[T0301] 设随机变量 \(\xi\) 取值于 \([0,1]\), 若 \(P\{x\le\xi<y\}\) 只与长度 \(y-x\) 有关 (对一切 \(0\le x\le y\le 1\)). 试证 \(\xi\sim U[0,1]\).
证 不妨设 \(P\{x\le\xi<y\}=f(y-x)\). 令 \(x=0\), 则有 \(P\{0\le\xi<y\}=f(y)\). 注意到对 \(\forall 0\le y_1<y_2\le1\), 有
\[f(y_1+y_2)=P\{0\le\xi<y_1+y_2\}=P\{0\le\xi<y_1\}+P\{y_1\le\xi<y_1+y_2\}=f(y_1)+f(y_2) \]于是 \(e^{f(y_1+y_2)}=e^{f(y_1)}\cdot e^{f(y_2)}\), 由 Cauchy 命题知 \(e^{f(x)}=a^x\Rightarrow f(x)=x\ln a\), 其中 \(a\) 待定. 又 \(f(1)=1\), 故 \(a=e\), 从而 \(f(x)=x\), 故
\[P\{0\le\xi<x\}=x, \quad x\in[0,1] \]即 \(\xi\sim U[0,1]\). #
[T0302] 设 \((\xi,\eta)\) 服从二元正态分布, 参数为 \(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,\rho\), 以 \(D(\lambda)\) 记下面椭圆的内部:
\[\frac{(x-\mu_1)^2}{\sigma_1^2}-\frac{2\rho(x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2}+\frac{(y-\mu_2)^2}{\sigma_2^2}=\lambda^2 \]试求 \(P\{(\xi,\eta)\in D(\lambda)\}\).
解 即求二重积分
\[\iint\limits_{(x,y)\in D(\lambda)}\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left\{-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_1)^2}{\sigma_1^2}-\frac{2\rho(x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2}+\frac{(y-\mu_2)^2}{\sigma_2^2}\right]\right\}\mathrm{d}x\mathrm dy \]坐标变换 \(T:\begin{cases}x=\mu_1+\sigma_1r\cos\theta\\ y=\mu_2+\sigma_2r\sin\theta\end{cases}\), 则 \(J=\sigma_1\sigma_2r\). 经过简单的计算, 上面的二重积分可化为
\[\frac{\sqrt{1-\rho^2}}{2\pi}\left(1-e^{-\frac{\lambda^2}{2(1-\rho^2)}}\right)\int_0^{2\pi}\frac{1}{1-2\rho\sin\theta\cos\theta}\mathrm{~d}\theta \]注意到 \(\lim\limits_{\lambda\to\infty}P\{(\xi,\eta)\in D(\lambda)\}=1\), 因此
\[\frac{\sqrt{1-\rho^2}}{2\pi}\int_0^{2\pi}\frac{1}{1-2\rho\sin\theta\cos\theta}\mathrm{~d}\theta=1\Rightarrow \int_0^{2\pi}\frac{1}{1-2\rho\sin\theta\cos\theta}\mathrm{~d}\theta=\frac{2\pi}{\sqrt{1-\rho^2}} \]故
\[P\{(\xi,\eta)\in D(\lambda)\}=1-e^{-\frac{\lambda^2}{2(1-\rho^2)}}.\quad\quad\# \][T0303] 若 \(\xi_1,\cdots,\xi_n\) 相互独立同分布于 \(N(0,1)\), 证明 \(\eta=\xi_1^2+\cdots+\xi_n^2\) 服从 \(\chi^2\) 分布.
