Let X and Y be jointly distributed with p.d.f:
f X , Y ( x , y ) = { 1 4 ( 1 + x y ) ∣ x ∣ < 1 , ∣ y ∣ < 1 0 Otherwise f_{X,Y}(x,y)= \begin{cases}
{1 \over 4}(1+xy) & |x|<1, |y|<1 \\
0 &\text{Otherwise }
\end{cases} f X , Y ( x , y ) = { 4 1 ( 1 + x y ) 0 ∣ x ∣ < 1 , ∣ y ∣ < 1 Otherwise Check
∫ − ∞ ∞ ∫ − ∞ ∞ f X , Y ( x , y ) d y d x = \displaystyle\int_{-\infin}^{\infin}\displaystyle\int_{-\infin}^{\infin}f_{X,Y}(x,y)dydx= ∫ − ∞ ∞ ∫ − ∞ ∞ f X , Y ( x , y ) d y d x =
= ∫ − 1 1 ∫ − 1 1 1 4 ( 1 + x y ) d y d x = =\displaystyle\int_{-1}^{1}\displaystyle\int_{-1}^{1} {1 \over 4}(1+xy)dydx= = ∫ − 1 1 ∫ − 1 1 4 1 ( 1 + x y ) d y d x =
= ∫ − 1 1 1 4 [ ( y + x y 2 2 ) ] 1 − 1 d x = =\displaystyle\int_{-1}^{1}{1 \over 4}\bigg[(y+{xy^2 \over 2})\bigg]\begin{matrix}
1 \\
-1
\end{matrix}dx= = ∫ − 1 1 4 1 [ ( y + 2 x y 2 ) ] 1 − 1 d x =
= ∫ − 1 1 1 4 ( 1 + x ( 1 ) 2 2 − ( − 1 + x ( − 1 ) 2 2 ) ) d x = =\displaystyle\int_{-1}^{1}{1 \over 4}(1+{x(1)^2 \over 2}-(-1+{x(-1)^2 \over 2}))dx= = ∫ − 1 1 4 1 ( 1 + 2 x ( 1 ) 2 − ( − 1 + 2 x ( − 1 ) 2 )) d x =
= ∫ − 1 1 1 2 d x = 1 2 [ x ] 1 − 1 =\displaystyle\int_{-1}^{1}{1 \over 2}dx={1 \over 2}\big[x\big]\begin{matrix}
1 \\
-1
\end{matrix} = ∫ − 1 1 2 1 d x = 2 1 [ x ] 1 − 1 Marginal distribution of X
g ( x ) = ∫ − ∞ ∞ f X , Y ( x , y ) d y = g(x)=\displaystyle\int_{-\infin}^{\infin}f_{X,Y}(x,y)dy= g ( x ) = ∫ − ∞ ∞ f X , Y ( x , y ) d y =
= ∫ − 1 1 1 4 ( 1 + x y ) d y = 1 4 [ ( y + x y 2 2 ) ] 1 − 1 = =\displaystyle\int_{-1}^{1}{1 \over4}(1+xy)dy={1 \over 4}\bigg[(y+{xy^2 \over 2})\bigg]\begin{matrix}
1 \\
-1
\end{matrix}= = ∫ − 1 1 4 1 ( 1 + x y ) d y = 4 1 [ ( y + 2 x y 2 ) ] 1 − 1 =
= 1 4 ( 1 + x ( 1 ) 2 2 − ( − 1 + x ( − 1 ) 2 2 ) ) = 1 2 ={1 \over 4}(1+{x(1)^2 \over 2}-(-1+{x(-1)^2 \over 2}))={1 \over 2} = 4 1 ( 1 + 2 x ( 1 ) 2 − ( − 1 + 2 x ( − 1 ) 2 )) = 2 1 Marginal distribution of Y
