f_X(x)=P(X=x_i)\\ &&\displaystyle\sum_{i=1}^{\infty} f_X(x_i)=1\\ 累積分布凾数(cumulative\,distribution\,function,\,CDF)&:&\displaystyle F_X(x)=\sum_{i:x_i \leq x} f_X(x_i)\\ 同時確率質量凾数(joint\,probability\,mass\,function)&:&f_X(x, y)=P(X=x_i,\,Y=y_j)\\ &&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} f_{XY}(x_i, y_j)=1\\ 周辺確率質量凾数(marginal\, probability\, mass\, function)&:&\displaystyle f_X(x)=\sum_{j=1}^{\infty} f_{XY}(x, y_i)\\ \end{array}$$ $$\begin{array}{rcl} V[X]&\equiv&E[(X-E[X])^2]\\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i-E[X])^2 f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i^2-2E[X]x_i+E[X]^2) f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i^2f_X(x_i)-2E[X]x_if_X(x_i)+E[X]^2f_X(x_i)) \\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i^2) f_X(x_i) -2E[X]\displaystyle\sum_{i=1}^{\infty} (x_i) f_X(x_i) +E[X]^2\displaystyle\sum_{i=1}^{\infty} f_X(x_i)\\ &=&E[X^2] -2E[X]E[X] +E[X]^2 1\\ &=&E[X^2] -2E[X]^2 +E[X]^2\\ &=&E[X^2]-E[X]^2\\ V[cX] &=&E[(cX-E[cX])^2]\,\dotso\,cは定数\\ &=&\displaystyle\sum_{i=1}^{\infty} (c x_i - E[c X])^2 f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} (c x_i - c E[X])^2 f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} (c (x_i - E[X]))^2 f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} c^2 (x_i-E[X])^2 f_X(x_i) \\ &=&c^2\displaystyle\sum_{i=1}^{\infty} (x_i-E[X])^2 f_X(x_i) \\ &=&c^2 V[X] \\ V[X \pm t]&=&E[((X \pm t)-E[X \pm t])^2]\,\dotso\,tは定数\\ &=&\displaystyle\sum_{i=1}^{\infty} ((x_i \pm t)-E[X \pm t])^2 f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} ((x_i \pm t)-(E[X] \pm t))^2 f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i \pm t - E[X] \mp t))^2 f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i-E[X])^2 f_X(x_i) \\ &=&V[X] \\ V[X \pm Y]&=&E[((X \pm Y)-E[X \pm Y])^2]\\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i \pm y_j)-E[X \pm Y])^2 f_{XY}(x_i, y_j) \\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i \pm y_j)-(E[X] \pm E[Y]))^2 f_{XY}(x_i, y_j) \\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i - E[X]) \pm (y_j - E[Y]))^2 f_{XY}(x_i, y_j) \\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i - E[X])^2 \pm 2(x_i - E[X])(y_j - E[Y]) +(y_j - E[Y])^2) f_{XY}(x_i, y_j) \\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i - E[X])^2 f_{XY}(x_i, y_j) \pm 2(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j) +(y_j - E[Y])^2 f_{XY}(x_i, y_j)) \\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (x_i - E[X])^2 f_{XY}(x_i, y_j) \pm 2\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j) + \displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (y_j - E[Y])^2 f_{XY}(x_i, y_j) \\ &=&\displaystyle\sum_{i=1}^{\infty}(x_i - E[X])^2\sum_{j=1}^{\infty}f_{XY}(x_i, y_j) \pm 2\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j) + \displaystyle\sum_{j=1}^{\infty} (y_j - E[Y])^2 \sum_{i=1}^{\infty} f_{XY}(x_i, y_j) \\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i - E[X])^2 f_{X}(x_i) \pm 2\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j) + \displaystyle\sum_{j=1}^{\infty} (y_j - E[Y])^2 f_{Y}(y_j)\,\dotso\,周辺確率質量凾数を適用 \\ &=&V[X] \pm 2Cov[X,Y] +V[Y]\,\dotso\,Cov[X,Y]=\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j)\\ \end{array}$$

