$$
\begin{eqnarray}
f(x;m,n)&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\frac{x^{(m-1)}(1-x)^{(n-1)}}{B(m,n)}\;\cdots\;ベータ分布}
\\B(m,n)&=&\int_0^1x^{(m-1)}(1-x)^{(n-1)} \mathrm{d}x\;\cdots\;ベータ凾数
\\&=&\frac{(m-1)!\;(n-1)!}{\left\{(m-1)+(n-1)+1\right\}!}
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\int_\alpha^\beta(x-\alpha)^p(\beta-x)^q \mathrm{d}x=\frac{p!\;q!}{(p+q+1)!}(\beta-\alpha)^{(p+q+1)}(第一種オイラー積分)}
\\&=&\frac{(m-1)!\;(n-1)!}{(m+n-1)!}
\end{eqnarray}
$$
連続型確率変数(continuous random variable) の一様分布(uniform distribution)の分散(variance)
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
積率母凾数の二階微分
$$\begin{array}{rcl} \displaystyle M_X^{(2)} &=& \displaystyle \frac{\mathrm{d}^2}{\mathrm{d}t^2}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform.html}{\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (t^{-2})'\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\}' \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[\left\{(tb-1)\mathrm{e}^{tb}\right\}'-\left\{(ta-1)\mathrm{e}^{ta}\right\}'\right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ (tb-1)'\mathrm{e}^{tb}+(tb-1)(\mathrm{e}^{tb})' \right\} -\left\{ (ta-1)'\mathrm{e}^{ta}+(ta-1)(\mathrm{e}^{ta})' \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[\frac{ \displaystyle -2\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +t\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right]}{t^3} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle -2(tb-1)\mathrm{e}^{tb}+2(ta-1)\mathrm{e}^{ta} \displaystyle +tb\mathrm{e}^{tb}+tb(tb-1)\mathrm{e}^{tb} \displaystyle -ta\mathrm{e}^{ta}-ta(ta-1)\mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle \left\{-2(tb-1)+tb+tb(tb-1)\right\} \mathrm{e}^{tb} \displaystyle +\left\{2(ta-1)-ta-ta(ta-1)\right\} \mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} \displaystyle -((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\}\\ &=& \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)}\\ \end{array}$$原点周りの二次モーメント
$$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)} \displaystyle \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} - ((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}+\frac{(tb)^3}{3!}\right) \displaystyle -((ta-1)^2+1) \left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}+\frac{(ta)^3}{3!}\right) \displaystyle \right\} \displaystyle \right]\\ && \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb} (マクローリン展開), t^3が分母にあるのでt^4の項以上は分子にtが残ることになるのでt^3の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle (t^2b^2-2tb+2) \left(1+tb+t^2\frac{b^2}{2}+t^3\frac{b^3}{6}\right) \displaystyle -(t^2a^2-2ta+2) \left(1+ta+t^2\frac{a^2}{2}+t^3\frac{a^3}{6}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left[ \left\{ \left( t^2b^2 +t^3b^3 +t^4\frac{b^4}{2} +t^5\frac{b^5}{6} \right) +\left( -2tb -2t^2b^2 -t^3b^3 -t^4\frac{b^4}{3} \right) +\left(2+2tb + t^2b^2 +t^3\frac{b^3}{3} \right) \right\} -\left\{ \left( t^2a^2 +t^3a^3 +t^4\frac{a^4}{2} +t^5\frac{a^5}{6} \right) +\left( -2ta -2t^2a^2 -t^3a^3 -t^4\frac{a^4}{3} \right) +\left(2+2ta + t^2a^2 +t^3\frac{a^3}{3} \right) \right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \left( 2+t^3\frac{b^3}{3}+t^4\frac{b^4}{6}+t^5\frac{b^5}{6} \right) -\left( 2+t^3\frac{a^3}{3}+t^4\frac{a^4}{6}+t^5\frac{a^5}{6} \right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{b^3-a^3}{3} +t^4\frac{b^4-a^4}{6} +t^5\frac{b^5-a^5}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{(b-a)(b^2+ab+a^2)}{3} +t^4\frac{(b-a)(b+a)(a^2+b^2)}{6} +t^5\frac{(b-a)\frac{b^5-a^5}{b-a}}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{b^2+ab+a^2}{3} +t \frac{(b+a)(a^2+b^2)}{6} +t^2\frac{\left(\frac{b^5-a^5}{b-a}\right)}{6} \displaystyle \right\}\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}=\frac{a^2+ab+b^2}{3} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$分散(二次の中心モーメント)
