类高斯联合密度

注意

$$f_{X,Y}(x,y)=c\,e^{-x^2/8}\,e^{-(x-y)^2/2}.$$

(1) 对固定 $x$，有

$$\int_{-\infty}^{\infty} e^{-(x-y)^2/2}\,dy=\sqrt{2\pi}.$$

所以边缘密度

$$f_X(x)=\int f_{X,Y}(x,y)\,dy=c\sqrt{2\pi}\,e^{-x^2/8}.$$

再对 $x$ 积分归一化：

$$\int_{-\infty}^{\infty} e^{-x^2/8}\,dx=\sqrt{8\pi}=2\sqrt{2\pi}.$$

因此

$$1=\int f_X(x)dx=c\sqrt{2\pi}\cdot 2\sqrt{2\pi}=c\cdot 4\pi \Rightarrow \boxed{c=\frac{1}{4\pi}}.$$

(2) 代回得

$$\boxed{f_X(x)=\frac{1}{2\sqrt{2\pi}}e^{-x^2/8}},$$

即 $X\sim N(0,4)$。

(3) 由 $e^{-(x-y)^2/2}$ 可视作 $Y\mid X=x\sim N(x,1)$，于是 $Y=X+Z$（$Z\sim N(0,1)$ 独立），故 $Y\sim N(0,5)$，边缘为

$$\boxed{f_Y(y)=\frac{1}{\sqrt{10\pi}}e^{-y^2/10}}.$$

(4) 若独立则应有 $f_{X,Y}(x,y)=f_X(x)f_Y(y)$，但此处指数项含有 $x-y$ 的耦合；等价地由 $Y=X+Z$ 可得 $\mathrm{Cov}(X,Y)=\mathrm{Var}(X)=4\ne 0$。

因此 $\boxed{X,Y\text{ 不独立}}$。

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英文解析

CAUTION

$$f_{X,Y}(x,y)=c\,e^{-x^2/8}\,e^{-(x-y)^2/2}.$$

• • (1) * * Fixed pair $x$, yes

$$\int_{-\infty}^{\infty} e^{-(x-y)^2/2}\,dy=\sqrt{2\pi}.$$

So the edge density

$$f_X(x)=\int f_{X,Y}(x,y)\,dy=c\sqrt{2\pi}\,e^{-x^2/8}.$$

Then normalize the $x$ points:

$$\int_{-\infty}^{\infty} e^{-x^2/8}\,dx=\sqrt{8\pi}=2\sqrt{2\pi}.$$

Therefore,

$$1=\int f_X(x)dx=c\sqrt{2\pi}\cdot 2\sqrt{2\pi}=c\cdot 4\pi \Rightarrow \boxed{c=\frac{1}{4\pi}}.$$

• • (2) * * Returned on behalf of

$$\boxed{f_X(x)=\frac{1}{2\sqrt{2\pi}}e^{-x^2/8}},$$

That's $X\sim N(0,4)$.

• • (3) * * From $e^{-(x-y)^2/2}$ can be seen as $Y\mid X=x\sim N(x,1)$, so $Y=X+Z$($Z\sim N(0,1)$independent), so$Y\sim N(0,5)$, the edge is

$$\boxed{f_Y(y)=\frac{1}{\sqrt{10\pi}}e^{-y^2/10}}.$$

• • (4) * * If independent, there should be $f_{X,Y}(x,y)=f_X(x)f_Y(y)$, but the index term here contains a coupling of $x-y$; equivalent to $\mathrm{Cov}(X,Y)=\mathrm{Var}(X)=4\ne 0$ from $Y=X+Z$.

So $\boxed{X,Y\text{ not independent}}$.