X,Yが独立の場合

$$\begin{array}{rcl} V[X \pm Y]&=&\displaystyle V[X]+V[Y]\,\dotso\,X,Yが独立なのでCov[X,Y]=0\\ V[XY]&=&\displaystyle E[(XY)^2]-E[XY]^2\\ &=&\displaystyle E[X^2Y^2]-E[XY]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle E[X^2]E[Y^2]-(E[X]E[Y])^2\,\dotso\,X,Yが独立なのでE[XY]=E[X]E[Y]\\ &=&\displaystyle E[X^2]E[Y^2]-E[X]^2E[Y]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle (V[X]+E[X]^2)(V[Y]+E[Y]^2)-E[X]^2E[Y]^2\,\dotso\,V[X]=E[X^2]-E[X]^2よりE[X^2]=V[X]+E[X]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2+E[X]^2E[Y]^2-E[X]^2E[Y]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2\,\dotso\,X,Yが独立でもV[X]V[Y]とはならない\\ \end{array}$$

離散型確率変数(discrete random variable) / 期待値(expected value)

$$\begin{array}{rcl} 母集団(population)の確率変数&:&X,Y\\ 確率質量凾数(probability\,mass\,function,\,PMF)&:&f_X(x)=P(X=x_i)\\ &&\displaystyle\sum_{i=1}^{\infty} f_X(x_i)=1\\ 累積分布凾数(cumulative\,distribution\,function,\,CDF)&:&\displaystyle F_X(x)=\sum_{i:x_i \leq x} f_X(x_i)\\ 同時確率質量凾数(joint\,probability\,mass\,function)&:&f_X(x, y)=P(X=x_i,\,Y=y_j)\\ &&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} f_{XY}(x_i, y_j)=1\\ 周辺確率質量凾数(marginal\, probability\, mass\, function)&:&\displaystyle f_X(x)=\sum_{j=1}^{\infty} f_{XY}(x, y_i)\\ \end{array}$$ $$\begin{array}{rcl} E[g(X)]&\equiv&\displaystyle\sum_{i=1}^{\infty} g(x_i) f_X(x_i) \\ E[X]&=&\displaystyle\sum_{i=1}^{\infty} (x_i) f_X(x_i) \\ E[cX] &=&\displaystyle\sum_{i=1}^{\infty} (c x_i) f_X(x_i) \,\dotso\,cは定数\\ &=&c\displaystyle\sum_{i=1}^{\infty} x_i f_X(x_i) \\ &=&c E[X] \\ E[X \pm t]&=&\displaystyle\sum_{i=1}^{\infty} (x_i \pm t) f_X(x_i) \,\dotso\,tは定数\\ &=&\displaystyle\sum_{i=1}^{\infty} (x_i f_X(x_i) \pm t f_X(x_i)) \\ &=&\displaystyle\sum_{i=1}^{\infty} x_i f_X(x_i) \pm \displaystyle\sum_{i=1}^{\infty} t f_X(x_i) \\ &=&\displaystyle\sum_{i=1}^{\infty} x_i f_X(x_i) \pm t \displaystyle\sum_{i=1}^{\infty} f_X(x_i) \\ &=&E[X] \pm t\\ E[X \pm Y]&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (x_i \pm y_j) f_{XY}(x_i, y_j) \\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (x_i f_{XY}(x_i, y_j) \pm y_j f_{XY}(x_i, y_j)) \\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} x_i f_{XY}(x_i, y_j) \pm \displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} y_j f_{XY}(x_i, y_j)\\ &=&\displaystyle\sum_{i=1}^{\infty} x_i \sum_{j=1}^{\infty} f_{XY}(x_i, y_j) \pm \displaystyle\sum_{j=1}^{\infty} y_j \sum_{i=1}^{\infty}f_{XY}(x_i, y_j)\\ &=&\displaystyle\sum_{i=1}^{\infty} x_i f_{X}(x_i) \pm \displaystyle\sum_{j=1}^{\infty} y_j f_{Y}(y_j)\,\dotso\,周辺確率質量凾数を適用\\ &=& E[X] \pm E[Y] \\ \end{array}$$

