UA MATH564 概率論 QE練習題1

• 第一題
• 第二題
• 第三題

2014年1月理論的1-3題。

第一題

Part (a)
Marginal density of $Y$ is
$$f_Y(y)={\int }_{?\infty }^{\infty }f(x,y)dx={\int }_{?\infty }^{\infty }\frac{e^{?y{x}^2/2}}{\sqrt{2\pi /y}}y{e}^{?y}dx=\frac{y{e}^{?y}}{\sqrt{2\pi /y}}{\int }_{?\infty }^{\infty }{e}^{?y{x}^2/2}dx$$

Recall the density of normal distribution $N(0,1/y)$,
$${\int }_{?\infty }^{\infty }\frac{1}{\sqrt{2\pi /y}}{e}^{?y{x}^2/2}dx=1?{\int }_{?\infty }^{\infty }{e}^{?y{x}^2/2}dx=\sqrt{2\pi /y}$$

So
$$f_Y(y)=\frac{y{e}^{?y}}{\sqrt{2\pi /y}}{\int }_{?\infty }^{\infty }{e}^{?y{x}^2/2}dx=\frac{y{e}^{?y}}{\sqrt{2\pi /y}}\sqrt{2\pi /y}=y{e}^{?y},y>0$$

Conditional density of $X$ given $Y$ is
$$f(x|y)=\frac{f(x,y)}{f_Y(y)}=\frac{\frac{e^{?y{x}^2/2}}{\sqrt{2\pi /y}}y{e}^{?y}}{y{e}^{?y}}=\frac{e^{?y{x}^2/2}}{\sqrt{2\pi /y}},x\in \mathbb{R}$$

Part (b)
$X|Y～N(0,1/y)$, so $E[X|Y]=0$

Part (c )
$X|Y～N(0,1/y)$, so $Var[X|Y]=\frac{1}{Y}$

Part (d)
$$Var(X)=E[Var(X|Y)]+Var[E(X|Y)]=E[1/Y]+0={\int }_0^{\infty }\frac{1}{y}y{e}^{?y}dy=1$$

第二題

Part (a)
Let $U=X+Y,V=X/Y$, and then $Y=\frac{U}{V+1}$. $X=\frac{UV}{V+1}$,
$$\frac{?(X,Y)}{?(U,V)}=\left|\begin{array}{cc}\frac{V}{V+1}& \frac{1}{V+1}\\ \frac{U}{(V+1{)}^2}& ?\frac{U}{(V+1{)}^2}\end{array}\right|=?\frac{U}{(V+1{)}^2}$$

Joint density of $X,Y$ is
$$f(x,y)=f_X(x)f_Y(y)=\frac{1}{{\lambda }^2}{e}^{?\frac{1}{\lambda }(x+y)},x>0,y>0$$

So joint density of $U,V$ is
$$f(u,v)=f(x(u,v),y(u,v))\left|\frac{?(X,Y)}{?(U,V)}\right|=\frac{u}{{\lambda }^2(v+1{)}^2}{e}^{?\frac{u}{\lambda }},u>0,v>0$$

By additivity of Gamma distribution, $U～\Gamma (2,\lambda )$,
$$f_U(u)=\frac{u{e}^{?u/\lambda }}{\Gamma (2){\lambda }^2}=\frac{u}{{\lambda }^2}{e}^{?\frac{u}{\lambda }}$$

Notice $$f(v|u)=\frac{f(u,v)}{f_U(u)}=\frac{1}{(v+1{)}^2},v>0$$

which is independent of $u$, so $X+Y$ and $X/Y$ are indepdendent. Check
$${\int }_0^{\infty }\frac{1}{(v+1{)}^2}dv=?\frac{1}{v+1}{|}_0^{\infty }=1$$

So marginal density of $V$ is
$$f(v)=\frac{1}{(v+1{)}^2},v>0$$

Part (b)
Notice $Z=\frac{X}{X+Y}=1?\frac{Y}{X+Y}=1?\frac{X}{X+Y}\frac{Y}{X}=1?\frac{Z}{V}?Z=\frac{V}{1+V},Z\in (0,1)$,
$$\begin{gathered} P(Z\le z)=P(\frac{V}{1+V}\le z)=P(V\le \frac{z}{1?z}) \\ ={\int }_0^{\frac{z}{1?z}}\frac{1}{(v+1{)}^2}dv=?\frac{1}{v+1}{|}_0^{\frac{z}{1?z}}=z \end{gathered}$$

第三題

$$\begin{gathered} Y_{n+1}=\lambda Y_n+X_{n+1}=\lambda (\lambda Y_{n?1}+X_n)+X_{n+1} \\ =\lambda (\lambda (\lambda Y_{n?2}+X_{n?1})+X_n)+X_{n+1}=?={\lambda }^{n+1}{Y}_0+\sum _{i=1}^{n+1}{\lambda }^{n+1?i}{X}_i \end{gathered}$$

Since $X_i{～}_{iid}N(\mu ,{\sigma }^2)$,
$$\begin{gathered} E{Y}_{n+1}={\lambda }^{n+1}x+\mu \sum _{i=1}^{n+1}{\lambda }^{n+1?i}={\lambda }^{n+1}x+\frac{\mu (1?{\lambda }^{n+1})}{1?\lambda }\to \frac{\mu }{1?\lambda },asn\to \infty \\ Var(Y_{n+1})=\sum _{i=1}^{n+1}{\lambda }^{2(n+1?i)}{\sigma }^2=\frac{{\sigma }^2(1?{\lambda }^{2(n+1)})}{1?{\lambda }^2}\to \frac{{\sigma }^2}{1?{\lambda }^2},asn\to \infty \end{gathered}$$

Above, $Y_n{\to }_{d}N(\frac{\mu }{1?\lambda },\frac{{\sigma }^2}{1?{\lambda }^2})$

總結

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