由于 χ 2 = ∑ i = 1 n X i 2 \chi^2 = \sum\limits_{i=1}^{n}X_i^2 χ2=i=1∑n​Xi2​，其中 X i ∼ N ( 0 , 1 ) X_i \sim N(0,1) Xi​∼N(0,1) ,因此有

E ( X i 2 ) = D ( X i ) + ( E X i ) 2 = 1 + 0 = 1 E(X_i^2) = D(X_i)+(EX_i)^2 = 1+0=1 E(Xi2​)=D(Xi​)+(EXi​)2=1+0=1

因此 E ( χ 2 ) = n ∗ E ( X i 2 ) = n E(\chi^2) = n*E(X_i^2) = n E(χ2)=n∗E(Xi2​)=n

D ( χ 2 ) = n ∗ D ( X i 2 ) = n ∗ [ E ( X i 2 ) 2 − E ( X i 2 ) ] = n ∗ [ E ( X i 4 ) − E ( X i 2 ) ] = n E ( X i 4 ) − n \begin{aligned} D(\chi^2) &= n*D(X_i^2) \\&= n*[E(X_i^2)^2-E(X_i^2)] \\&= n*[E(X_i^4)-E(X_i^2)] \\&=nE(X_i^4)-n\end{aligned} D(χ2)​=n∗D(Xi2​)=n∗[E(Xi2​)2−E(Xi2​)]=n∗[E(Xi4​)−E(Xi2​)]=nE(Xi4​)−n​

我们需要知道 E ( X i 4 ) E(X_i^4) E(Xi4​)， 目前没有更好的方式，我们尝试使用期望的定义进行计算

E ( X i 4 ) = ∫ − ∞ + ∞ x 4 1 2 π e − x 2 2 d x = 1 2 π ∫ − ∞ + ∞ x 4 e − x 2 2 d x = 1 2 π [ − x 3 e − x 2 2 ∣ − ∞ + ∞ + ∫ − ∞ + ∞ 3 x 2 e − x 2 2 d x ] = 0 + 3 ∫ − ∞ + ∞ 1 2 π x 2 e − x 2 2 d x = 3 ∫ − ∞ + ∞ x 2 1 2 π e − x 2 2 d x \begin{aligned} E(X_i^4) &= \int_{-\infty}^{+\infty}x^4\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx \\&= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}x^4e^{-\frac{x^2}{2}}dx \\&= \frac{1}{\sqrt{2\pi}}\bigg[-x^3e^{-\frac{x^2}{2}}\bigg|_{-\infty}^{+\infty}+\int_{-\infty}^{+\infty}3x^2e^{-\frac{x^2}{2}}dx\bigg] \\&=0+3\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}x^2e^{-\frac{x^2}{2}}dx \\&= 3\int_{-\infty}^{+\infty}x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx\end{aligned} E(Xi4​)​=∫−∞+∞​x42π ​1​e−2x2​dx=2π ​1​∫−∞+∞​x4e−2x2​dx=2π ​1​[−x3e−2x2​∣∣∣∣​−∞+∞​+∫−∞+∞​3x2e−2x2​dx]=0+3∫−∞+∞​2π ​1​x2e−2x2​dx=3∫−∞+∞​x22π ​1​e−2x2​dx​

这里对后面的积分项可继续采用分部积分方法进行处理，但是这地方其实有个技巧 ，根据期望定义有

E ( X i 2 ) = ∫ − ∞ + ∞ x 2 1 2 π e − x 2 2 d x \begin{aligned} E(X_i^2) = \int_{-\infty}^{+\infty}x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx \end{aligned} E(Xi2​)=∫−∞+∞​x22π ​1​e−2x2​dx​

前面我们已经算出 E ( X i 2 ) = 1 E(X_i^2)=1 E(Xi2​)=1

因此有 E ( X i 2 ) = ∫ − ∞ + ∞ x 2 1 2 π e − x 2 2 d x = 1 \begin{aligned} E(X_i^2) = \int_{-\infty}^{+\infty}x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=1 \end{aligned} E(Xi2​)=∫−∞+∞​x22π ​1​e−2x2​dx=1​

所以有 E ( X i 4 ) = 3 ∗ 1 = 3 E(X_i^4) = 3*1 = 3 E(Xi4​)=3∗1=3

∴ D ( χ 2 ) = n E ( X i 4 ) − n = 3 n − n = 2 n \therefore D(\chi^2) = nE(X_i^4)-n = 3n-n=2n ∴D(χ2)=nE(Xi4​)−n=3n−n=2n