latex 编译问题，求助！！！新手，第一次编，不知道怎么回事就是运行不了，报错也看不懂，大家帮帮忙吧
发布网友
共3个回答
热心网友
主要有四方面的问题：
1. because 和therefore 不加"\",因为它们的不是命令；
2. 中括号不加"\"，直接键盘输入[ ]即可；
3.数组命令要有*{llllll}，里面"l" 的个数是你有几列就写几个，当然为了简单，我就随便按了几个，但一定要大于你的列数；
4.代码里不要用“######”,那是注释用的，正文中不用。
这样一改，就可以运行了，但还有一些问题，因为不知道你的原文，所以我不好改。你再看看吧。
下面是代码
1\ because \[f(x_{1})=\int x_{1}^18x_{1}x_{2} dx_{2}=4x_{1}-4(x_{1})^{3}\]
\[f(x_{2})=\int_0^\infty f(x_{1},x_{2}) dx_{1}=\int 0^x_{2} 8x_{1}x_{2} dx_{1}=4(x_{2})^{3}\]
\[f(x_{1})f(x_{2})\neq f(x_{1},x_{2})\]
therefore $x_{1},x_{2}$不相互\\
2\ \[because Ex_{1}=1\ast0.4+2\ast0.1+3\ast0.5=2.1;Ex_{2}=0.2
therefore E(x)=\binom{Ex_{1}}{Ex_{2}}=\binom{2.1}{0.2}\]
3\ \[because x\sim N_{3}(\mu,2I_{3})\];y=Ax+d
\[therefore E(y)=AE(x)+d=A\mu+d=\binom{2}{1};
Var(y)=AVar(x)A'=\left(\begin{array}{lllllllll}3& -1\\-1&1 \\ \end{array} \right)
[therefore y\sim N_{2}(E(y),Var(y))\]
4\ \[E(x)=\binom{Ex_{1}}{Ex_{2}}=\binom{\mu _{1}}{\mu _{2}}\];
[because $x_{1},x_{2}$ ,therefore $Cov(x_{1},x_{2})=0$]
[therefore Var(x)=$\left(\begin{array}{lllllllllll}\Sigma_{11}&0\\0&Sigma_{22}\\ \end{array} \right)]$
[therefore $x\sim N_{p+q}(E(x),Var(x))$
5\ (1) \[ a=(1,1),b=(1.-1),\] therefore \[x_{1}+x_{2}=ax,x_{1}-x_{2}=bx\]
\[therefore Cov(x_{1}+x_{2},x_{1}-x_{2})=Cov(ax,bx)=aCov(x,x)b'=0\]
(2)\[E(x_{1}+x_{2})=E(x_{1})+E(x_{2})=\mu_{1}+\mu_{2};Var(x_{1}+x_{2})=aVar(x)a'=2(\sigma)^{2}(1+\rho)\];
\[E(x_{1}-x_{2})=E(x_{1})-E(x_{2})=\mu_{1}-\mu_{2};Var(x_{1}-x_{2})=bVar(x)b'=2(\sigma)^{2}(1-\rho)\]
6\ (1)\[a=(I_{p},I_{p}),b=(I_{p},-I_{p})
therefore Cov(x^{1}+x^{2},x^{1}-x^{2})=Cov(ax,bx)=aCov(x)b'=0\]
(2)\[E(x^{1}+x^{2})=(\mu^{1})+(\mu^{2});Var(x^{1}+x^{2})=aVar(x)a'=2(\Sigma_{1}+\Sigma_{2});\]
\[E(x^{1}-x^{2})=(\mu^{1})-(\mu^{2});Var(x^{1}-x^{2})=bVar(x)b'=2(\Sigma_{1}-\Sigma_{2});\]
热心网友
那个因为所以的符号需要调用宏包才行，而不是上面那位所说的不是命令
热心网友
看不懂
