由\( f\left(x + \frac{1}{x}\right) = \sin\left[\pi \left(x + \frac{1}{x}\right)^2\right] = \sin\left[\pi \left(x^2 + \frac{1}{x^2} + 2\right)\right] = \sin\left[\pi \left(x^2 + \frac{1}{x^2}\right)\right] \)，故\( f\left(x + \frac{1}{x}\right) - f(x) = \sin\left[\pi \left(x^2 + \frac{1}{x^2}\right)\right] - \sin(\pi x^2) = \pi \cdot \frac{1}{x^2} \cos \xi \)，其中\( \pi x^2 < \xi < \pi x^2 + \pi \frac{1}{x^2} \)，故\( \lim_{x \to \infty} x\left[ f\left(x + \frac{1}{x}\right) - f(x) \right] = \lim_{x \to \infty} \left( \pi \cdot \frac{1}{x} \cos \xi \right) = 0 \)，由归结原则，原式\( = 0 \).

令\( \frac{m}{x} = y \)，则\( x = \frac{m}{y} \)，\( \mathrm{d}x = -\frac{m}{y^2} \mathrm{d}y \)，故\( \int_{0}^{n} \sqrt{x} \left[ \frac{m}{x} \right] \mathrm{d}x = \int_{+\infty}^{\frac{m}{n}} \sqrt{\frac{m}{y}} [y] \left(-\frac{m}{y^2}\right) \mathrm{d}y = m^{\frac{3}{2}} \int_{\frac{m}{n}}^{+\infty} \frac{[y]}{y^{\frac{5}{2}}} \mathrm{d}y \)。由于\( [y] \leq y \)，故\( \frac{[y]}{y^{\frac{5}{2}}} \leq \frac{1}{y^{\frac{3}{2}}} \)。又\( \int_{\frac{m}{n}}^{+\infty} \frac{1}{y^{\frac{3}{2}}} \mathrm{d}y \)收敛，由比较判别法，有\( \int_{\frac{m}{n}}^{+\infty} \frac{[y]}{y^{\frac{5}{2}}} \mathrm{d}y \)收敛，故\( \int_{0}^{n} \sqrt{x} \left[ \frac{m}{x} \right] \mathrm{d}x \)的敛散性与\( m \)，\( n \)均无关.

由题图可知，\( f(0) > 0 \)，故①不正确。题图中\( f'(x_1) > 0 \)，而②中\( f'(x_1) = -\frac{1}{2} < 0 \)，故②不正确。因在\( x = x_2 \)附近，函数\( f(x) \)的图形为凸的，故\( f''(x_2) < 0 \)，而③中\( f''(x_2) = \frac{2}{3} > 0 \)，故③不正确.

由于\( \frac{\partial^2 (xv)}{\partial y^2} = \frac{\partial \left[ \frac{\partial (xv)}{\partial y} \right]}{\partial y} = \frac{\partial \left( x \frac{\partial v}{\partial y} \right)}{\partial y} = x \frac{\partial^2 v}{\partial y^2} \)，\( \frac{\partial^2 (xv)}{\partial x^2} = \frac{\partial \left[ \frac{\partial (xv)}{\partial x} \right]}{\partial x} = \frac{\partial \left( v + x \frac{\partial v}{\partial x} \right)}{\partial x} = \frac{\partial v}{\partial x} + \frac{\partial v}{\partial x} + x \frac{\partial^2 v}{\partial x^2} = 2 \frac{\partial v}{\partial x} + x \frac{\partial^2 v}{\partial x^2} = x \frac{\partial^2 v}{\partial y^2} \)，故\( \frac{\partial^2 (xv)}{\partial x^2} - \frac{\partial^2 (xv)}{\partial y^2} = 0 \)，即\( u = xv \)，\( v = \frac{u(x, y)}{x} \).

设\( n = 2k + 1 (k = 0, 1, 2, \cdots) \)，二次型\( f \)的矩阵为\( \boldsymbol{A} = \begin{pmatrix} a & & & & b \\ & \ddots & & & \vdots \\ & & a & b & \\ & & b & a & \\ & \vdots & & & \ddots \\ b & & & & a \end{pmatrix}_{2k + 1} \)。由\( |\lambda \boldsymbol{E} - \boldsymbol{A}| = \begin{pmatrix} \lambda - a & & & & -b \\ & \ddots & & & \vdots \\ & & \lambda - a & -b & \\ & & -b & \lambda - a - b & \\ & \vdots & & & \ddots \\ -b & & & & \lambda - a \end{pmatrix}_{2k + 1} = (\lambda - a - b)^{k + 1} (\lambda - a + b)^k \)，得\( \boldsymbol{A} \)的特征值为\( a + b (k + 1 \text{ 重}) \)，\( a - b (k \text{ 重}) \)，于是当\( a + b > 0 \)，\( a - b > 0 \)，即\( a > |b| \)时，\( \boldsymbol{A} \)正定，从而\( f \)正定.