h ( y ) = ∫ − ∞ ∞ f X , Y ( x , y ) d x = h(y)=\displaystyle\int_{-\infin}^{\infin}f_{X,Y}(x,y)dx= h ( y ) = ∫ − ∞ ∞ f X , Y ( x , y ) d x =
= ∫ − 1 1 1 4 ( 1 + x y ) d x = 1 4 [ ( x + x 2 y 2 ) ] 1 − 1 = =\displaystyle\int_{-1}^{1}{1 \over4}(1+xy)dx={1 \over 4}\bigg[(x+{x^2y \over 2})\bigg]\begin{matrix}
1 \\
-1
\end{matrix}= = ∫ − 1 1 4 1 ( 1 + x y ) d x = 4 1 [ ( x + 2 x 2 y ) ] 1 − 1 =
= 1 4 ( 1 + y ( 1 ) 2 2 − ( − 1 + y ( − 1 ) 2 2 ) ) = 1 2 ={1 \over 4}(1+{y(1)^2 \over 2}-(-1+{y(-1)^2 \over 2}))={1 \over 2} = 4 1 ( 1 + 2 y ( 1 ) 2 − ( − 1 + 2 y ( − 1 ) 2 )) = 2 1
f X ( x , y ) = { 1 2 ∣ x ∣ < 1 , ∣ y ∣ < 1 0 Otherwise f_{X}(x,y)= \begin{cases}
{1 \over 2} & |x|<1, |y|<1 \\
0 &\text{Otherwise }
\end{cases} f X ( x , y ) = { 2 1 0 ∣ x ∣ < 1 , ∣ y ∣ < 1 Otherwise f Y ( x , y ) = { 1 2 ∣ x ∣ < 1 , ∣ y ∣ < 1 0 Otherwise f_{Y}(x,y)= \begin{cases}
{1 \over 2} & |x|<1, |y|<1 \\
0 &\text{Otherwise }
\end{cases} f Y ( x , y ) = { 2 1 0 ∣ x ∣ < 1 , ∣ y ∣ < 1 Otherwise The support set is a rectangle, so we need to check if it is true that f X Y ( x , y ) = f X ( x ) f Y ( y ) , f_{XY}(x,y)=f_{X}(x)f_{Y}(y), f X Y ( x , y ) = f X ( x ) f Y ( y ) , ( x , y ) . (x, y). ( x , y ) .
f X , Y ( 1 2 , 1 2 ) = 1 4 ( 1 + 1 2 ( 1 2 ) ) = 5 16 f_{X,Y}({1 \over 2},{1 \over 2})={1 \over 4}(1+{1 \over 2}({1 \over 2}))={5 \over 16} f X , Y ( 2 1 , 2 1 ) = 4 1 ( 1 + 2 1 ( 2 1 )) = 16 5
f X ( 1 2 ) f Y ( 1 2 ) = 1 2 ( 1 2 ) = 1 4 f_{X}({1 \over 2})f_{Y}({1 \over 2})={1 \over 2}({1 \over 2})={1 \over 4} f X ( 2 1 ) f Y ( 2 1 ) = 2 1 ( 2 1 ) = 4 1
f X , Y ( 1 2 , 1 2 ) ≠ f X ( 1 2 ) f Y ( 1 2 ) f_{X,Y}({1 \over 2},{1 \over 2})\not=f_{X}({1 \over 2})f_{Y}({1 \over 2}) f X , Y ( 2 1 , 2 1 ) = f X ( 2 1 ) f Y ( 2 1 ) Therefore, X and Y are not independent.
