二項分布の歪度と尖度の導出

二項分布の歪度の導出

$$\begin{equation*}\begin{split}E\left[ \left( X-\mu\right) ^{3}\right] &=E\left( X^{3}\right) -3μE\left( X^{2}\right) +3\mu^{2}E\left( X\right) -E\left( \mu^{3}\right) \\ &=E\left( X^{3}\right) -3μE\left( X^{2}\right) +2\mu^{3}\end{split}\end{equation*}$$

また、

$$\begin{equation*}\begin{split}E\left( X^{3}\right) &=\frac {d^{3}M_{X}\left( t\right) }{dt^{3}}\left( t=0\right) \\ &=\{n\left( n-1\right) \left( e^{t}p+q\right) ^{n-2}\left( pe^t\right) ^{2}+np\left( e^{t}p+q\right) ^{n-1}e^{t}\}'\left( t=0\right) \\ &=n\left( n-1\right) \left( n-2\right) p^{3}+3n\left( n-1\right) p^{2}+np\end{split}\end{equation*}$$

となるので、

$$\begin{equation*}\begin{split}E\left[ \left( X-\mu\right) ^{3}\right] &=E\left( X^{3}\right) -3μE\left( X^{2}\right) +2{\mu}^3\\ &=M\left( N-1\right) \left( n-2\right) p^{3}+3n\left( n-1\right) p^{2}+np -3np\left\{ n\left( n-1\right) p^{2}+np\right\} +2n^{3}p^{3}\\ &=np\left( p-1\right) \left( 2p-1\right)\end{split}\end{equation*}$$

よって、歪度は

$$\begin{equation*}\begin{split}\dfrac {E\left[ \left( x-\mu\right) ^{3}\right] }{\sigma^{3}}&=\dfrac {np\left( p-1\right) \left( 2p-1\right) }{np\left( 1-p\right) \sqrt {np\left( 1-p\right) }}\\ &=\dfrac {1-2p}{\left( 1-p\right) \sqrt {np\left( 1-p\right) }}\end{split}\end{equation*}$$

二項分布の尖度の導出

$$\begin{equation*}\begin{split}E\left[ \left( x-\mu\right)^4\right] =E\left( X^{4}\right) -4μE\left( X^{3}\right) +6μE\left( X^{2}\right) -3\mu^4\end{split}\end{equation*}$$

また、

$$\begin{equation*}\begin{split}E\left( X^{4}\right) &=\dfrac {d^{4}M_{x}\left( t\right) }{dt^{4}}\left( t=0\right) \\ &=n\left( n-1\right) \left( n-2\right) \left( n-3\right) p^{4}+6n\left( n-1\right) \left( n-2\right) p^{3}+7n\left( n-1\right) p^{2}+np\end{split}\end{equation*}$$

となるので、

$$\begin{equation*}\begin{split}E\left[ \left( x-\mu\right) ^{4}\right] &=3n\left( n-2\right) p^{4}-6n\left( n-2\right) p^{3}+n\left( 3n-7\right) p^{2}+np\\ &=np\left( 3np^{3}-6p^{3}-6np^{2}+12p^{2}+3np-7p+1\right) \\ &=np\left( 1-p\right) \left\{ 1+3\left( n-2\right) p\left( 1-p\right) \right\} \end{split}\end{equation*}$$

よって、尖度は

$$\begin{equation*}\begin{split}\dfrac {E\left[ \left( x-\mu\right)^4\right] }{\sigma^{4}}-3&=\dfrac {np\left( 1-p\right) \left( 1+3(n-2\right) p\left( 1-P\right) }{n^{2}p^{2}\left( 1-p\right) ^{2}}-3\\ &=\dfrac {1-6p\left( 1-p\right) }{np\left( 1-p\right) ^{2}}\end{split}\end{equation*}$$

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