Kurtoza

Niech:

${\displaystyle X\colon \Omega \to \mathbb {R} ,}$

${\displaystyle {\mbox{Kurt}}_{E}={\frac {\mu _{4}(X)}{\sigma ^{4}}}-3,}$

${\displaystyle \mu _{n}(X)=\operatorname {E} ((X-\operatorname {E} X)^{n})}$ – moment centralny n–tego rzędu,

${\displaystyle m_{n}(X)=\operatorname {E} ((X)^{n})}$ – moment zwykły n–tego rzędu,

Dystrybuanta rozkładu normalnego oraz gęstość prawdopodobieństwa rozkładu normalnego to odpowiednio:

${\displaystyle P_{x}(A)=\int _{A}f(x)dx,\qquad f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{\frac {-(x-m)^{2}}{2\sigma ^{2}}},dla\qquad x\in \mathbb {R} .}$

Wiadomo, że w rozkładzie normalnym:

${\displaystyle m=\operatorname {E} X=m_{1}(X),}$

${\displaystyle \sigma ^{2}=\operatorname {D} ^{2}X=\operatorname {E} ((X-\operatorname {E} X)^{2})=\mu _{2}(X).}$

Mamy:

a)

${\displaystyle m_{n}(X)=\int _{\Omega }X^{n}(w)P(dw)=\int _{\mathbb {R} }x^{n}f(x)dx,}$

b)

${\displaystyle {\begin{aligned}\mu _{4}(X)&=\operatorname {E} ((X-\operatorname {E} X)^{4})\\&=\operatorname {E} (X^{4}-4(\operatorname {E} X)X^{3}+6(\operatorname {E} X)^{2}X^{2}-4(\operatorname {E} X)^{3}X+(\operatorname {E} X)^{4})\\&=\operatorname {E} (X^{4})-4m\operatorname {E} (X^{3})+6\operatorname {E} (X^{2})(\operatorname {E} X)^{2}-4(\operatorname {E} X)^{3}m+m^{4}\\&=\operatorname {E} (X^{4})-4m\operatorname {E} (X^{3})+6m^{2}\operatorname {E} (X^{2})-3m^{4}.\end{aligned}}}$

Obliczamy momenty zwykłe:

${\displaystyle \operatorname {E} (X^{2})=\int _{-\infty }^{+\infty }x^{2}{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{\frac {-(x-m)^{2}}{2\sigma ^{2}}}dx={\frac {1}{\sigma {\sqrt {2\pi }}}}\int _{-\infty }^{+\infty }(z{\sqrt {2}}\sigma +m)^{2}e^{-z^{2}}{\sqrt {2}}\sigma dz=^{*}}$

${\displaystyle z={\frac {x-m}{{\sqrt {2}}\sigma }},\quad x=z{\sqrt {2}}\sigma +m}$

${\displaystyle dz={\frac {1}{{\sqrt {2}}\sigma }},\quad dx={\sqrt {2}}\sigma dz}$

${\displaystyle =^{*}{\frac {1}{\sqrt {\pi }}}\int _{-\infty }^{+\infty }(2z^{2}\sigma ^{2}+2{\sqrt {2}}z\sigma m+m^{2})e^{-z^{2}}dz=}$

${\displaystyle ={\frac {1}{\sqrt {\pi }}}{\Bigg (}2\sigma ^{2}\left(\int _{-\infty }^{+\infty }z^{2}e^{-z^{2}}dz\right)+2{\sqrt {2}}\sigma m\left(\int _{-\infty }^{+\infty }ze^{-z^{2}}dz\right)_{=0}+m^{2}\left(\int _{-\infty }^{+\infty }e^{-z^{2}}dz\right)_{={\sqrt {\pi }}}{\Bigg )}=}$

${\displaystyle ={\frac {1}{\sqrt {\pi }}}\left(2\sigma ^{2}\int _{-\infty }^{+\infty }z^{2}e^{-z^{2}}dz+m^{2}{\sqrt {\pi }}\right)=^{*}}$

${\displaystyle \int _{-\infty }^{+\infty }z^{2}e^{-z^{2}}dz=\int _{-\infty }^{+\infty }z\cdot ze^{-z^{2}}dz=-{\frac {1}{2}}ze^{-z^{2}}{\Bigg |}_{-\infty }^{+\infty }+{\frac {1}{2}}\int _{-\infty }^{+\infty }e^{-z^{2}}dz=}$

