例 3.1.7hard一级题目发布者: ai-batch题干设二维随机变量 (X,Y)∼N(μ1,μ2,σ12,σ22,ρ)(X, Y) \sim N(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho)(X,Y)∼N(μ1,μ2,σ12,σ22,ρ)，求 (X,Y)(X, Y)(X,Y) 落在区域 D={(x,y):(x−μ1)2σ12−2ρ(x−μ1)(y−μ2)σ1σ2+(y−μ2)2σ22⩽λ2}D = \left\{(x, y) : \frac{(x - \mu_1)^2}{\sigma_1^2} - 2\rho\frac{(x - \mu_1)(y - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y - \mu_2)^2}{\sigma_2^2} \leqslant \lambda^2 \right\}D={(x,y):σ12(x−μ1)2−2ρσ1σ2(x−μ1)(y−μ2)+σ22(y−μ2)2⩽λ2} 内的概率。答案点击展开后可查看解析解析所求概率为 p=12πσ1σ21−ρ2∬Dexp{−12(1−ρ2)[(x−μ1σ1)2−2ρ(x−μ1)(y−μ2)σ1σ2+(y−μ2σ2)2]}dxdy.p = \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \iint_D \exp\left\{-\frac{1}{2(1-\rho^2)}\left[\left(\frac{x-\mu_1}{\sigma_1}\right)^2 - 2\rho\frac{(x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2} + \left(\frac{y-\mu_2}{\sigma_2}\right)^2\right]\right\} \mathrm{d}x\mathrm{d}y.p=2πσ1σ21−ρ21∬Dexp{−2(1−ρ2)1[(σ1x−μ1)2−2ρσ1σ2(x−μ1)(y−μ2)+(σ2y−μ2)2]}dxdy. 作变换 {u=x−μ1σ1−ρy−μ2σ2,v=y−μ2σ21−ρ2,\begin{cases} u = \dfrac{x - \mu_1}{\sigma_1} - \rho\dfrac{y - \mu_2}{\sigma_2}, \\[6pt] v = \dfrac{y - \mu_2}{\sigma_2}\sqrt{1 - \rho^2}, \end{cases}⎩⎨⎧u=σ1x−μ1−ρσ2y−μ2,v=σ2y−μ21−ρ2, 则可得 J−1=∂(u,v)∂(x,y)=∣1σ1−ρσ201−ρ2σ2∣=1−ρ2σ1σ2,J^{-1} = \frac{\partial(u, v)}{\partial(x, y)} = \begin{vmatrix} \dfrac{1}{\sigma_1} & -\dfrac{\rho}{\sigma_2} \\[6pt] 0 & \dfrac{\sqrt{1-\rho^2}}{\sigma_2} \end{vmatrix} = \frac{\sqrt{1-\rho^2}}{\sigma_1\sigma_2},J−1=∂(x,y)∂(u,v)=σ110−σ2ρσ21−ρ2=σ1σ21−ρ2, 由此得 p=12π(1−ρ2)∬u2+v2⩽λ2exp{−u2+v22(1−ρ2)}dudv.p = \frac{1}{2\pi(1-\rho^2)} \iint_{u^2+v^2 \leqslant \lambda^2} \exp\left\{-\frac{u^2 + v^2}{2(1-\rho^2)}\right\} \mathrm{d}u\mathrm{d}v.p=2π(1−ρ2)1∬u2+v2⩽λ2exp{−2(1−ρ2)u2+v2}dudv. 再作极坐标变换 {u=rsinα,v=rcosα,\begin{cases} u = r\sin\alpha, \\ v = r\cos\alpha, \end{cases}{u=rsinα,v=rcosα, 则可得 J=∂(u,v)∂(r,α)=∣sinαrcosαcosα−rsinα∣=−r(sin2α+cos2α)=−r,J = \frac{\partial(u, v)}{\partial(r, \alpha)} = \begin{vmatrix} \sin\alpha & r\cos\alpha \\ \cos\alpha & -r\sin\alpha \end{vmatrix} = -r(\sin^2\alpha + \cos^2\alpha) = -r,J=∂(r,α)∂(u,v)=sinαcosαrcosα−rsinα=−r(sin2α+cos2α)=−r, 最后得 p=12π(1−ρ2)∫02πdα∫0λrexp{−r22(1−ρ2)}drp = \frac{1}{2\pi(1-\rho^2)} \int_0^{2\pi} \mathrm{d}\alpha \int_0^{\lambda} r\exp\left\{-\frac{r^2}{2(1-\rho^2)}\right\} \mathrm{d}rp=2π(1−ρ2)1∫02πdα∫0λrexp{−2(1−ρ2)r2}dr =∫0λexp{−r22(1−ρ2)}d(r22(1−ρ2))= \int_0^{\lambda} \exp\left\{-\frac{r^2}{2(1-\rho^2)}\right\} \mathrm{d}\left(\frac{r^2}{2(1-\rho^2)}\right)=∫0λexp{−2(1−ρ2)r2}d(2(1−ρ2)r2) =−exp{−r22(1−ρ2)}∣0λ=1−exp{−λ22(1−ρ2)}.= -\exp\left\{-\frac{r^2}{2(1-\rho^2)}\right\}\bigg|_0^{\lambda} = 1 - \exp\left\{-\frac{\lambda^2}{2(1-\rho^2)}\right\}.=−exp{−2(1−ρ2)r2}0λ=1−exp{−2(1−ρ2)λ2}.