充分性：令\( \boldsymbol{e}_i = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \)（第\( i \)行），且由\( \boldsymbol{\alpha} \)，\( \boldsymbol{\beta} \)的任意性，取\( \boldsymbol{\alpha} = \boldsymbol{\beta} = \boldsymbol{e}_i \)，则\( \boldsymbol{\alpha}^T \boldsymbol{A}^T \boldsymbol{A} \boldsymbol{\beta} = \boldsymbol{e}_i^T \boldsymbol{A}^T \boldsymbol{A} \boldsymbol{e}_i = \boldsymbol{e}_i^T \boldsymbol{e}_i = 1 \)，又\( \boldsymbol{e}_i^T \boldsymbol{A}^T \boldsymbol{A} \boldsymbol{e}_i = [\boldsymbol{A} \boldsymbol{e}_i, \boldsymbol{A} \boldsymbol{e}_i] = \|\boldsymbol{A} \boldsymbol{e}_i\|^2 \)，故\( \|\boldsymbol{A} \boldsymbol{e}_i\| = 1 \)。于是\( \boldsymbol{A} \)的第\( i \)列向量是单位向量，\( i = 1, 2, \cdots, n \)。再取\( \boldsymbol{\alpha} = \boldsymbol{e}_i \)，\( \boldsymbol{\beta} = \boldsymbol{e}_j \)，\( i \neq j \)，则\( \boldsymbol{\alpha}^T \boldsymbol{A}^T \boldsymbol{A} \boldsymbol{\beta} = [\boldsymbol{A} \boldsymbol{\alpha}, \boldsymbol{A} \boldsymbol{\beta}] = [\boldsymbol{A} \boldsymbol{e}_i, \boldsymbol{A} \boldsymbol{e}_j] = [\boldsymbol{e}_i, \boldsymbol{e}_j] = 0 \)，于是\( \boldsymbol{A} \)的第\( i \)列向量与第\( j \)列向量正交，故\( \boldsymbol{A} \)为正交矩阵。必要性：若\( \boldsymbol{A} \)为正交矩阵，令\( \boldsymbol{A} = (\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \cdots, \boldsymbol{\alpha}_n) \)，\( \boldsymbol{\alpha} = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \)，\( \boldsymbol{\beta} = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} \)，则\( \boldsymbol{\alpha}^T \boldsymbol{A}^T \boldsymbol{A} \boldsymbol{\beta} = [\boldsymbol{A} \boldsymbol{\alpha}, \boldsymbol{A} \boldsymbol{\beta}] = [a_1 \boldsymbol{\alpha}_1 + \cdots + a_n \boldsymbol{\alpha}_n, b_1 \boldsymbol{\alpha}_1 + \cdots + b_n \boldsymbol{\alpha}_n] = a_1 b_1 \|\boldsymbol{\alpha}_1\|^2 + a_2 b_2 \|\boldsymbol{\alpha}_2\|^2 + \cdots + a_n b_n \|\boldsymbol{\alpha}_n\|^2 = a_1 b_1 + a_2 b_2 + \cdots + a_n b_n = \boldsymbol{\alpha}^T \boldsymbol{\beta} \)成立.

由题图可知，\( r(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3) = 2 \)，\( r(\boldsymbol{\beta}_1, \boldsymbol{\beta}_2, \boldsymbol{\beta}_3) = 3 \)，故\( r(\boldsymbol{\beta}_1, \boldsymbol{\beta}_3) = 2 \)，又\( \pi_1 \)与\( \pi_2 \)平行，故\( r(\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_3) = 2 \)。若\( r(\boldsymbol{\beta}_1, \boldsymbol{\beta}_2) = 1 \)，即\( \pi_1 \)与\( \pi_2 \)重合，不符合题意，故C错误.