证
\[P\{\eta<x\}=P\{\xi_1^2+\cdots+\xi_n^2<x\}=\frac{1}{(2\pi)^{n/2}}{\int\cdots\int}_{x_1^2+\cdots+x_n^2<x}\exp\left\{-\frac{x_1^2+\cdots+x_n^2}{2}\right\}\mathrm dx_1\cdots\mathrm dx_n \]作变换
\[\begin{cases} x_1=\rho\sin\theta_1\sin\theta_2\cdots\sin\theta_{n-1}\\ x_2=\rho\cos\theta_1\sin\theta_2\cdots\sin\theta_{n-1}\\ \cdots\cdots\\ x_{n-1}=\rho\cos\theta_{n-2}\sin\theta_{n-1}\\ x_n=\rho\cos\theta_{n-1} \end{cases} \]则 \(J=\frac{\partial(x_1,\cdots,x_n)}{\partial(\rho,\theta_1,\cdots,\theta_{n-1})}=\rho^{n-1}D(\theta_1,\cdots,\theta_{n-1})\), 其中 \(D(\theta_1,\cdots,\theta_{n-1})\) 是 \(\theta_1,\cdots,\theta_{n-1}\) 的函数, 与 \(\rho\) 无关. 于是
\[\begin{aligned} F(x)&=\frac{1}{(2\pi)^{n/2}}\int_{0}^{\sqrt x}e^{-\frac{\rho^2}{2}}\rho^{n-1}\mathrm d\rho \ \int_0^{\pi}\cdots\int_0^{\pi}\int_0^{2\pi}D(\theta_1,\cdots,\theta_{n-1})\mathrm d\theta_1\cdots\mathrm d\theta_{n-1}\\ &=C_n\int_{0}^{\sqrt x}e^{-\frac{\rho^2}{2}}\rho^{n-1}\mathrm d\rho \end{aligned} \]其中
\[C_n=\frac{1}{(2\pi)^{n/2}} \int_0^{\pi}\cdots\int_0^{\pi}\int_0^{2\pi}D(\theta_1,\cdots,\theta_{n-1})\mathrm d\theta_1\cdots\mathrm d\theta_{n-1} \]令 \(\rho=\sqrt{t}\), 得
\[F(x)=\frac{C_n}{2}\int_0^xe^{-\frac12t}\cdot t^{\frac n2-1}\mathrm dt \]再令 \(x\to+\infty\), 得
\[1=\frac{C_n}{2}\int_0^{+\infty}e^{-\frac12t}\cdot t^{\frac n2-1}\mathrm dt\xlongequal{t=2u}2^{\frac n2-1}C_n\int_0^{+\infty}e^{-u}\cdot u^{\frac n2-1}\mathrm du=2^{\frac n2-1}C_n\Gamma\left(\frac n2\right) \]故
\[C_n=\frac{1}{2^{\frac n2-1}\Gamma\left(\frac n2\right)} \]于是 \(\eta\) 的分布函数为
\[F(x)=\frac{1}{2^{\frac n2-1}\Gamma\left(\frac n2\right)}\int_0^xe^{-\frac12t}\cdot t^{\frac n2-1}\mathrm dt, \quad x>0. \]即 \(\eta\) 服从 \(\chi^2\) 分布. #
[T0304] 设随机变量 \(\xi,\eta\) 独立同分布于 \(N(0,1)\). 试求 \(U=\xi^2+\eta^2\) 与 \(V=\frac{\xi}{\eta}\) 的分布密度函数, 并证明他们是独立的.
解 易知
\[p_{\xi\eta}(x,y)=\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}} \]设 \(u=x^2+y^2, \ v=\frac{x}{y}\), 则 \(J=-\frac{1}{2}\frac{1}{1+v^2}\). 注意到 \((x,y)\) 与 \((-x,-y)\) 对应于同一个 \((u,v)\), 故
\[\begin{aligned} p_{UV}(u,v)&=p(x(u,v),y(u,v))|J|\binom{-\infty<x<\infty}{y\ge0}+p(x(u,v),y(u,v))|J|\binom{-\infty<x<\infty}{y<0}\\ &=\frac{1}{2}e^{-\frac u2}\cdot\frac{1}{\pi(1+v^2)}:=p_U(u)\cdot p_V(v) \end{aligned} \]故 \(U\) 与 \(V\) 相互独立, 且
\[p_U(u)=\frac{1}{2}e^{-\frac u2},\quad u>0\\ p_V(v)=\frac{1}{\pi(1+v^2)}, -\infty<v<\infty \]即 \(U\) 服从参数为 \(\frac12\) 的指数分布, \(V\) 服从 Cauchy 分布. #
[T0305] 若 \((\xi,\eta)\) 服从参数为 \(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,\rho\) 的二元正态分布, 试找出 \(\xi+\eta\) 与 \(\xi-\eta\) 相互独立的充要条件.
