[1]
\(Y\)の周辺分布は、
\(\displaystyle f_Y(y)=\int^\infty_{-\infty} f(x,y)dx\)
\(\displaystyle =\int^\infty_{-\infty}f_X(x)f_{Y|X}(y|x)dx\)
\(\displaystyle =\int^\infty_{-\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{\sqrt{2\pi}\sqrt{1-\rho^2}}e^{-\frac{(y-\rho x)^2}{2(1-\rho^2)}}dx\)
\(\displaystyle =\frac{1}{2\pi\sqrt{1-\rho^2}}\int^\infty_{-\infty}e^{-\frac{x^2}{2}-\frac{(y-\rho x)^2}{2(1-\rho^2)}}dx\)
\(\displaystyle =\frac{1}{2\pi\sqrt{1-\rho^2}}\int^\infty_{-\infty}e^{-\frac{(x-\rho y)^2+(1-\rho^2)y^2}{2(1-\rho^2)}}dx\)
\(\displaystyle =\frac{1}{2\pi\sqrt{1-\rho^2}}\int^\infty_{-\infty}e^{-\frac{(x-\rho y)^2}{2(1-\rho^2)}-\frac{y^2}{2}}dx\)
\(\displaystyle = \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\int^\infty_{-\infty}\frac{1}{\sqrt{2\pi}\sqrt{1-\rho^2}}e^{-\frac{(x-\rho y)^2}{2(1-\rho^2)}}dx\)
\(\displaystyle = \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\)
よって\(Y\)の周辺分布は\(N(0,1)\)
別解
を利用すると、
\(E_Y[Y]=E_X\left[E_{Y|X}[Y|X]\right]\)
\(=E_X[\rho X]=0\)
\(V_Y[Y]\)\(=E_X\left[V_{Y|X}[Y|X]\right]\)\(+V_X\left[E_{Y|X}[Y|X]\right]\)
\(=E_X[1-\rho^2]+V_X[\rho X]\)
\(=1-\rho^2+\rho^2=1\)
より、\(Y\sim N(0,1)\)
[2]
公式の解答がよく分からなかったので、確率密度関数を使ってゴリ押す解き方します。
まず、\(f_{X_{t+1}|X_t}(x_{t+1}|x_t)\)に\(Y_t\)の条件を入れ込みたいので、
\(f_{X_{t+1}|X_t}(x_{t+1}|x_t)\)
\(\displaystyle = \int f_{X_{t+1},Y_t|X_t}(x_{t+1},y_t|x_t)dy_t\)
\(\displaystyle = \int \frac{f_{X_{t+1},Y_t,X_t}(x_{t+1},y_t,x_t)}{f_{X_t}(x_t)}dy_t\)
\(\displaystyle = \int \frac{f_{X_{t+1}|Y_t,X_t}(x_{t+1}|y_t,x_t)f_{Y_t,X_t}(y_t,x_t)}{f_{X_t}(x_t)}dy_t\)
\(\displaystyle = \int f_{X_{t+1}|Y_t,X_t}(x_{t+1}|y_t,x_t)\frac{f_{Y_t,X_t}(y_t,x_t)}{f_{X_t}(x_t)}dy_t\)
\(\displaystyle = \int f_{X_{t+1}|Y_t,X_t}(x_{t+1}|y_t,x_t)f_{Y_t|X_t}(y_t|x_t)dy_t\)
\(X_{t+1}\)はマルコフ性から、\(Y_t\)のみに依存するので、
\(\displaystyle = \int f_{X_{t+1}|Y_t}(x_{t+1}|y_t)f_{Y_t|X_t}(y_t|x_t)dy_t\)
\(\displaystyle = \int^\infty_{-\infty} \frac{1}{\sqrt{2\pi} \sqrt{1-\rho^2}}e^{-\frac{(x_{t+1}-\rho y_t)^2}{2(1-\rho^2)}} \)\(\displaystyle \frac{1}{\sqrt{2\pi} \sqrt{1-\rho^2}} \)\(\displaystyle e^{-\frac{(y_{t}-\rho x_t)^2}{2(1-\rho^2)}}dy_t\)
\(\displaystyle = \frac{1}{2\pi (1-\rho^2)} \)\(\displaystyle \int^\infty_{-\infty} \exp\left[-\frac{1+\rho^2}{2(1-\rho^2)} \right.\)\(\displaystyle \left(y_t-\frac{\rho}{1+\rho^2}(x_{t+1}+x_t)^2\right) \)\(\displaystyle \left. -\frac{(x_{t+1}-\rho^2 x_t)^2}{2(1-\rho^4)}\right]dy_t\)