记\( N \)为出错字符的总个数，则\( N \sim B(n, p) \)，\( p = 0.001 \)。该信息传输出错的概率为\( 1 - P\{N = 0\} = 1 - (1 - p)^n \)。由题设有\( 1 - (1 - 0.001)^n \leq 0.01 \)，即\( n \leq \frac{\ln(1 - 0.01)}{\ln(1 - 0.001)} \)，又\( \ln(1 - 0.01) = -0.01 - \frac{1}{2}(-0.01)^2 + \cdots \)，\( \ln(1 - 0.001) = -0.001 - \frac{1}{2}(-0.001)^2 + \cdots \)，\( -11\ln(1 - 0.001) > -\ln(1 - 0.01) > -10\ln(1 - 0.001) \)，于是\( 10 < \frac{\ln(1 - 0.01)}{\ln(1 - 0.001)} < 11 \)，所以\( n \)最大为\( 10 \).

令$W = \min\{X,Y\}$，则$W \sim E(2\lambda)$。 由$Z + W = X + Y$知，$E(Z) + E(W) = E(X) + E(Y)$，于是$E(Z) = E(X) + E(Y) - E(W) = \frac{3}{2\lambda}$，而选项A，B，C中的期望均不为$\frac{3}{2\lambda}$，排除，对于选项D，有$E(\frac{2X + Y}{2}) = \frac{3}{2\lambda}$。 【注】本题亦可计算$F_Z(z) = (1 - e^{-\lambda z})^2$，$F_{\frac{2X + Y}{2}}(z) = (1 - e^{-\lambda z})^2$，计算过程稍复杂。

对于①：由已知得， $X_1 - X_2 \sim N(0,2\sigma^2)$，$X_3 + X_4 - 2 \sim N(0,2\sigma^2)$。 故 $\frac{X_1 - X_2}{\sqrt{2}\sigma} \sim N(0,1)$，$\frac{X_3 + X_4 - 2}{\sqrt{2}\sigma} \sim N(0,1)$， 从而$(\frac{X_3 + X_4 - 2}{\sqrt{2}\sigma})^2 \sim \chi^2(1)$，且$X_1 - X_2$与$X_3 + X_4 - 2$相互独立，所以 $\frac{\frac{X_1 - X_2}{\sqrt{2}\sigma}}{\sqrt{\frac{(\frac{X_3 + X_4 - 2}{\sqrt{2}\sigma})^2}{1}}} = \frac{X_1 - X_2}{|X_3 + X_4 - 2|} \sim t(1)$。 对于②： $\overline{X} \sim N(1,\frac{\sigma^2}{4})$，$\frac{\overline{X} - 1}{\frac{\sigma}{2}} \sim N(0,1)$， 所以 $(\frac{\overline{X} - 1}{\frac{\sigma}{2}})^2 = \frac{4(\overline{X} - 1)^2}{\sigma^2} \sim \chi^2(1)$。 对于③： $\frac{(4 - 1)S^2}{\sigma^2} = \frac{3S^2}{\sigma^2} \sim \chi^2(3)$， 所以 $\frac{\frac{4(\overline{X} - 1)^2}{\sigma^2}}{\frac{3S^2}{\sigma^2} \cdot \frac{1}{3}} = \frac{4(\overline{X} - 1)^2}{S^2} \sim F(1,3)$。

（1）$\mathrm{e}^{x}=1 + x + \frac{\mathrm{e}^{\xi}}{2}x^{2}$，$\xi$介于$0$，$x$之间。 （2）被积函数没有初等函数形式的原函数，考虑其麦克劳林展开式： $\mathrm{e}^{-\frac{x^{2}}{n}} = 1 - \frac{x^{2}}{n} + \frac{\mathrm{e}^{\xi_{n}}}{2} \cdot \frac{x^{4}}{n^{2}}$， 其中$\xi_{n} \in \left(-\frac{x^{2}}{n}, 0\right)$，则 $\int_{0}^{1}\mathrm{e}^{-\frac{x^{2}}{n}}\mathrm{d}x = 1 - \frac{1}{n}\int_{0}^{1}x^{2}\mathrm{d}x + \frac{1}{2n^{2}}\int_{0}^{1}\mathrm{e}^{\xi_{n}}x^{4}\mathrm{d}x$ $= 1 - \frac{1}{3n} + \frac{\int_{0}^{1}\mathrm{e}^{\xi_{n}}x^{4}\mathrm{d}x}{2n^{2}}$。 又$\mathrm{e}^{\xi_{n}} \in (0, 1)$，故$\int_{0}^{1}\mathrm{e}^{\xi_{n}}x^{4}\mathrm{d}x$有界，故 $\lim\limits_{n\to\infty}\frac{\frac{\int_{0}^{1}\mathrm{e}^{\xi_{n}}x^{4}\mathrm{d}x}{2n^{2}}}{\frac{1}{n}} = 0$，$\frac{\int_{0}^{1}\mathrm{e}^{\xi_{n}}x^{4}\mathrm{d}x}{2n^{2}} = o\left(\frac{1}{n}\right)$。 于是原式$= \lim\limits_{n\to\infty}\left[1 - \frac{1}{3n} + o\left(\frac{1}{n}\right)\right]^{\frac{1}{n}} = \mathrm{e}^{\lim\limits_{n\to\infty}\left[-\frac{1}{3n} + o\left(\frac{1}{n}\right)\right]} = 1$。

