UA MATH564 概率论 QE练习题1

UA MATH564 概率论 QE练习题1

• 第一题
• 第二题
• 第三题

2014年1月理论的1-3题。

第一题

Part (a)
Marginal density of $Y$Y is
$f_Y(y) = \int_{-\infty}^{\infty} f(x,y)dx = \int_{-\infty}^{\infty} \frac{e^{-yx^2/2}}{\sqrt{2\pi/y}}ye^{-y}dx = \frac{ye^{-y}}{\sqrt{2\pi/y}} \int_{-\infty}^{\infty} e^{-yx^2/2}dx$fY​(y)=∫−∞∞​f(x,y)dx=∫−∞∞​2π/y​e−yx2/2​ye−ydx=2π/y​ye−y​∫−∞∞​e−yx2/2dx

Recall the density of normal distribution $N(0,1/y)$N(0,1/y),
$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi /y}}e^{-yx^2/2}dx = 1 \Rightarrow \int_{-\infty}^{\infty} e^{-yx^2/2}dx=\sqrt{2\pi /y}$∫−∞∞​2π/y​1​e−yx2/2dx=1⇒∫−∞∞​e−yx2/2dx=2π/y​

So
$f_Y(y) =\frac{ye^{-y}}{\sqrt{2\pi/y}} \int_{-\infty}^{\infty} e^{-yx^2/2}dx = \frac{ye^{-y}}{\sqrt{2\pi/y}} \sqrt{2\pi /y} = ye^{-y},y>0$fY​(y)=2π/y​ye−y​∫−∞∞​e−yx2/2dx=2π/y​ye−y​2π/y​=ye−y,y>0

Conditional density of $X$X given $Y$Y is
$f(x|y) = \frac{f(x,y)}{f_Y(y)} = \frac{\frac{e^{-yx^2/2}}{\sqrt{2\pi/y}}ye^{-y}}{ye^{-y}} = \frac{e^{-yx^2/2}}{\sqrt{2\pi/y}},x\in \mathbb{R}$f(x∣y)=fY​(y)f(x,y)​=ye−y2π/y​e−yx2/2​ye−y​=2π/y​e−yx2/2​,x∈R

Part (b)
$X|Y \sim N(0,1/y)$X∣Y∼N(0,1/y), so $E[X|Y]=0$E[X∣Y]=0

Part (c )
$X|Y \sim N(0,1/y)$X∣Y∼N(0,1/y), so $Var[X|Y]=\frac{1}{Y}$Var[X∣Y]=Y1​

Part (d)
$Var(X) = E[Var(X|Y)] + Var[E(X|Y)] = E[1/Y] + 0 = \int_{0}^{\infty} \frac{1}{y}ye^{-y}dy = 1$Var(X)=E[Var(X∣Y)]+Var[E(X∣Y)]=E[1/Y]+0=∫0∞​y1​ye−ydy=1

第二题

Part (a)
Let $U = X+Y,V = X/Y$U=X+Y,V=X/Y, and then $Y = \frac{U}{V+1}$Y=V+1U​. $X = \frac{UV}{V+1}$X=V+1UV​,
$\frac{\partial (X,Y)}{\partial (U,V)} = \left| \begin{matrix} \frac{V}{V+1} & \frac{1}{V+1} \\ \frac{U}{(V+1)^2} & -\frac{U}{(V+1)^2}\end{matrix} \right| = -\frac{U}{(V+1)^2}$∂(U,V)∂(X,Y)​=∣∣∣∣∣​V+1V​(V+1)2U​​V+11​−(V+1)2U​​∣∣∣∣∣​=−(V+1)2U​

Joint density of $X,Y$X,Y is
$f(x,y) = f_X(x)f_Y(y) = \frac{1}{\lambda^2}e^{-\frac{1}{\lambda}(x+y)},x>0,y>0$f(x,y)=fX​(x)fY​(y)=λ21​e−λ1​(x+y),x>0,y>0