\(\displaystyle =\frac{1}{\sqrt{2\pi}\sqrt{1-\rho^4}} \)\(\displaystyle \exp\left[-\frac{(x_{t+1}-\rho^2 x_t)^2}{2(1-\rho^4)}\right] \)\(\displaystyle \int^\infty_{-\infty} \)\(\displaystyle \frac{1}{\sqrt{2\pi}\sqrt{\frac{1-\rho^2}{1+\rho^2}}} \)\(\displaystyle \exp\left[-\frac{\left(y_t-\frac{\rho}{1+\rho^2}(x_{t+1}+x_t)\right)^2}{2\frac{1-\rho^2}{1+\rho^2}}\right]dy_t\)
\(\displaystyle =\frac{1}{\sqrt{2\pi}\sqrt{1-\rho^4}}\exp\left[-\frac{(x_{t+1}-\rho^2 x_t)^2}{2(1-\rho^4)}\right]\)
よって、\(X_{t+1}|X_t \sim N(\rho^2x_t,1-\rho^4)\)
別解
記号を簡単のため、\(X=X_t,Y=Y_t,Z=X_{t+1}\)とする。
\(E_{Z|X}[Z|X]\)の導出1
\(\displaystyle E_{Z|X}\left[Z|X\right]=\int_z z f_{Z|X}(z|x)dz\)
\(\displaystyle =\int_z f_{Z|X}(z|x)dz\)
\(\displaystyle =\int_z z \frac{f_{X,Z}(x,z)}{f_X(x)}dz\)
\(\displaystyle =\int_z z \frac{\int_y f_{X,Y,Z}(x,y,z)dy}{f_X(x)}dz\)
\(\displaystyle =\int_z z \int_y \frac{f_{X,Y,Z}(x,y,z)}{f_X(x)}dydz\)
\(\displaystyle =\int_z z \int_y \frac{f_{X,Y}(x,y)}{f_X(x)}\frac{f_{X,Y,Z}(x,y,z)}{f_{X,Y}(x,y)}dydz\)
\(\displaystyle =\int_z z \int_y f_{Y|X}(y|x)f_{Z|X,Y}(z|x,y)dydz\)
\(\displaystyle =\int_y f_{Y|X}(y|x)\int_z z f_{Z|X,Y}(z|x,y)dzdy\)
\(\displaystyle =\int_y f_{Y|X}(y|x)\int_z z f_{Z|Y}(z|y)dzdy \quad \)(マルコフ性のため)
\(\displaystyle =\int_y f_{Y|X}(y|x)E_{Z|Y}[Z|Y]dy\)
\(\displaystyle =E_{Y|X}\left[E_{Z|Y}[Z|Y]\right]\)
\(\displaystyle =E_{Y|X}\left[\rho Y \right]\)
\(\displaystyle =\rho^2 x_t\)
これよりもっと簡単な証明として、次の導出2を紹介する。
\(E_{Z|X}[Z|X]\)の導出2
を利用して、
\(\displaystyle E_{Z|X}\left[Z|X\right]\)\(= E_{Y|X}\left[E_{(Z|X)|(Y|X)}[(Z|X)|(Y|X)]\right]\)
\(= E_{Y|X}\left[E_{Z|Y}[Z|Y]\right] \quad\)(マルコフ性のため)
\(V_{Z|X}[Z|X]\)の導出
を利用して、
\(V_{Z|X}[Z|X]=V_{Y|X}\left[E_{(Z|X)|(Y|X)}[(Z|X)|(Y|X)]\right]\)\(+E_{Y|X}\left[V_{(Z|X)|(Y|X)}[(Z|X)|(Y|X)]\right]\)
\(=V_{Y|X}\left[E_{Z|Y}[Z|Y]\right]\)\(+E_{Y|X}\left[V_{Z|Y}[Z|Y]\right]\quad\)(マルコフ性のため)
\(=V_{Y|X}\left[\rho Y \right]\)\(+E_{Y|X}\left[1-\rho^2\right]\)
\(=\rho^2 V_{Y|X}\left[ Y \right]\)\(+1-\rho^2\)
\(=\rho^2 (1-\rho^2)\)\(+1-\rho^2\)\(=1-\rho^4\)
[3]
ある\(t\)で\(X_t|X_0 \sim N(\rho^{2t} x_0, 1-\rho^{4t})\)を仮定すると、[2]と同様に、
\(E_{X_{t+1}|X_0}[X_{t+1}|X_0]=E_{X_t|X_0}\left[E_{X_{t+1}|X_t}[{X_{t+1}|X_t}]\right]\)
\(=E_{X_t|X_0}\left[\rho^2 X_t \right]\)
\(=\rho^{2(t+1)} x_0\quad\)(仮定より)
また、
\(V_{X_{t+1}|X_0}[X_{t+1}|X_0]\)\(=V_{X_t|X_0}\left[E_{X_{t+1}|X_t}[X_{t+1}|X_t]\right]\)\(+E_{X_t|X_0}\left[V_{X_{t+1}|X_t}[X_{t+1}|X_t]\right]\)
\(=V_{X_t|X_0}\left[\rho^2 X_t \right]\)\(+E_{X_t|X_0}\left[1-\rho^4]\right]\)
\(=\rho^4 V_{X_t|X_0}\left[ X_t \right]\)\(+1-\rho^4\)
\(=\rho^4 (1-\rho^{4t})\)\(+1-\rho^4 \quad\)(仮定より)
\(=1-\rho^{4(t+1)}\)
よって、\(X_{t+1}|X_0 \sim N\left(\rho^{2(t+1)} x_0,1-\rho^{4(t+1)}\right)\)なので帰納的に示された。
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