X,Yが独立の場合

$$\begin{array}{rcl} E[XY]&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i y_j)f_{XY}(x_i, y_j)\\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}x_iy_jf_{X}(x_i)f_{Y}(y_i) \,\dotso\,X,Yが独立\,f_{XY}(x, y)=f_{X}(x)f_{Y}(y)\\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}x_if_{X}(x_i)y_jf_{Y}(y_i)\\ &=&\displaystyle\left(\sum_{i=1}^{\infty}x_if_{X}(x_i)\right)\left(\sum_{j=1}^{\infty} y_jf_{Y}(y_i)\right) \,\dotso\,\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_ib_j=\sum_{i=1}^{\infty}a_i\sum_{j=1}^{\infty}b_j\\ &=&E[X]E[Y] \end{array}$$

ガウス積分(Gaussian integral)

$$\begin{array}{rcl} \displaystyle \int_{\mathbb{R}}\mathrm{e}^{-(x^2+y^2)}\mathrm{d}A &=& \displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathrm{e}^{-(x^2+y^2)}\mathrm{d}x\mathrm{d}y\dotso \mathrm{d}A=\mathrm{d}x\mathrm{d}y\\ &=& \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-x^2}\mathrm{d}x\int_{-\infty}^{\infty}\mathrm{e}^{-y^2}\mathrm{d}y\\ &=& \displaystyle \left(\int_{-\infty}^{\infty}\mathrm{e}^{-t^2}\mathrm{d}t\right)^2\\ \displaystyle \int_{\mathbb{R}}\mathrm{e}^{-(x^2+y^2)}\mathrm{d}A &=& \displaystyle \int_{0}^{2\pi}\int_{0}^{\infty}\mathrm{e}^{-r^2}r\mathrm{d}r\mathrm{d}\theta\dotso \mathrm{d}A=r\mathrm{d}r\mathrm{d}\theta\\ &=& \displaystyle 2\pi\int_{0}^{\infty}\mathrm{e}^{-r^2}r\mathrm{d}r\\ &=& \displaystyle 2\pi\int_{0}^{\infty}\mathrm{e}^{-z} r\frac{1}{2r}\mathrm{d}z \dotso z=r^2,\frac{\mathrm{d}z}{\mathrm{d}r}=2r,\mathrm{d}r=\frac{1}{2r}\mathrm{d}z\\ &=& \displaystyle \pi\int_{0}^{\infty}\mathrm{e}^{-z}\mathrm{d}z\\ &=& \displaystyle \pi\left[-\mathrm{e}^{-z}\right]_0^{\infty} = \displaystyle \pi\left[\left(-\mathrm{e}^{-\infty}\right)-\left(-\mathrm{e}^{-0}\right)\right] = \pi\left[0-\left(-1\right)\right]\\ &=& \displaystyle \pi\\ \displaystyle \left(\int_{-\infty}^{\infty}\mathrm{e}^{-t^2}\mathrm{d}t\right)^2 &=& \displaystyle \pi\\ \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-t^2}\mathrm{d}t &=& \displaystyle \sqrt{\pi} \end{array}$$

\(-t^2\)でなく\(-\left(\frac{x+b}{c}\right)^2\)の場合(\(b,cは定数\))

$$\begin{array}{rcl} \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-\left(\frac{x+b}{c}\right)^2}\mathrm{d}x &=& \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-z^2}\,c\,\mathrm{d}z \,\dotso\,\frac{x+b}{c}=z,\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{1}{c},\mathrm{d}x=c\mathrm{d}z\\ &=& \displaystyle c\int_{-\infty}^{\infty}\mathrm{e}^{-z^2}\mathrm{d}z\\ &=& \displaystyle c\sqrt{\pi}\\ \end{array}$$