${\displaystyle =-{\frac {1}{2}}ze^{-z^{2}}{\Bigg |}_{-\infty }^{+\infty }+{\frac {1}{2}}\int _{-\infty }^{+\infty }e^{-z^{2}}dz={\frac {\sqrt {\pi }}{2}}}$

${\displaystyle =^{*}{\frac {1}{\sqrt {\pi }}}(2\sigma ^{2}{\frac {\sqrt {\pi }}{2}}+m^{2}{\sqrt {\pi }})=\sigma ^{2}+m^{2}}$

${\displaystyle \operatorname {E} (X^{3})=\int _{-\infty }^{+\infty }x^{3}{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{\frac {-(x-m)^{2}}{2\sigma ^{2}}}dx={\frac {1}{\sqrt {\pi }}}\int _{-\infty }^{+\infty }(z{\sqrt {2}}\sigma +m)^{3}e^{-z^{2}}dz=^{*}}$

${\displaystyle z={\frac {x-m}{{\sqrt {2}}\sigma }},\quad x=z{\sqrt {2}}\sigma +m}$

${\displaystyle dz={\frac {1}{{\sqrt {2}}\sigma }},\quad dx={\sqrt {2}}\sigma dz}$

${\displaystyle =^{*}{\frac {1}{\sqrt {\pi }}}\int _{-\infty }^{+\infty }(z^{3}{\sqrt {2^{3}}}\sigma ^{3}+3z^{2}2\sigma ^{2}m+3z{\sqrt {2}}\sigma m^{2}+m^{3})e^{-z^{2}}dz=}$

${\displaystyle ={\frac {2{\sqrt {2}}\sigma ^{3}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }z^{3}e^{-z^{2}}dz{\bigg )}_{=0}+{\frac {6\sigma ^{2}m}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }z^{2}e^{-z^{2}}dz{\bigg )}_{={\frac {\sqrt {\pi }}{2}}}+}$

${\displaystyle +{\frac {3{\sqrt {2}}\sigma m^{2}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }ze^{-z^{2}}dz{\bigg )}_{=0}+{\frac {m^{3}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }e^{-z^{2}}dz{\bigg )}_{={\sqrt {\pi }}}=}$

${\displaystyle ={\frac {6\sigma ^{2}m}{\sqrt {\pi }}}\cdot {\frac {\sqrt {\pi }}{2}}+m^{3}=3\sigma ^{2}m+m^{3}}$

${\displaystyle \operatorname {E} (X^{4})=\int _{-\infty }^{+\infty }x^{4}{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{\frac {-(x-m)^{2}}{2\sigma ^{2}}}dx={\frac {1}{\sqrt {\pi }}}\int _{-\infty }^{+\infty }(z{\sqrt {2}}\sigma +m)^{4}e^{-z^{2}}dz=^{*}}$

${\displaystyle z={\frac {x-m}{{\sqrt {2}}\sigma }},\quad x=z{\sqrt {2}}\sigma +m}$

${\displaystyle dz={\frac {1}{{\sqrt {2}}\sigma }},\quad dx={\sqrt {2}}\sigma dz}$

${\displaystyle =^{*}{\frac {1}{\sqrt {\pi }}}\int _{-\infty }^{+\infty }(4\sigma ^{4}z^{4}+32{\sqrt {2}}\sigma ^{3}mz^{3}+12\sigma ^{2}m^{2}z^{2}+4{\sqrt {2}}\sigma m^{3}z+m^{4})e^{-z^{2}}dz=}$

${\displaystyle ={\frac {4\sigma ^{4}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }z^{4}e^{-z^{2}}dz{\bigg )}+{\frac {32{\sqrt {2}}\sigma ^{3}m}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }z^{3}e^{-z^{2}}dz{\bigg )}_{=0}+}$

${\displaystyle +{\frac {12\sigma ^{2}m^{2}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }z^{2}e^{-z^{2}}dz{\bigg )}_{={\frac {\sqrt {\pi }}{2}}}+{\frac {4{\sqrt {2}}\sigma m^{3}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }ze^{-z^{2}}dz{\bigg )}_{=0}+{\frac {m^{4}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }e^{-z^{2}}dz{\bigg )}_{={\sqrt {\pi }}}=}$

