【Derivation】MarkDown Letex編碼 之 正態分佈特徵函式證明
阿新 • • 發佈:2019-01-07
**求證:$\varphi(u)=e^{jau-\frac{1}{2}u^2\sigma^2} \ \ \ , t\in R $** **證:** * * $$\varphi(u)=\int _ {-\infty} ^ {+\infty} e^{jux}f(x)dx$$ $$=\int_ {-\infty}^{+\infty} e^{jux} \frac{1}{\sqrt{2\pi\sigma^2}} e^{- \frac{(x-a)^2}{2\sigma^2}}dx$$ * 整理,得: * $$\varphi(u)= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}dx $$ * * beacuse $|jx e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}| \leq |x| e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}$ and $ \frac{1}{\sqrt{2\pi}}|x| e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}} < +\infty$ , $so $可以對$\varphi(u)$求$u$的一階導數, * 有: $$\varphi \prime(u)= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} {jx}\ e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}dx $$ 綜合可推: $$j{(u-j\frac{a}{\sigma^2})\varphi (u)}+\frac{j{\varphi \prime(u) } } {\sigma^2} $$$$=$$ $$ \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}dx =$$$$ \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux- \frac{(x-a)^2}{2\sigma^2}}dx $$ $$=\frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} 1de^{jux- \frac{(x-a)^2}{2\sigma^2}} $$$$=\frac{1}{\sqrt{2\pi\sigma^2}}[e^{jux- \frac{(x-a)^2}{2\sigma^2}}]|_{-\infty}^{+\infty}=0$$ 即得微分方程 $${u\varphi (u)-j\frac{a}{\sigma^2}\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2}=$$ $${(u\sigma^2 -ja)}{\varphi (u)}+{\varphi \prime(u) } =0$$ 即得微分方程 ++++分水嶺,從後往前推+++++++ $${\varphi (u)}+\frac{{\varphi \prime(u) } } {u\sigma^2 -ja} $$ $$={u\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2 -\frac{ja}{u}}=0 $$ 求解: $$\frac{\varphi\prime(u)}{\varphi(u)}=-u\sigma^2+ja$$ 解得:$$\ln\varphi (u)=-\frac{1}{2}u^2D(x)+jau+C$$ 進一步化簡: $$\varphi (u)=e^Ce^{-\frac{1}{2}u^2D(x)+jau}$$ 令$u=0,e^C=\varphi(0)=E[E^(j0X)]=E[e^0]=1$,故$C=0;$ 代入通解為: $$\varphi (u)=e^{jau-\frac{1}{2}u^2D(x)}$$ 由以上推導,**正態分佈特徵函式表示式** 得證
**求證:$\varphi(u)=e^{jau-\frac{1}{2}u^2\sigma^2} \ \ \ , t\in R $** **證:** * * $$\varphi(u)=\int _ {-\infty} ^ {+\infty} e^{jux}f(x)dx$$ $$=\int_ {-\infty}^{+\infty} e^{jux} \frac{1}{\sqrt{2\pi\sigma^2}} e^{- \frac{(x-a)^2}{2\sigma^2}}dx$$ * 整理,得: * $$\varphi(u)= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}dx $$ * * beacuse $|jx e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}| \leq |x| e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}$ and $ \frac{1}{\sqrt{2\pi}}|x| e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}} < +\infty$ , $so $可以對$\varphi(u)$求$u$的一階導數, * 有: $$\varphi \prime(u)= \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} {jx}\ e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}dx $$ 綜合可推: $$j{(u-j\frac{a}{\sigma^2})\varphi (u)}+\frac{j{\varphi \prime(u) } } {\sigma^2} $$$$=$$ $$ \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux} e^{- \frac{(x-a)^2}{2\sigma^2}}dx =$$$$ \frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} { ( ju-\frac{x-a}{\sigma^2} })\ e^{jux- \frac{(x-a)^2}{2\sigma^2}}dx $$ $$=\frac{1}{\sqrt{2\pi\sigma^2}}\int _ {-\infty} ^ {+\infty} 1de^{jux- \frac{(x-a)^2}{2\sigma^2}} $$$$=\frac{1}{\sqrt{2\pi\sigma^2}}[e^{jux- \frac{(x-a)^2}{2\sigma^2}}]|_{-\infty}^{+\infty}=0$$ 即得微分方程 $${u\varphi (u)-j\frac{a}{\sigma^2}\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2}=$$ $${(u\sigma^2 -ja)}{\varphi (u)}+{\varphi \prime(u) } =0$$ 即得微分方程 ++++分水嶺,從後往前推+++++++ $${\varphi (u)}+\frac{{\varphi \prime(u) } } {u\sigma^2 -ja} $$ $$={u\varphi (u)}+\frac{{\varphi \prime(u) } } {\sigma^2 -\frac{ja}{u}}=0 $$ 求解: $$\frac{\varphi\prime(u)}{\varphi(u)}=-u\sigma^2+ja$$ 解得:$$\ln\varphi (u)=-\frac{1}{2}u^2D(x)+jau+C$$ 進一步化簡: $$\varphi (u)=e^Ce^{-\frac{1}{2}u^2D(x)+jau}$$ 令$u=0,e^C=\varphi(0)=E[E^(j0X)]=E[e^0]=1$,故$C=0;$ 代入通解為: $$\varphi (u)=e^{jau-\frac{1}{2}u^2D(x)}$$ 由以上推導,**正態分佈特徵函式表示式** 得證