$$\begin{array}{rcl} \displaystyle V[X] &=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\left(\href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{b+a}{2}}\right)^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\frac{b^2+2ab+a^2}{4}\\ &=&\displaystyle \frac{4(b^2+ab+a^2)-3(b^2+2ab+a^2)}{12}\\ &=&\displaystyle \frac{4b^2+4ab+4a^2-3b^2-6ab-3a^2}{12}\\ &=&\displaystyle \frac{b^2-2ab+a^2}{12}\\ &=&\displaystyle \frac{(b-a)^2}{12}\\ \end{array}$$連続型確率変数(continuous random variable) の一様分布(uniform distribution)の期待値(expected value)
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
積率母凾数の一階微分
$$\begin{array}{rcl} \displaystyle M_X^{(1)} &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform.html}{\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle \left(t^{-1}\right)'\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right) \displaystyle +\left(t^{-1}\right)\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right)' \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle \left(-t^{-2}\right)\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right) \displaystyle +\left(t^{-1}\right)\left(b\mathrm{e}^{tb}-a\mathrm{e}^{ta}\right) \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left( \displaystyle -\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t^2} \displaystyle +\frac{b\mathrm{e}^{tb}-a\mathrm{e}^{ta}}{t} \displaystyle \right)\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle -\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t^2} \displaystyle +\frac{(b\mathrm{e}^{tb}-a\mathrm{e}^{ta})t}{t^2} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left( \displaystyle \frac{tb\mathrm{e}^{tb}-\mathrm{e}^{tb}-ta\mathrm{e}^{ta}+\mathrm{e}^{ta}}{t^2} \displaystyle \right)\\ &=& \displaystyle \frac{1}{t^2(b-a)} \displaystyle \left\{ \displaystyle (tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta} \displaystyle \right\}\\ &=& \displaystyle \frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)}\\ \end{array}$$原点周りの一次モーメント=期待値
$$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)} \displaystyle \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}\right) \displaystyle -(ta-1)\left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}\right) \displaystyle \right\} \displaystyle \right]\\ && \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\dotsb} (マクローリン展開), t^2が分母にあるのでt^3の項以上は分子にtが残ることになるのでt^2の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(1+tb+t^2\frac{b^2}{2}\right) \displaystyle -(ta-1)\left(1+ta+t^2\frac{a^2}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left[ \displaystyle \left\{\left(tb+t^2b^2+t^3\frac{b^3}{2}\right)-\left(1+tb+t^2\frac{b^2}{2}\right)\right\} \displaystyle -\left\{\left(ta+t^2a^2+t^3\frac{a^3}{2}\right)-\left(1+ta+t^2\frac{a^2}{2}\right)\right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t(b-b)+t^2(b^2-\frac{b^2}{2})+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t(a-a)+t^2(a^2-\frac{a^2}{2})+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t^2\frac{b^2}{2}+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t^2\frac{a^2}{2}+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{1}{t^2(b-a)}\left( \displaystyle t^2\frac{b^2-a^2}{2} \displaystyle +t^3\frac{b^3-a^3}{2} \displaystyle \right) \displaystyle \right\}\\ &=&\displaystyle \lim_{t \to 0} \left[\frac{1}{t^2(b-a)}\left\{ \displaystyle t^2 \frac{ (b-a)(b+a) }{2} \displaystyle +t^3 \frac{ (b-a)(b^2+ab+a^2) }{2} \displaystyle \right\}\right] \,\dotso\,b^2-a^2=(b-a)(b+a),\,b^3-a^3=(b-a)(b^2+ab+a^2)\\ &=&\displaystyle \lim_{t \to 0} \displaystyle \left\{ \displaystyle \frac{ (b+a) }{2} \displaystyle +t \frac{ (b^2+ab+a^2) }{2} \displaystyle \right\}\\ &=&\displaystyle \frac{b+a}{2}=\frac{a+b}{2} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$連続型確率変数(continuous random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)
$$f_X(x) =
\begin{cases}
\displaystyle \frac{1}{b-a} & \quad \left\{ a\leq x \leq b\right\}\\
\displaystyle 0 & \quad \left\{ x\lt a,\,b\lt x\right\}\\
\end{cases}
$$
$$\begin{array}{rcl}
\displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\
&=&\displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{tx}f_X(x)\mathrm{d}x\\