（1）令$F(x) = \left[\int_{0}^{x}f'(t)\mathrm{d}t\right]^{2} - x\int_{0}^{x}[f'(t)]^{2}\mathrm{d}t$，则 $F'(x) = 2\int_{0}^{x}f'(t)\mathrm{d}t \cdot f'(x) - \int_{0}^{x}[f'(t)]^{2}\mathrm{d}t - x[f'(x)]^{2}$ $= \int_{0}^{x}\left(2f'(t)f'(x) - [f'(t)]^{2} - [f'(x)]^{2}\right)\mathrm{d}t$ $= -\int_{0}^{x}[f'(x) - f'(t)]^{2}\mathrm{d}t \leq 0$。 故$F(x)$单调递减。又$F(0) = 0$，于是当$x \in [0,1]$时，$F(x) \leq 0$，证毕。 （2）由$f(0) = f(1) = 0$，得$f(x) = \int_{0}^{x}f'(t)\mathrm{d}t$，$f(x) = -\int_{x}^{1}f'(t)\mathrm{d}t$，故由（1）知 $[f(x)]^{2} = \left[\int_{0}^{x}f'(t)\mathrm{d}t\right]^{2} \leq x \cdot \int_{0}^{x}[f'(t)]^{2}\mathrm{d}t$， ① 同理， $[f(x)]^{2} = \left[\int_{x}^{1}f'(t)\mathrm{d}t\right]^{2} \leq (1 - x) \cdot \int_{x}^{1}[f'(t)]^{2}\mathrm{d}t$， ② ①$\times(1 - x)$与②$\times x$相加，有 $[f(x)]^{2} \leq x(1 - x) \cdot \int_{0}^{1}[f'(t)]^{2}\mathrm{d}t \leq \frac{1}{4}\int_{0}^{1}[f'(t)]^{2}\mathrm{d}t$。 于是$|f(x)| \leq \frac{1}{2}\sqrt{\int_{0}^{1}[f'(t)]^{2}\mathrm{d}t}$，$x \in [0,1]$，故$M \leq \frac{1}{2}\sqrt{\int_{0}^{1}[f'(t)]^{2}\mathrm{d}t}$，原命题得证。

