第一题

Part (a)
Marginal density of YYY is
fY(y)=∫−∞∞f(x,y)dx=∫−∞∞e−yx2/22π/yye−ydx=ye−y2π/y∫−∞∞e−yx2/2dxf_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}dxfY(y)=∫−∞∞f(x,y)dx=∫−∞∞2π/ye−yx2/2ye−ydx=2π/yye−y∫−∞∞e−yx2/2dx

Recall the density of normal distribution N(0,1/y)N(0,1/y)N(0,1/y),
∫−∞∞12π/ye−yx2/2dx=1⇒∫−∞∞e−yx2/2dx=2π/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/y1e−yx2/2dx=1⇒∫−∞∞e−yx2/2dx=2π/y

So
fY(y)=ye−y2π/y∫−∞∞e−yx2/2dx=ye−y2π/y2π/y=ye−y,y>0f_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/yye−y∫−∞∞e−yx2/2dx=2π/yye−y2π/y=ye−y,y>0

Conditional density of XXX given YYY is
f(x∣y)=f(x,y)fY(y)=e−yx2/22π/yye−yye−y=e−yx2/22π/y,x∈Rf(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/ye−yx2/2ye−y=2π/ye−yx2/2,x∈R

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

Part (c )
X∣Y∼N(0,1/y)X|Y \sim N(0,1/y)X∣Y∼N(0,1/y), so Var[X∣Y]=1YVar[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=∫0∞1yye−ydy=1Var(X) = E[Var(X|Y)] + Var[E(X|Y)] = E[1/Y] + 0 = \int_{0}^{\infty} \frac{1}{y}ye^{-y}dy = 1Var(X)=E[Var(X∣Y)]+Var[E(X∣Y)]=E[1/Y]+0=∫0∞y1ye−ydy=1

第二题

Part (a)
Let U=X+Y,V=X/YU = X+Y,V = X/YU=X+Y,V=X/Y, and then Y=UV+1Y = \frac{U}{V+1}Y=V+1U. X=UVV+1X = \frac{UV}{V+1}X=V+1UV,
∂(X,Y)∂(U,V)=∣VV+11V+1U(V+1)2−U(V+1)2∣=−U(V+1)2\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)2UV+11−(V+1)2U∣∣∣∣∣=−(V+1)2U

Joint density of X,YX,YX,Y is
f(x,y)=fX(x)fY(y)=1λ2e−1λ(x+y),x>0,y>0f(x,y) = f_X(x)f_Y(y) = \frac{1}{\lambda^2}e^{-\frac{1}{\lambda}(x+y)},x>0,y>0f(x,y)=fX(x)fY(y)=λ21e−λ1(x+y),x>0,y>0

So joint density of U,VU,VU,V is
f(u,v)=f(x(u,v),y(u,v))∣∂(X,Y)∂(U,V)∣=uλ2(v+1)2e−uλ,u>0,v>0f(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>0f(u,v)=f(x(u,v),y(u,v))∣∣∣∣∂(U,V)∂(X,Y)∣∣∣∣=λ2(v+1)2ue−λu,u>0,v>0

By additivity of Gamma distribution, U∼Γ(2,λ)U \sim \Gamma(2,\lambda)U∼Γ(2,λ),
fU(u)=ue−u/λΓ(2)λ2=uλ2e−uλf_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=λ2ue−λu

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

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

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

Part (b)
Notice Z=XX+Y=1−YX+Y=1−XX+YYX=1−ZV⇒Z=V1+V,Z∈(0,1)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+YXXY=1−VZ⇒Z=1+VV,Z∈(0,1),
P(Z≤z)=P(V1+V≤z)=P(V≤z1−z)=∫0z1−z1(v+1)2dv=−1v+1∣0z1−z=zP(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}} = zP(Z≤z)=P(1+VV≤z)=P(V≤1−zz)=∫01−zz(v+1)21dv=−v+11∣01−zz=z

第三题

Yn+1=λYn+Xn+1=λ(λYn−1+Xn)+Xn+1=λ(λ(λYn−2+Xn−1)+Xn)+Xn+1=⋯=λn+1Y0+∑i=1n+1λn+1−iXiY_{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+1λn+1−iXi

Since Xi∼iidN(μ,σ2)X_i \sim_{iid} N(\mu,\sigma^2)Xi∼iidN(μ,σ2),
EYn+1=λn+1x+μ∑i=1n+1λn+1−i=λn+1x+μ(1−λn+1)1−λ→μ1−λ,asn→∞Var(Yn+1)=∑i=1n+1λ2(n+1−i)σ2=σ2(1−λ2(n+1))1−λ2→σ21−λ2,asn→∞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 \inftyEYn+1=λn+1x+μi=1∑n+1λn+1−i=λn+1x+1−λμ(1−λn+1)→1−λμ, as n→∞Var(Yn+1)=i=1∑n+1λ2(n+1−i)σ2=1−λ2σ2(1−λ2(n+1))→1−λ2σ2, as n→∞

Above, Yn→dN(μ1−λ,σ21−λ2)Y_n \to_d N( \frac{\mu}{1-\lambda},\frac{\sigma^2}{1-\lambda^2})Yn→dN(1−λμ,1−λ2σ2)

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UA MATH564 概率论 QE练习题1

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