&=&\displaystyle \int_{a}^{b}\mathrm{e}^{tx}\left(\frac{1}{b-a}\right)\mathrm{d}x\\
&=&\displaystyle \frac{1}{b-a}\int_{a}^{b}\mathrm{e}^{tx}\mathrm{d}x\\
&=&\displaystyle \frac{1}{b-a}\int_{ta}^{tb}\mathrm{e}^{s}\frac{1}{t}\mathrm{d}s\,\dotso\,s=tx,\,\frac{\mathrm{d}s}{\mathrm{d}x}=t,\, \mathrm{d}x=\frac{1}{t}\mathrm{d}s,\,a\to ta,\,b\to tb\\
&=&\displaystyle \frac{1}{b-a}\frac{1}{t}\int_{ta}^{tb}\mathrm{e}^{s}\mathrm{d}s\\
&=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{s}\right]_{ta}^{tb}\\
&=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{tb}-\mathrm{e}^{ta}\right]\\
&=&\displaystyle \frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}\\
\end{array}$$
連続型確率変数(continuous random variable) / 期待値(expected value)
$$\begin{array}{rcl}
母集団(population)の確率変数&:&X,Y\\
確率密度凾数(probability \, density \, function, \, PDF)&:&\displaystyle\int_{a}^{b}f_X(x)\mathrm{d}x=P(a \leq x \leq b)\\
&&\displaystyle\int_{-\infty}^{\infty}f_X(x)\mathrm{d}x=1\\
累積分布凾数(cumulative \, distribution \, function, \, CDF)&:&\displaystyle F_X(x)=\int_{-\infty}^{x}f_X(t)\mathrm{d}t\\
&&\displaystyle f_X(x)=\frac{\mathrm{d}}{\mathrm{d}x}F_X(x)\\
同時確率密度凾数(joint\,probability\,density\,function)&:&\displaystyle\int_{a}^{b}\int_{c}^{d}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=P(a \leq x \leq b,\,c \leq y \leq d)\\
&&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=1\\
周辺確率密度凾数(marginal\, probability\, density\, function)&:&\displaystyle f_X(x)=\int_{-\infty}^{\infty}f_{XY}(x,y)\mathrm{d}y
\end{array}$$
$$\begin{array}{rcl}
E[g(X)]&\equiv&\displaystyle\int_{-\infty}^{\infty} g(x) f_X(x) \mathrm{d}x\\
E[X]&=&\displaystyle\int_{-\infty}^{\infty} (x) f_X(x) \mathrm{d}x\\
E[cX] &=&\displaystyle\int_{-\infty}^{\infty} (c x) f_X(x) \mathrm{d}x\,\dotso\,cは定数\\
&=&c\displaystyle\int_{-\infty}^{\infty} x f_X(x) \mathrm{d}x\\
&=&c E[X] \\
E[X \pm t]&=&\displaystyle\int_{-\infty}^{\infty} (x \pm t) f_X(x) \mathrm{d}x\,\dotso\,tは定数\\
&=&\displaystyle\int_{-\infty}^{\infty} (x f_X(x) \pm t f_X(x)) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} x f_X(x)
\pm \displaystyle\int_{-\infty}^{\infty} t f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} x f_X(x)
\pm t \displaystyle\int_{-\infty}^{\infty} f_X(x) \mathrm{d}x\\
&=&E[X] \pm t\\
E[X \pm Y]&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x \pm y) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x f_{XY}(x, y) \pm y f_{XY}(x, y)) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x f_{XY}(x, y) \mathrm{d}x\mathrm{d}y
\pm \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} y f_{XY}(x, y)\mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty} x \int_{-\infty}^{\infty} f_{XY}(x, y) \mathrm{d}y\,\mathrm{d}x
\pm \displaystyle\int_{-\infty}^{\infty} y \int_{-\infty}^{\infty} f_{XY}(x, y)\mathrm{d}x\,\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty} x f_{X}(x) \mathrm{d}x
\pm \displaystyle\int_{-\infty}^{\infty} y f_{Y}(y) \mathrm{d}y\,\dotso\,周辺確率密度凾数を適用\\
&=& E[X] \pm E[Y] \\
\end{array}$$
X,Yが独立の場合
$$\begin{array}{rcl} E[XY]&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x y)f_{XY}(x, y)\mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xyf_{X}(x)f_{Y}(y)\mathrm{d}x\mathrm{d}y \,\dotso\,X,Yが独立\,f_{XY}(x, y)=f_{X}(x)f_{Y}(y)\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xf_{X}(x)yf_{Y}(y)\mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\left(\int_{-\infty}^{\infty}xf_{X}(x)\mathrm{d}x\right)\left(\int_{-\infty}^{\infty}yf_{Y}(y)\mathrm{d}y\right) \,\dotso\,\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)g(y)\mathrm{d}x\mathrm{d}y=\left(\int_{-\infty}^{\infty}f(x)\mathrm{d}x\right)\left(\int_{-\infty}^{\infty}g(y)\mathrm{d}y\right)\\ &=&E[X]E[Y] \end{array}$$連続型確率変数(continuous random variable) / 分散(variance)
$$\begin{array}{rcl}
母集団(population)の確率変数&:&X,Y\\
確率密度凾数(probability \, density \, function, \, PDF)&:&\displaystyle\int_{a}^{b}f_X(x)\mathrm{d}x=P(a \leq x \leq b)\\
&&\displaystyle\int_{-\infty}^{\infty}f_X(x)\mathrm{d}x=1\\
