概率良定性

$$\int_0^{+\infty} f_n(x) \, dx = \frac{1}{2^{n/2} \Gamma(n/2)} \int_0^{+\infty} x^{n/2 - 1} e^{-x/2} \, dx$$

令 $y = \frac{x}{2}$，则 $x = 2y$，$dx = 2 dy$

$$= \frac{1}{2^{n/2} \Gamma(n/2)} \int_0^{+\infty} (2y)^{n/2 - 1} e^{-y} \cdot 2 \, dy = \frac{2^{n/2}}{2^{n/2} \Gamma(n/2)} \int_0^{+\infty} y^{n/2 - 1} e^{-y} \, dy = \frac{\Gamma(n/2)}{\Gamma(n/2)} = 1$$

矩母函数

$$M_{\chi^2}(t) = E[e^{t \chi^2}] = \int_0^{+\infty} e^{t x} f_n(x) \, dx = \frac{1}{2^{n/2} \Gamma(n/2)} \int_0^{+\infty} x^{n/2 - 1} e^{-x/2 + t x} \, dx$$

令 $u = \left(\frac{1}{2} - t\right) x$，则 $x = \frac{u}{\frac{1}{2} - t}$，$dx = \frac{du}{\frac{1}{2} - t}$

$$= \frac{1}{2^{n/2} \Gamma(n/2)} \int_0^{+\infty} \left(\frac{u}{\frac{1}{2} - t}\right)^{n/2 - 1} e^{-u} \cdot \frac{du}{\frac{1}{2} - t} = \frac{1}{2^{n/2} \Gamma(n/2)} \cdot \frac{1}{\left(\frac{1}{2} - t\right)^{n/2}} \int_0^{+\infty} u^{n/2 - 1} e^{-u} \, du = \frac{1}{2^{n/2} \Gamma(n/2)} \cdot \frac{1}{\left(\frac{1}{2} - t\right)^{n/2}} \Gamma\left(\frac{n}{2}\right) = (1 - 2t)^{-n/2}$$

期望值

$$E[\chi^2] = M_{\chi^2}'(0) = \left. \frac{n}{2} (1 - 2t)^{-n/2 - 1} \cdot 2 \right|_{t=0} = n$$

方差

$$E[\chi^4] = M_{\chi^2}''(0) = \left. \frac{n(n+2)}{4} (1 - 2t)^{-n/2 - 2} \cdot 4 \right|_{t=0} = n(n+2)$$

$$V[\chi^2] = E[\chi^4] - (E[\chi^2])^2 = n(n+2) - n^2 = 2n$$

$\square$

由于 $X_i \sim N(0,1)$，其平方 $X_i^2$ 的矩母函数为：

$$M_{X_i^2}(t) = E[e^{tX_i^2}] = \int_{-\infty}^\infty e^{tx^2} \frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{1-2t}{2}x^2} dx = (1-2t)^{-1/2}$$

由于 $X_i$ 相互独立，和的矩母函数等于矩母函数的积：

$$M_{\sum X_i^2}(t) = \prod_{i=1}^n (1-2t)^{-1/2} = (1-2t)^{-n/2}$$

这正是 $\chi^2(n)$ 的矩母函数。
$\square$

直接对 PDF 进行积分非常繁琐。利用 $F$ 分布的构造定义证明更为简洁。
由定义知，若 $U \sim \chi^2(m), V \sim \chi^2(n)$ 且独立，则 $F = \frac{U/m}{V/n}$。
先计算 $\chi^2(k)$ 的倒数期望。设 $Y \sim \chi^2(k)$：

$$E[Y^r] = \int_0^\infty x^r \frac{1}{2^{k/2}\Gamma(k/2)} x^{k/2-1} e^{-x/2} dx = \frac{2^r \Gamma(k/2+r)}{\Gamma(k/2)}$$

所以

$$E[V^{-1}] = \frac{2^{-1} \Gamma(n/2-1)}{\Gamma(n/2)} = \frac{1}{2} \frac{1}{n/2 - 1} = \frac{1}{n-2} \quad (n>2)$$

$$E[V^{-2}] = \frac{2^{-2} \Gamma(n/2-2)}{\Gamma(n/2)} = \frac{1}{4} \frac{1}{(n/2-1)(n/2-2)} = \frac{1}{(n-2)(n-4)} \quad (n>4)$$

期望值
由于 $U, V$ 独立

$$E[F] = \frac{n}{m} E[U] E[V^{-1}] = \frac{n}{m} \cdot m \cdot \frac{1}{n-2} = \frac{n}{n-2}$$

方差

$$E[F^2] = \frac{n^2}{m^2} E[U^2] E[V^{-2}]$$

已知 $E[U^2] = V[U] + (E[U])^2 = 2m + m^2 = m(m+2)$

$$E[F^2] = \frac{n^2}{m^2} \cdot m(m+2) \cdot \frac{1}{(n-2)(n-4)} = \frac{n^2(m+2)}{m(n-2)(n-4)}$$

$$V[F] = E[F^2] - (E[F])^2 = \frac{n^2(m+2)}{m(n-2)(n-4)} - \left(\frac{n}{n-2}\right)^2$$

通分整理后得

$$V[F] = \frac{2 n^2 (m+n-2)}{m (n-2)^2 (n-4)}$$

$\square$

利用 $t$ 分布的构造定义证明。
由定义知，若 $Z \sim N(0,1), V \sim \chi^2(n)$ 且独立，则 $t = \frac{Z}{\sqrt{V/n}}$。

期望值
由于 $Z$ 与 $V$ 独立，且 $E[Z]=0$

$$E[t] = E[Z] E\left[\frac{1}{\sqrt{V/n}}\right] = 0 \cdot E\left[\frac{1}{\sqrt{V/n}}\right] = 0$$

方差

$$V[t] = E[t^2] - (E[t])^2 = E\left[\frac{Z^2}{V/n}\right] - 0 = n E[Z^2] E[V^{-1}]$$

已知 $E[Z^2] = V[Z] + (E[Z])^2 = 1$，且此前已证 E[V^{-1}] = \frac{1}

$$V[t] = n \cdot 1 \cdot \frac{1}{n-2} = \frac{n}{n-2}$$

$\square$