U = X 2 , V = Y 2 , 0 ≤ u < 1 , ≤ v < 1 U=X^2, V=Y^2, 0\leq u<1, \leq v<1 U = X 2 , V = Y 2 , 0 ≤ u < 1 , ≤ v < 1
If − 1 < x < 0 a n d − 1 < y < 0 -1<x<0 \ and -1<y<0 − 1 < x < 0 an d − 1 < y < 0
X = − U , Y = − V X=-\sqrt{U}, Y=-\sqrt{V} X = − U , Y = − V
The Jacobian
J ( u , v ) = ∣ ∂ x / ∂ u ∂ x / ∂ v ∂ y / ∂ u ∂ y / ∂ v ∣ = J(u, v)=\begin{vmatrix}
\partial x/\partial u & \partial x/\partial v \\
\partial y/\partial u & \partial y/\partial v
\end{vmatrix}= J ( u , v ) = ∣ ∣ ∂ x / ∂ u ∂ y / ∂ u ∂ x / ∂ v ∂ y / ∂ v ∣ ∣ = = ∣ − 1 / ( 2 u ) 0 0 − 1 / ( 2 v ) ∣ = 1 / ( 4 u v ) =\begin{vmatrix}
-1/(2\sqrt{u}) & 0 \\
0 & -1/(2\sqrt{v})
\end{vmatrix}=1/(4\sqrt{uv}) = ∣ ∣ − 1/ ( 2 u ) 0 0 − 1/ ( 2 v ) ∣ ∣ = 1/ ( 4 uv ) If 0 ≤ x < 1 a n d 0 ≤ y < 1 0\leq x<1 \ and\ 0\leq y<1 0 ≤ x < 1 an d 0 ≤ y < 1
X = U , Y = V X=\sqrt{U}, Y=\sqrt{V} X = U , Y = V
J ( u , v ) = ∣ ∂ x / ∂ u ∂ x / ∂ v ∂ y / ∂ u ∂ y / ∂ v ∣ = J(u, v)=\begin{vmatrix}
\partial x/\partial u & \partial x/\partial v \\
\partial y/\partial u & \partial y/\partial v
\end{vmatrix}= J ( u , v ) = ∣ ∣ ∂ x / ∂ u ∂ y / ∂ u ∂ x / ∂ v ∂ y / ∂ v ∣ ∣ = = ∣ 1 / ( 2 u ) 0 0 1 / ( 2 v ) ∣ = 1 / ( 4 u v ) =\begin{vmatrix}
1/(2\sqrt{u}) & 0 \\
0 & 1/(2\sqrt{v})
\end{vmatrix}=1/(4\sqrt{uv}) = ∣ ∣ 1/ ( 2 u ) 0 0 1/ ( 2 v ) ∣ ∣ = 1/ ( 4 uv )
1 4 ( 1 + u v ) ( 1 4 u v ) = 1 16 ( 1 + 1 u v ) {1 \over 4}(1+\sqrt{uv})({1 \over 4\sqrt{uv}})= {1 \over 16}(1+{1 \over \sqrt{uv}}) 4 1 ( 1 + uv ) ( 4 uv 1 ) = 16 1 ( 1 + uv 1 )
If − 1 < x < 0 a n d 0 ≤ y < 1 -1<x<0 \ and\ 0\leq y<1 − 1 < x < 0 an d 0 ≤ y < 1 0 ≤ x < 1 a n d 0 ≤ y < 1 \ 0\leq x<1 \ and\ 0\leq y<1 0 ≤ x < 1 an d 0 ≤ y < 1
X = − U , Y = V X=-\sqrt{U}, Y=\sqrt{V} X = − U , Y = V
J ( u , v ) = ∣ ∂ x / ∂ u ∂ x / ∂ v ∂ y / ∂ u ∂ y / ∂ v ∣ = J(u, v)=\begin{vmatrix}
\partial x/\partial u & \partial x/\partial v \\
\partial y/\partial u & \partial y/\partial v
\end{vmatrix}= J ( u , v ) = ∣ ∣ ∂ x / ∂ u ∂ y / ∂ u ∂ x / ∂ v ∂ y / ∂ v ∣ ∣ = = ∣ − 1 / ( 2 u ) 0 0 1 / ( 2 v ) ∣ = − 1 / ( 4 u v ) =\begin{vmatrix}
-1/(2\sqrt{u}) & 0 \\
0 & 1/(2\sqrt{v})
\end{vmatrix}=-1/(4\sqrt{uv}) = ∣ ∣ − 1/ ( 2 u ) 0 0 1/ ( 2 v ) ∣ ∣ = − 1/ ( 4 uv ) If 0 ≤ x < 1 a n d − 1 < y < 0 0\leq x<1 \ and\ -1< y<0 0 ≤ x < 1 an d − 1 < y < 0
