标准正态分布k阶原点矩公式

今天在看 χ 2 chi^2 χ2分布，计算其方差时，遇到了标准正态分布的四阶原点矩。书上直接写 E ( X i 4 ) = 3 E(X_i^4)=3 E(Xi4)=3，很好奇。设 X i ∼ N ( 0 , 1 ) X_isimmathcal{N}(0,1) Xi∼N(0,1)想根据定义计算： E ( X i 4 ) = ∫ − ∞ + ∞ x 4 1 2 π e − x 2 2 d x E(X_i^4)=int_{-infty}^{+infty}x^4frac{1}{sqrt{2pi}}e^{-frac{x^2}{2}}dx E(Xi4)=∫−∞+∞x42π 1e−2x2dx 计算起来复杂度有点高，K阶就更不敢想。 想着估计结果是 k k k的递推关系式。所有直接计算： E ( X k ) = ∫ − ∞ + ∞ x k 1 2 π e − x 2 2 d x = 1 k + 1 x k + 1 1 2 π e − x 2 2 ∣ − ∞ + ∞ + 1 k + 1 ∫ − ∞ + ∞ x k + 2 1 2 π e − x 2 2 d x = 0 + 1 k + 1 E ( X k + 2 ) egin{aligned} E(X^k) &= int_{-infty}^{+infty}x^kfrac{1}{sqrt{2pi}}e^{-frac{x^2}{2}}dx \ & = frac{1}{k+1}x^{k+1}frac{1}{sqrt{2pi}}e^{-frac{x^2}{2}} |_{-infty}^{+infty} + frac{1}{k+1}int_{-infty}^{+infty}x^{k+2}frac{1}{sqrt{2pi}}e^{-frac{x^2}{2}}dx \ & = 0 + frac{1}{k+1}E(X^{k+2}) end{aligned} E(Xk)=∫−∞+∞xk2π 1e−2x2dx=k+11xk+12π 1e−2x2∣−∞+∞+k+11∫−∞+∞xk+22π 1e−2x2dx=0+k+11E(Xk+2) 从上式看，结果已经很明显了，其递推关系为： E ( X k ) = ( k − 1 ) E ( X k − 2 ) ， k = 2 , 3 , 4 ⋯ E(X^k) = (k-1)E(X^{k-2})，k=2,3,4cdots E(Xk)=(k−1)E(Xk−2)，k=2,3,4⋯ 其中: E ( X 0 ) = ∫ − ∞ + ∞ x 0 1 2 π e − x 2 2 d x = 1 E(X^0)=int_{-infty}^{+infty}x^0frac{1}{sqrt{2pi}}e^{-frac{x^2}{2}}dx=1 E(X0)=∫−∞+∞x02π 1e−2x2dx=1 E ( X 1 ) = 0 E(X^1)=0 E(X1)=0 所以，当 k = 2 i k=2i k=2i为偶数时： E ( X k ) = ( k − 1 ) ( k − 3 ) ⋯ 3 × 1 × E ( X 0 ) = ∏ i = 1 k / 2 ( 2 i − 1 ) egin{aligned} E(X^k) &= (k-1)(k-3)cdots3 imes1 imes E(X^0) \ &=prod_{i=1}^{k/2}(2i-1) end{aligned} E(Xk)=(k−1)(k−3)⋯3×1×E(X0)=i=1∏k/2(2i−1) 当 k = 2 i − 1 k=2i-1 k=2i−1为奇数时： E ( X k ) = ( k − 1 ) ( k − 3 ) ⋯ 3 × 1 × E ( X 1 ) = 0 egin{aligned} E(X^k) &= (k-1)(k-3)cdots3 imes1 imes E(X^1) \ &=0 end{aligned} E(Xk)=(k−1)(k−3)⋯3×1×E(X1)=0 综上有： E ( X k ) = { ∏ i = 1 k / 2 ( 2 i − 1 ) k = 2 i , i = 1 , 2 , 3 ⋯ 0 k = 2 i − 1 E(X^k) = left { egin{array}{ll} prod_{i=1}^{k/2}(2i-1) & k=2i, i=1,2,3cdots \ & \ 0 & k=2i-1 end{array} ight. E(Xk)=⎩⎨⎧∏i=1k/2(2i−1)0k=2i,i=1,2,3⋯k=2i−1