So joint density of $U,V$U,V is
$f(u,v) = f(x(u,v),y(u,v)) \left| \frac{\partial (X,Y)}{\partial (U,V)} \right| = \frac{u}{\lambda^2(v+1)^2}e^{-\frac{u}{\lambda}},u>0,v>0$f(u,v)=f(x(u,v),y(u,v))∣∣∣∣​∂(U,V)∂(X,Y)​∣∣∣∣​=λ2(v+1)2u​e−λu​,u>0,v>0

By additivity of Gamma distribution, $U \sim \Gamma(2,\lambda)$U∼Γ(2,λ),
$f_U(u) = \frac{ ue^{-u/\lambda}}{\Gamma(2)\lambda^2} = \frac{u}{\lambda^2}e^{-\frac{u}{\lambda}}$fU​(u)=Γ(2)λ2ue−u/λ​=λ2u​e−λu​

Notice $f(v|u) = \frac{f(u,v)}{f_U(u)} = \frac{1}{(v+1)^2},v>0$f(v∣u)=fU​(u)f(u,v)​=(v+1)21​,v>0

which is independent of $u$u, so $X+Y$X+Y and $X/Y$X/Y are indepdendent. Check
$\int_{0}^{\infty} \frac{1}{(v+1)^2}dv = -\frac{1}{v+1}|_0^{\infty} = 1$∫0∞​(v+1)21​dv=−v+11​∣0∞​=1

So marginal density of $V$V is
$f(v) = \frac{1}{(v+1)^2},v>0$f(v)=(v+1)21​,v>0

Part (b)
Notice $Z = \frac{X}{X+Y} = 1 - \frac{Y}{X+Y} = 1 - \frac{X}{X+Y} \frac{Y}{X}= 1 - \frac{Z}{V} \Rightarrow Z = \frac{V}{1+V}, Z \in (0,1)$Z=X+YX​=1−X+YY​=1−X+YX​XY​=1−VZ​⇒Z=1+VV​,Z∈(0,1),
$P(Z \le z) = P(\frac{V}{1+V} \le z) = P(V \le \frac{z}{1-z}) \\=\int_{0}^{\frac{z}{1-z}} \frac{1}{(v+1)^2}dv = -\frac{1}{v+1}|_0^{\frac{z}{1-z}} = z$P(Z≤z)=P(1+VV​≤z)=P(V≤1−zz​)=∫01−zz​​(v+1)21​dv=−v+11​∣01−zz​​=z

第三题

$Y_{n+1} = \lambda Y _n + X_{n+1} = \lambda(\lambda Y_{n-1} + X_n) + X_{n+1} \\= \lambda(\lambda (\lambda Y_{n-2} + X_{n-1})+ X_n) + X_{n+1} = \cdots =\lambda^{n+1} Y_0 + \sum_{i=1}^{n+1} \lambda^{n+1-i}X_{i}$Yn+1​=λYn​+Xn+1​=λ(λYn−1​+Xn​)+Xn+1​=λ(λ(λYn−2​+Xn−1​)+Xn​)+Xn+1​=⋯=λn+1Y0​+i=1∑n+1​λn+1−iXi​

Since $X_i \sim_{iid} N(\mu,\sigma^2)$Xi​∼iid​N(μ,σ2),
$EY_{n+1} =\lambda^{n+1} x + \mu \sum_{i=1}^{n+1} \lambda^{n+1-i} = \lambda^{n+1} x + \frac{\mu(1-\lambda^{n+1})}{1-\lambda} \to \frac{\mu}{1-\lambda},\ as\ n \to \infty \\ Var(Y_{n+1}) = \sum_{i=1}^{n+1} \lambda^{2(n+1-i)} \sigma^2 = \frac{\sigma^2(1-\lambda^{2(n+1)})}{1-\lambda^2} \to \frac{\sigma^2}{1-\lambda^2},\ as\ n \to \infty$EYn+1​=λn+1x+μi=1∑n+1​λn+1−i=λn+1x+1−λμ(1−λn+1)​→1−λμ​, as n→∞Var(Yn+1​)=i=1∑n+1​λ2(n+1−i)σ2=1−λ2σ2(1−λ2(n+1))​→1−λ2σ2​, as n→∞

Above, $Y_n \to_d N( \frac{\mu}{1-\lambda},\frac{\sigma^2}{1-\lambda^2})$Yn​→d​N(1−λμ​,1−λ2σ2​)