${\displaystyle ={\frac {4\sigma ^{4}}{\sqrt {\pi }}}{\bigg (}\int _{-\infty }^{+\infty }z^{4}e^{-z^{2}}dz{\bigg )}+0+{\frac {12\sigma ^{2}m^{2}}{\sqrt {\pi }}}(\sigma ^{2}+m^{2})+0+{\frac {m^{4}}{\sqrt {\pi }}}{\sqrt {\pi }}=^{**}}$

${\displaystyle \int _{-\infty }^{+\infty }z^{4}e^{-z^{2}}dz=\int _{-\infty }^{+\infty }z\cdot z^{3}e^{-z^{2}}dz=^{*}}$

${\displaystyle u=z,\quad u'=1,\quad v'=z^{3}e^{-z^{2}},\quad v=?}$

${\displaystyle v=\int z^{3}e^{-z^{2}}dz={\frac {1}{2}}\int te^{-t}dt={\frac {1}{2}}(-te^{-t})+\int e^{-t}dt={\frac {1}{2}}(-z^{2}e^{-z^{2}}-e^{-z^{2}})}$

${\displaystyle t=z^{2}\quad u=t,\quad u'=1,\quad dz={\frac {1}{2z}}dt,\quad v'=e^{-t},\quad v=-e^{-t}}$

${\displaystyle =^{*}z\cdot {\frac {1}{2}}(-z^{2}e^{-z^{2}}-e^{-z^{2}}){\Big |}_{-\infty }^{+\infty }-\int _{-\infty }^{+\infty }{\frac {1}{2}}(-z^{2}e^{-z^{2}}-e^{-z^{2}})dz={\frac {1}{2}}\cdot 0+{\frac {1}{2}}\int _{-\infty }^{+\infty }z^{2}e^{-z^{2}}dz+{\frac {1}{2}}\int _{-\infty }^{+\infty }e^{-z^{2}}dz=}$

${\displaystyle ={\frac {1}{2}}\cdot {\frac {\sqrt {\pi }}{2}}+{\frac {1}{2}}\cdot {\sqrt {\pi }}={\frac {3{\sqrt {\pi }}}{4}}}$

${\displaystyle =^{**}{\frac {4\sigma ^{4}}{\sqrt {\pi }}}\cdot {\frac {3{\sqrt {\pi }}}{4}}+{\frac {12\sigma ^{2}m^{2}}{\sqrt {\pi }}}\cdot {\frac {\sqrt {\pi }}{2}}+{\frac {m^{4}}{\sqrt {\pi }}}\cdot {\sqrt {\pi }}=3\sigma ^{4}+6\sigma ^{2}m^{2}+m^{4}\quad (=\operatorname {E} (X^{4}))}$

Obliczone wartości:

${\displaystyle \operatorname {E} (X^{2})=\sigma ^{2}+m^{2}}$

${\displaystyle \operatorname {E} (X^{3})=3\sigma ^{2}m+m^{3}}$

${\displaystyle \operatorname {E} (X^{4})=3\sigma ^{4}+6\sigma ^{2}m^{2}+m^{4}}$

podstawiamy do wzoru na czwarty moment centralny z punktu b):

${\displaystyle {\begin{aligned}\mu _{4}(X)&=\operatorname {E} (X^{4})-4m\operatorname {E} (X^{3})+6m^{2}\operatorname {E} (X^{2})-3m^{4}\\&=(3\sigma ^{4}+6\sigma ^{2}m^{2}+m^{4})-4m(3\sigma ^{2}m+m^{3})+6m^{2}(\sigma ^{2}+m^{2})-3m^{4}\\&=3\sigma ^{4}+6\sigma ^{2}m^{2}+m^{4}-12\sigma ^{2}m^{2}-4m^{4}+6\sigma ^{2}m^{2}+6m^{4}-3m^{4}=3\sigma ^{4}\end{aligned}}}$

Stąd kurtoza jest równa:

${\displaystyle \operatorname {Kurt} _{E}={\frac {(\mu _{4}(X))}{(\sigma ^{2})^{2}}}-3={\frac {3\sigma ^{4}}{\sigma ^{4}}}-3=3-3=0.}$