累積分布凾数(cumulative \, distribution \, function, \, CDF)&:&\displaystyle F_X(x)=\int_{-\infty}^{x}f_X(t)\mathrm{d}t\\
&&\displaystyle f_X(x)=\frac{\mathrm{d}}{\mathrm{d}x}F_X(x)\\
同時確率密度凾数(joint\,probability\,density\,function)&:&\displaystyle\int_{a}^{b}\int_{c}^{d}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=P(a \leq x \leq b,\,c \leq y \leq d)\\
&&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=1\\
周辺確率密度凾数(marginal\, probability\, density\, function)&:&\displaystyle f_X(x)=\int_{-\infty}^{\infty}f_{XY}(x,y)\mathrm{d}y
\end{array}$$
$$\begin{array}{rcl}
V[X]&\equiv&E[(X-E[X])^2]\\
&=&\displaystyle\int_{-\infty}^{\infty} (x-E[X])^2 f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} (x^2-2E[X]x+E[X]^2) f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} (x^2f_X(x)-2E[X]xf_X(x)+E[X]^2f_X(x)) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} (x^2) f_X(x)\mathrm{d}x
-2E[X]\displaystyle\int_{-\infty}^{\infty} (x) f_X(x)\mathrm{d}x
+E[X]^2\displaystyle\int_{-\infty}^{\infty} f_X(x)\mathrm{d}x\\
&=&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\int_{-\infty}^{\infty} (c x - E[c X])^2 f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} (c x - c E[X])^2 f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} (c (x - E[X]))^2 f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} c^2 (x-E[X])^2 f_X(x) \mathrm{d}x\\
&=&c^2\displaystyle\int_{-\infty}^{\infty} (x-E[X])^2 f_X(x) \mathrm{d}x\\
&=&c^2 V[X]\\
V[X \pm t]&=&E[((X \pm t)-E[X \pm t])^2]\,\dotso\,tは定数\\
&=&\displaystyle\int_{-\infty}^{\infty} ((x \pm t)-E[X \pm t])^2 f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} ((x \pm t)-(E[X] \pm t))^2 f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} (x \pm t - E[X] \mp t))^2 f_X(x) \mathrm{d}x\\
&=&\displaystyle\int_{-\infty}^{\infty} (x-E[X])^2 f_X(x) \mathrm{d}x\\
&=&V[X]\\
V[X \pm Y]&=&E[((X \pm Y)-E[X \pm Y])^2]\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x \pm y)-E[X \pm Y])^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x \pm y)-(E[X] \pm E[Y]))^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x - E[X]) \pm (y - E[Y]))^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x - E[X])^2 \pm 2(x - E[X])(y - E[Y]) +(y - E[Y])^2) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x - E[X])^2 f_{XY}(x, y) \pm 2(x - E[X])(y - E[Y]) f_{XY}(x, y) +(y - E[Y])^2 f_{XY}(x, y)) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x - E[X])^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y
\pm 2\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y
+ \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (y - E[Y])^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty} (x - E[X])^2 \int_{-\infty}^{\infty} f_{XY}(x, y) \mathrm{d}y\, \mathrm{d}x
\pm 2\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y
+ \displaystyle\int_{-\infty}^{\infty} (y - E[Y])^2 \int_{-\infty}^{\infty} f_{XY}(x, y) \mathrm{d}x\,\mathrm{d}y\\
&=&\displaystyle\int_{-\infty}^{\infty} (x - E[X])^2 f_{X}(x) \mathrm{d}x
\pm 2\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y
+ \displaystyle\int_{-\infty}^{\infty} (y - E[Y])^2 f_{Y}(y) \mathrm{d}y \,\dotso\,周辺確率密度凾数を適用\\
&=&V[X] \pm 2Cov[X,Y] +V[Y]\,\dotso\,Cov[X,Y]=\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\
\end{array}$$
X,Yが独立の場合
$$\begin{array}{rcl} V[X \pm Y]&=&\displaystyle V[X]+V[Y]\,\dotso\,X,Yが独立なのでCov[X,Y]=0\\ V[XY]&=&\displaystyle E[(XY)^2]-E[XY]^2\\ &=&\displaystyle E[X^2Y^2]-E[XY]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle E[X^2]E[Y^2]-(E[X]E[Y])^2\,\dotso\,X,Yが独立なのでE[XY]=E[X]E[Y]\\ &=&\displaystyle E[X^2]E[Y^2]-E[X]^2E[Y]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle (V[X]+E[X]^2)(V[Y]+E[Y]^2)-E[X]^2E[Y]^2\,\dotso\,V[X]=E[X^2]-E[X]^2よりE[X^2]=V[X]+E[X]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2+E[X]^2E[Y]^2-E[X]^2E[Y]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2\,\dotso\,X,Yが独立でもV[X]V[Y]とはならない\\ \end{array}$$
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