证 显然 \(\xi+\eta\) 与 \(\xi-\eta\) 相互独立 \(\Longleftrightarrow\) \(U=(\xi-\mu_1)+(\eta-\mu_2)\) 与 \(V=(\xi-\mu_1)-(\eta-\mu_2)\) 相互独立. 设
\[\binom uv=\pmatrix{1&1\\1&-1}\binom{x-\mu_1}{y-\mu_2}\Rightarrow\binom{x-\mu_1}{y-\mu_2}=\frac12\pmatrix{1&1\\1&-1}\binom uv \]则 \(J=\frac{\partial(x,y)}{\partial(u,v)}=\frac12\). 于是 \((U,V)\) 的密度函数为
\[\begin{aligned} q(u,v)&=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\cdot\frac12\cdot\exp\left\{-\frac12(u,v)\left[\pmatrix{1&1\\1&-1}\pmatrix{\sigma_1^2&\rho\sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2}\pmatrix{1&1\\1&-1}\right]^{-1}\binom uv\right\}\\ &=\frac{1}{4\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left\{-\frac12(u,v)\pmatrix{\sigma_1^2+2\rho\sigma_1\sigma_2+\sigma_2^2&\sigma_1^2-\sigma_2^2\\\sigma_1^2-\sigma_2^2&\sigma_1^2-2\rho\sigma_1\sigma_2+\sigma_2^2}^{-1}\binom uv\right\} \end{aligned} \]故 \(\xi+\eta\) 与 \(\xi-\eta\) 相互独立 \(\Longleftrightarrow\) \(\sigma_1^2-\sigma_2^2=0 \Longleftrightarrow \sigma_1=\sigma_2\). #
[T0306] 设随机变量 \(\xi\) 与 \(\eta\) 独立同分布, 其密度函数不为 \(0\) 且有二阶导数, 试证: 若 \(\xi+\eta\) 与 \(\xi-\eta\) 相互独立, 则随机变量 \(\xi,\eta,\xi+\eta,\xi-\eta\) 均服从正态分布.
证 设 \(U=\xi+\eta, \ V=\xi-\eta\), 则
\[p_U(u)p_V(v)=p_{UV}(u,v)=\frac12p_{\xi,\eta}(\frac{u+v}{2},\frac{u-v}{2})=\frac12p(\frac{u+v}{2})p(\frac{u-v}{2}) \]其中 \(p\) 为 \(\xi\) 和 \(\eta\) 的密度. 两边取对数得
\[\ln p(\frac{u+v}{2})+\ln p(\frac{u-v}{2})-\ln2=\ln p_U(u)+\ln p_V(v) \]记 \(f(x)=\ln p(x)\), 则 \(f(\frac{u+v}{2})+f(\frac{u-v}{2})-\ln2=\ln p_U(u)+\ln p_V(v)\). 两边对 \(u\) 求偏导, 得
\[\frac12f'(\frac{u+v}{2})+\frac12f'(\frac{u-v}{2})=\frac{p_U'(u)}{p_U(u)} \]两边再对 \(v\) 求偏导, 得
\[\frac14f''(\frac{u+v}{2})-\frac14f''(\frac{u-v}{2})=0 \]令 \(u=v=x\) 则 \(f''(x)=f''(0):=-\lambda\), 得 \(f(x)=-\frac{\lambda}{2}x^2+c_1x+c_2\), 于是
\[\ln p(x)=-\frac{\lambda}{2}x^2+c_1x+c_2\Rightarrow p(x)=e^{c_2+\frac{c_1^2}{2\lambda}}e^{-\frac{\lambda}{2}(x-\frac{c_1}{\lambda})^2}:=Ke^{-\frac{\lambda}{2}(x-\frac{c_1}{\lambda})^2} \]因 \(p(x)\) 是密度函数, 故 \(\lambda>0\). 且 \(c_1=f'(0)=\frac{p'(0)}{p(0)}, \ c_2=f(0)=\ln p(0)\). 故 \(\xi,\eta\) 服从正态分布, 从而 \(\xi+\eta,\xi-\eta\) 也服从正态分布. #
[T0307] 设随机变量 \(\xi\) 和 \(\eta\) 相互独立, 且 \(\xi\sim B(n,p), \ \eta\sim U(0,1)\), 试求 \(\xi+\eta\) 的分布函数和密度函数.