（1）由题知， \(A = \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} - 2\begin{pmatrix}\frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2}\end{pmatrix}(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})\) \(= \begin{pmatrix}\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\end{pmatrix}\)， 则矩阵\(A\)的特征多项式为 \(\vert\lambda E - A\vert = \begin{vmatrix}\lambda - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \lambda - \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \lambda - \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \lambda - \frac{1}{2}\end{vmatrix} = \frac{1}{16}\begin{vmatrix}2\lambda - 1 & 1 & 1 & 1 \\ 1 & 2\lambda - 1 & 1 & 1 \\ 1 & 1 & 2\lambda - 1 & 1 \\ 1 & 1 & 1 & 2\lambda - 1\end{vmatrix}\) \(= \frac{1}{16}\begin{vmatrix}0 & 2 - 2\lambda & 2 - 2\lambda & 1 - (2\lambda - 1)^{2} \\ 0 & 2\lambda - 2 & 0 & 2 - 2\lambda \\ 0 & 0 & 2\lambda - 2 & 2 - 2\lambda \\ 1 & 1 & 1 & 2\lambda - 1\end{vmatrix}\) \(= -\frac{1}{16}\begin{vmatrix}2 - 2\lambda & 2 - 2\lambda & 1 - (2\lambda - 1)^{2} \\ 2\lambda - 2 & 0 & 2 - 2\lambda \\ 0 & 2\lambda - 2 & 2 - 2\lambda\end{vmatrix}\) \(= -\frac{1}{16}\begin{vmatrix}2 - 2\lambda & 2 - 2\lambda & 1 - (2\lambda - 1)^{2} \\ 0 & 2 - 2\lambda & 1 - (2\lambda - 1)^{2} + 2 - 2\lambda \\ 0 & 2\lambda - 2 & 2 - 2\lambda\end{vmatrix}\) \(= -\frac{1}{16}(2 - 2\lambda)\begin{vmatrix}2 - 2\lambda & 3 - 2\lambda - (2\lambda - 1)^{2} \\ 2\lambda - 2 & 2 - 2\lambda\end{vmatrix}\) \(= -\frac{1}{16}(2 - 2\lambda)\{[(2 - 2\lambda)^{2} + (2 - 2\lambda)[3 - 2\lambda - (2\lambda - 1)^{2}]\}\) \(= -\frac{1}{16}(2 - 2\lambda)^{2}[5 - 4\lambda - (4\lambda^{2} - 4\lambda + 1)]\) \(= -\frac{1}{16}(2 - 2\lambda)^{2}(-4\lambda^{2} + 4) = (\lambda - 1)^{3}(\lambda + 1)\)， 故\(A\)的特征值为\(\lambda_{1} = \lambda_{2} = \lambda_{3} = 1\)，\(\lambda_{4} = -1\)。 由于\(E - A = \begin{pmatrix}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{pmatrix} \to \begin{pmatrix}1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}\)， 故线性方程组\((E - A)x = 0\)的基础解系为\(\begin{pmatrix}-1 \\ 1 \\ 0 \\ 0\end{pmatrix}\)，\(\begin{pmatrix}-1 \\ 0 \\ 1 \\ 0\end{pmatrix}\)，\(\begin{pmatrix}-1 \\ 0 \\ 0 \\ 1\end{pmatrix}\)，将其进行正交化、单位化，可得\(A\)属于特征值\(1\)的单位正交的特征向量为\(\gamma_{1} = \frac{1}{\sqrt{2}}\begin{pmatrix}-1 \\ 1 \\ 0 \\ 0\end{pmatrix}\)，\(\gamma_{2} = \frac{1}{\sqrt{6}}\begin{pmatrix}1 \\ 1 \\ -2 \\ 0\end{pmatrix}\)，\(\gamma_{3} = \frac{1}{2\sqrt{3}}\begin{pmatrix}1 \\ 1 \\ 1 \\ -3\end{pmatrix}\)。 类似可以求得，矩阵\(A\)属于特征值\(-1\)的一个单位特征向量为\(\gamma_{4} = \frac{1}{2}\begin{pmatrix}1 \\ 1 \\ 1 \\ 1\end{pmatrix}\)。 于是取\(Q = (\gamma_{1}, \gamma_{2}, \gamma_{3}, \gamma_{4})\)，\(\Lambda = \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1\end{pmatrix}\)，则\(Q\)是正交矩阵，\(\Lambda\)是对角矩阵，且\(Q^{T}AQ = \Lambda\)。 （2）由（1）可知，二次型\(x^{T}Ax\)的正惯性指数为\(3\)，负惯性指数为\(1\)，其规范形为\(y_{1}^{2} + y_{2}^{2} + y_{3}^{2} - y_{4}^{2}\)，若\(y_{1} \to +\infty\)，\(y_{2} = y_{3} = y_{4} = 0\)，则\(x^{T}Ax \to +\infty\)；若\(y_{4} \to +\infty\)，\(y_{1} = y_{2} = y_{3} = 0\)，则\(x^{T}Ax \to -\infty\)，所以\(f(x)\)既没有最大值，也没有最小值。 【注】\(f(x) = x^{T}Ax\)不一定有最值，要看条件。