X = U , Y = − V X=\sqrt{U}, Y=-\sqrt{V} X = U , Y = − V
J ( u , v ) = ∣ ∂ x / ∂ u ∂ x / ∂ v ∂ y / ∂ u ∂ y / ∂ v ∣ = J(u, v)=\begin{vmatrix}
\partial x/\partial u & \partial x/\partial v \\
\partial y/\partial u & \partial y/\partial v
\end{vmatrix}= J ( u , v ) = ∣ ∣ ∂ x / ∂ u ∂ y / ∂ u ∂ x / ∂ v ∂ y / ∂ v ∣ ∣ = = ∣ 1 / ( 2 u ) 0 0 − 1 / ( 2 v ) ∣ = − 1 / ( 4 u v ) =\begin{vmatrix}
1/(2\sqrt{u}) & 0 \\
0 & -1/(2\sqrt{v})
\end{vmatrix}=-1/(4\sqrt{uv}) = ∣ ∣ 1/ ( 2 u ) 0 0 − 1/ ( 2 v ) ∣ ∣ = − 1/ ( 4 uv )
1 4 ( 1 − u v ) ( − 1 4 u v ) = 1 16 ( 1 − 1 u v ) {1 \over 4}(1-\sqrt{uv})(-{1 \over 4\sqrt{uv}})= {1 \over 16}(1-{1 \over \sqrt{uv}}) 4 1 ( 1 − uv ) ( − 4 uv 1 ) = 16 1 ( 1 − uv 1 )
If − 1 < x < 0 a n d − 1 < y < 0 -1<x<0 \ and -1<y<0 − 1 < x < 0 an d − 1 < y < 0
f U , V ( u , v ) = { 1 16 ( 1 + 1 u v ) 0 < u < 1 , 0 < v < 1 0 Otherwise f_{U,V}(u,v)= \begin{cases}
{1 \over 16}(1+{1 \over \sqrt{uv}}) & 0<u<1,0<v<1 \\
0 &\text{Otherwise }
\end{cases} f U , V ( u , v ) = { 16 1 ( 1 + uv 1 ) 0 0 < u < 1 , 0 < v < 1 Otherwise q ( u ) = ∫ − ∞ ∞ f U , V ( u , v ) d v = q(u)=\displaystyle\int_{-\infin}^{\infin}f_{U,V}(u,v)dv= q ( u ) = ∫ − ∞ ∞ f U , V ( u , v ) d v = = ∫ 0 1 1 16 ( 1 + 1 u v ) d v = 1 16 ( 1 + 1 u ) =\displaystyle\int_{0}^{1} {1 \over 16}(1+{1 \over \sqrt{uv}})dv={1 \over 16}(1+{1 \over \sqrt{u}}) = ∫ 0 1 16 1 ( 1 + uv 1 ) d v = 16 1 ( 1 + u 1 )
w ( v ) = ∫ − ∞ ∞ f U , V ( u , v ) d u = w(v)=\displaystyle\int_{-\infin}^{\infin}f_{U,V}(u,v)du= w ( v ) = ∫ − ∞ ∞ f U , V ( u , v ) d u = = ∫ 0 1 1 16 ( 1 + 1 u v ) d u = 1 16 ( 1 + 1 v ) =\displaystyle\int_{0}^{1} {1 \over 16}(1+{1 \over \sqrt{uv}})du={1 \over 16}(1+{1 \over \sqrt{v}}) = ∫ 0 1 16 1 ( 1 + uv 1 ) d u = 16 1 ( 1 + v 1 )
f U ( v ) f V ( u ) = 1 16 ( 1 + 1 u ) ( 1 16 ) ( 1 + 1 v ) ≠ f_{U}(v)f_{V}(u)={1 \over 16}(1+{1 \over \sqrt{u}})({1 \over 16})(1+{1 \over \sqrt{v}})\not= f U ( v ) f V ( u ) = 16 1 ( 1 + u 1 ) ( 16 1 ) ( 1 + v 1 ) = ≠ 1 16 ( 1 + 1 u v ) = f U , V ( u , v ) \not= {1 \over 16}(1+{1 \over \sqrt{uv}})=f_{U,V}(u,v) = 16 1 ( 1 + uv 1 ) = f U , V ( u , v ) Therefore, X2 and Y2 are not independent.
#332078 on Oct 2022
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