解 对 \(x\le 0\), 有 \(P\{\xi+\eta<x\}=0\); 对 \(x>n+1\), 有 \(P\{\xi+\eta<x\}=1\); 对 \(k<x\le k+1, \ k=0,1,2,\cdots,n\), 有
\[\begin{aligned} P\{\xi+\eta<x\}&=P\{\xi=k,\xi+\eta<x\}+P\{\xi\le k-1,\xi+\eta<x\}+P\{\xi\ge k+1,\xi+\eta<x\}\\ &=P\{\xi=k,\eta<x-k\}+P\{\xi\le k-1\}\\ &=\binom{n}{k}p^k(1-p)^{n-k}(x-k)+\sum_{\mathscr{l}=0}^{k-1}\binom{n}{\mathscr{l}}p^{\mathscr{l}}(1-p)^{n-\mathscr{l}} \end{aligned} \]于是
\[F(x)=\begin{cases} 0,& x\le0\\ \binom{n}{k}p^k(1-p)^{n-k}(x-k)+\sum_{\mathscr{l}=0}^{k-1}\binom{n}{\mathscr{l}}p^{\mathscr{l}}(1-p)^{n-\mathscr{l}}, &k<x\le k+1, \ k=0,1,2,\cdots,n\\ 1, & x>n+1 \end{cases} \]从而密度函数为
\[p(x)=\begin{cases} \binom{n}{k}p^k(1-p)^{n-k},&k<x\le k+1, \ k=0,1,2,\cdots,n\\ 0,& x\le0, \ x>n+1. \end{cases}\quad\quad\# \][T0308] 试求顺序统计量 \(\xi_k^*\) 与 \(\xi_l^* \ (k<l)\) 的联合密度函数.
解 对 \(x<y\), 有
\[\begin{aligned} &P\{x\le\xi_k^*<x+\Delta x, y\le\xi_l^*<y+\Delta y\}\\ =&\frac{n!}{(k-1)!1!(l-k-1)!1!(n-l)!}F(x)^{k-1}\cdot p(x)\Delta x\cdot[F(y)-F(x+\Delta x)]^{l-k-1}\cdot p(y)\Delta y\cdot[1-F(y+\Delta y)]^{n-l} \end{aligned} \]于是 \(\xi_k^*\) 与 \(\xi_l^* \ (k<l)\) 的联合密度函数为
\[p(x,y)= \begin{cases} \frac{n!}{(k-1)!(l-k-1)!(n-l)!}F(x)^{k-1}\cdot [F(y)-F(x)]^{l-k-1}\cdot[1-F(y)]^{n-l}\cdot p(x)\cdot p(y),&x<y\\ 0,&x\ge y \end{cases} \]#
标签：xi,frac,函数,sigma,eta,rho,theta,习题,随机变量 From： https://www.cnblogs.com/hznudmh/p/18070212