（1）由题意知，\(X_{1},X_{2}\stackrel{ind}{\sim}N(1,\sigma_{1}^{2})\)，令\(U = X_{1}-X_{2}\)，\(U\sim N(\mu,\sigma^{2})\)，则 \(\mu = E(X_{1}-X_{2}) = 0\)，\(\sigma^{2}=D(X_{1}-X_{2}) = 2\sigma_{1}^{2}\)， 即\(U\sim N(0,2\sigma_{1}^{2})\)。 于是 \[ \begin{align*} E(\vert U\vert)&=\int_{-\infty}^{+\infty}\vert u\vert\frac{1}{2\sqrt{\pi}\sigma_{1}}e^{-\frac{u^{2}}{4\sigma_{1}^{2}}}du\\ &=2\int_{0}^{+\infty}u\cdot\frac{1}{2\sqrt{\pi}\sigma_{1}}e^{-\frac{u^{2}}{4\sigma_{1}^{2}}}du\\ &=\frac{2\sigma_{1}}{\sqrt{\pi}\sigma_{1}}\int_{0}^{+\infty}\frac{u}{2\sigma_{1}}e^{-(\frac{u}{2\sigma_{1}})^{2}}d(\frac{u}{2\sigma_{1}})=\frac{2}{\sqrt{\pi}}\sigma_{1}. \end{align*} \] 故 \[ \begin{align*} D(\vert U\vert)&=E(U^{2}) - [E(\vert U\vert)]^{2}=D(U)+[E(U)]^{2}-[E(\vert U\vert)]^{2}\\ &=2\sigma_{1}^{2}-\frac{4}{\pi}\sigma_{1}^{2}=2\sigma_{1}^{2}(1 - \frac{2}{\pi}). \end{align*} \] （2）由题设可知，似然函数为 \[ L(\sigma_{1}^{2},\sigma_{2}^{2})=\frac{1}{(\sqrt{2\pi})^{5}\sigma_{1}^{3}\sigma_{2}^{2}}e^{-\frac{(x_{1}-1)^{2}+(x_{2}-1)^{2}+(x_{5}-1)^{2}}{2\sigma_{1}^{2}}-\frac{(x_{3}-1)^{2}+(x_{4}-1)^{2}}{2\sigma_{2}^{2}}}, \] 取对数，得 \[ \begin{align*} \ln L&=-5\ln\sqrt{2\pi}-\frac{3}{2}\ln\sigma_{1}^{2}-\ln\sigma_{2}^{2}-\frac{(x_{1}-1)^{2}+(x_{2}-1)^{2}+(x_{5}-1)^{2}}{2\sigma_{1}^{2}}-\\ &\frac{(x_{3}-1)^{2}+(x_{4}-1)^{2}}{2\sigma_{2}^{2}}, \end{align*} \] 又由 \[ \frac{\partial[\ln L(\sigma_{1}^{2},\sigma_{2}^{2})]}{\partial\sigma_{1}^{2}}=-\frac{3}{2}\frac{1}{\sigma_{1}^{2}}+\frac{1}{2}\cdot\frac{1}{(\sigma_{1}^{2})^{2}}[(x_{1}-1)^{2}+(x_{2}-1)^{2}+(x_{5}-1)^{2}]\stackrel{令}{=}0, \] 解得\(\sigma_{1}^{2}\)的最大似然估计量为 \[ \hat{\sigma}_{1}^{2}=\frac{(X_{1}-1)^{2}+(X_{2}-1)^{2}+(X_{5}-1)^{2}}{3}, \] 故 \[ \begin{align*} E(\hat{\sigma}_{1}^{2})&=E[(X_{1}-1)^{2}]=D(X_{1}-1)+[E(X_{1}-1)]^{2}\\ &=\sigma_{1}^{2}+(1 - 1)^{2}=\sigma_{1}^{2}, \end{align*} \] \[ \begin{align*} D(\hat{\sigma}_{1}^{2})&=E[(\hat{\sigma}_{1}^{2})^{2}]-[E(\hat{\sigma}_{1}^{2})]^{2}\\ &=\frac{3E[(X_{1}-1)^{4}]+6E[(X_{1}-1)^{2}(X_{2}-1)^{2}]}{9}-\sigma_{1}^{4},\\ &=\frac{E[(X_{1}-1)^{4}]}{3}+\frac{2}{3}\sigma_{1}^{4}-\sigma_{1}^{4}. \end{align*} \] 令 \(X = X_{1}-1\sim N(0,\sigma_{1}^{2})\)， 则有 \[ E(X^{4})=\int_{-\infty}^{+\infty}x^{4}\cdot\frac{1}{\sqrt{2\pi}\sigma_{1}}e^{-\frac{x^{2}}{2\sigma_{1}^{2}}}dx = 3\sigma_{1}^{4}, \] 故\(D(\hat{\sigma}_{1}^{2})=\sigma_{1}^{4}+\frac{2}{3}\sigma_{1}^{4}-\sigma_{1}^{4}=\frac{2}{3}\sigma_{1}^{4}\)。