Chapter 6

4.24 更新：注释了部分会导致渲染错误的公式，请自行脑补。

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例 3.20

f(x,y)=\frac14 (0\leq x,y\leq2)

$$
f(x,z)=(f(x,x-z)+f(x,x+z))|\frac{\partial(x,y)}{\partial(x,z)}|=
\begin{cases}
\frac12 &(z\leq x,z\leq 2-x)\\
\frac14 &(z\leq x\cap z> 2-x \cup z>x\cap z\leq2-x)\\
0 &(z> x,z> 2-x)
\end{cases}
$$

$$
f_Z(z)=\int_{0}^{2}f(x,z)dx=\begin{cases}
(2-2z)\cdot\frac12+2z\cdot\frac14 &0\leq z\leq1\\
(2z-2)\cdot0+(4-2z)\cdot\frac14 &1< z\leq2
\end{cases}
$$

\therefore f_Z(z)=1-\frac z2 (0\leq z\leq2)

例 3.21

f(x,y)=1 (x<y<2-x,0<x<1)

f_X(x)=\int_0^{min{x,2-x}}f(x,y)dy=min{x,2-x}

f_Y(y)=\int_y^{2-y}f(x,y)dx=2-2y

f_{X|Y}(x|y)=\frac1{2-2y} (y<x<2-y,0\leq y<1)

例 3.29

XY>0 时：

f_{XY}(t)=p\cdot N(\mu,\sigma^2)=p\frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}}

XY=0 时：

P(XY=0)=1-p

例 3.35

(2)

f(x,y)=A\cdot e^{-3x}\cdot e^{-4y}

故 X,Y 独立

(1)

\iint_{x,y>0}f(x,y)dxdy=A\cdot\int_0^{\infin}e^{-3x}dx\cdot\int_0^{\infin}e^{-4y}dy=1\Rightarrow A=12

(3)

f_X(x)=3e^{-3x}

f_Y(y)=4e^{-4y}

f_Z(z)=\int_0^zf_X(x)f_Y(z-x)dx=12e^{-4z}(e^z-1)

\int_0^z3e^{-3x}4e^{-4z+4x}dx=12e^{-4z}\int_0^ze^xdx=12e^{-4z}(e^z-1)

(4)

\begin{align}
P(X>0.5|X+Y=1)&=
\frac{\int_{0.5}^1f_X(x)f_Y(1-x)dx}{\int_0^1f_X(x)f_Y(1-x)dx}\
&=\frac{\int_{0.5}^13e^{-3x}4e^{-4+4x}dx}{\int_0^13e^{-3x}4e^{-4+4x}dx}\
&=\frac{e-e^{\frac12}}{e-1}
\end{align}

例 3.39

f(x,y)=\frac12 (|x|-1<y<1-|x|,-1<x<1)

f_X(x)=\frac12\cdot2(1-|x|)=1-|x|

f_Y(y)=\frac12\cdot2(1-|y|)=1-|y|

(2)

不相互独立，因为 f(x,y)\neq f_X(x)f_Y(y)

例 3.40

(1)

f_X(x)=3x\cdot x=3x^2 (0<x<1)

f_Y(y)=\int_y^13xdx=\frac32(1-y^2) (0<y<1)

(2)

不相互独立，因为 f(x,y)\neq f_X(x)f_Y(y)

(3)

\begin{align}
P(X+Y\leq1)
&=\int_0^1dx\cdot 3x\cdot min{x,1-x}\
&=\int_0^{0.5}3x^2dx
+\int_{0.5}^13x(1-x)dx\
&=\frac38
\end{align}

例 3.41

(1)

F_X(x)=lim_{y->\infin}F(x,y)=1-(x+1)e^{-x}

F_y(y)=lim_{x->\infin}F(x,y)=\frac y{1+y}

(2)

$f(x,y)=\frac{\part F(x,y)}{\part x\part y}=xe^{-x}\cdot\frac1{(1+y)^2}$

f_X(x)=xe^{-x}

f_Y(y)=\frac1{(1+y)^2}

(3)

F(x,y)=F_X(x)F_Y(y)

X,Y 独立

例 3.43

f_X(x)=f_Y(y)=\frac12

f(x,y)\neq f_X(x)f_Y(y), 故 X,Y 不互相独立。

记 U=X2,V=Y^2

$f(u,v)=\frac{1+\sqrt{uv}+1-\sqrt{uv}+1+\sqrt{uv}+1-\sqrt{uv}}4\frac{\part (x,y)}{\part(u,v)}=\frac{1}{4\sqrt{uv}}$ $(0<u,v<1)$

f_U(u)=\frac1{2\sqrt u}

f_V(v)=\frac1{2\sqrt v}

f(u,v)=f_U(u)f_V(v)

故 U,V 独立。

例 3.47

记 m=\frac{x-\mu_1}{\sigma_1},n=\frac{y-\mu_2}{\sigma_2}

f(x,y)=\frac1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}exp{-\frac{m^2-2\rho mn+n^2}{2(1-\rho^2)}}

例 4.2

(1)

\begin{align}
E(X)&=
\Sigma_{n=1}^{\infin}nP(X=n)\
&=\Sigma_{n=1}^{\infin}n(P(X\leq n)-P(X\leq n+1))\
&=\Sigma_{n=1}^{\infin}P(X\leq n)
\end{align}

(2)

\begin{align}
E(X)&=\int_0^{\infin}xdF(x)\
&=\int_0^{\infin}xdF(x)\
&=\int_0^{\infin}d(xF(x))-F(x)dx\
&=-\int_0^{\infin}xd(1-F(x))\
&=\int_0^{\infin}(-d(x(1-F(x)))+(1-F(x))dx)\
&=-(x(1-F(x)))|_0^{\infin}+\int_0^{\infin}(1-F(x))dx\
&=\int_0^{\infin}(1-F(x))dx
\end{align}

(3)

\begin{align}
E(X)&=\int_0^{\infin}xdF(x)\
&=\int_0^{\infin}dF(x)\cdot\int_0^xdy\
&=\int_0^{\infin}dy\cdot\int_y^{\infin}dF(x)\
&=\int_0^{\infin}dy(F(\infin)-F(y))\
&=\int_0^{\infin}(1-F(x))dx
\end{align}

或者：

\begin{align}
E(X)&=\Sigma_xxP(X=x)\
&=\Sigma_x(P(X=x)\int_x^{\infin}dy)\
&=\int_0^{\infin}dxP(X\geq x)\
&=\int_0^{\infin}(1-F(x))dx
\end{align}