Find the joint pdf $f_Z$ given $f_{X,Y}, Z=XY$
$begingroup$
According to @Did in his answer, given the joint pdf $f_{X,Y}$ and $Z=XY$,
$$
f_Z(z)=int f_{X,Y}(y^{-1}z,y)y^{-1}mathrm dy.
$$
because $J= begin{bmatrix}
frac{d y^{-1}z}{dz} & frac{d y^{-1}z}{dy} \
frac{d y}{dz} & frac{d y}{dy} end{bmatrix} = begin{bmatrix}
1/y & −z/y^2 \
0 & 1 end{bmatrix} = 1/y$
But how is $f_Z(z)$ reduced from $f_Z(z)=frac{d}{dZ}F_Z(z)=intint_{xyleq Z} f_{X,Y}(x,y) dx dy$?
calculus probability integration change-of-variable
asked Dec 9 '18 at 23:35
drerDdrerD
1559
$endgroup$
add a comment |
$begingroup$
According to @Did in his answer, given the joint pdf $f_{X,Y}$ and $Z=XY$,
$$
f_Z(z)=int f_{X,Y}(y^{-1}z,y)y^{-1}mathrm dy.
$$
because $J= begin{bmatrix}
frac{d y^{-1}z}{dz} & frac{d y^{-1}z}{dy} \
frac{d y}{dz} & frac{d y}{dy} end{bmatrix} = begin{bmatrix}
1/y & −z/y^2 \
0 & 1 end{bmatrix} = 1/y$
But how is $f_Z(z)$ reduced from $f_Z(z)=frac{d}{dZ}F_Z(z)=intint_{xyleq Z} f_{X,Y}(x,y) dx dy$?
calculus probability integration change-of-variable
asked Dec 9 '18 at 23:35
drerDdrerD
1559
$endgroup$
$begingroup$
$$F_Z(z) = iint_{xyle z} f_{X,Y}(x,y),mathrm dxmathrm dy = int_{0}^z int_0^infty y^{-1}, f_{X,Y}(y^{-1}z,y),mathrm dymathrm dz .$$
$endgroup$
– Kanu Kim
Dec 10 '18 at 0:36
add a comment |
$begingroup$
According to @Did in his answer, given the joint pdf $f_{X,Y}$ and $Z=XY$,
$$
f_Z(z)=int f_{X,Y}(y^{-1}z,y)y^{-1}mathrm dy.
$$
because $J= begin{bmatrix}
frac{d y^{-1}z}{dz} & frac{d y^{-1}z}{dy} \
frac{d y}{dz} & frac{d y}{dy} end{bmatrix} = begin{bmatrix}
1/y & −z/y^2 \
0 & 1 end{bmatrix} = 1/y$
But how is $f_Z(z)$ reduced from $f_Z(z)=frac{d}{dZ}F_Z(z)=intint_{xyleq Z} f_{X,Y}(x,y) dx dy$?
calculus probability integration change-of-variable
asked Dec 9 '18 at 23:35
drerDdrerD
1559
$endgroup$
According to @Did in his answer, given the joint pdf $f_{X,Y}$ and $Z=XY$,
$$
f_Z(z)=int f_{X,Y}(y^{-1}z,y)y^{-1}mathrm dy.
$$
because $J= begin{bmatrix}
frac{d y^{-1}z}{dz} & frac{d y^{-1}z}{dy} \
frac{d y}{dz} & frac{d y}{dy} end{bmatrix} = begin{bmatrix}
1/y & −z/y^2 \
0 & 1 end{bmatrix} = 1/y$
But how is $f_Z(z)$ reduced from $f_Z(z)=frac{d}{dZ}F_Z(z)=intint_{xyleq Z} f_{X,Y}(x,y) dx dy$?
calculus probability integration change-of-variable
calculus probability integration change-of-variable
asked Dec 9 '18 at 23:35
drerDdrerD
1559
asked Dec 9 '18 at 23:35
drerDdrerD
1559
asked Dec 9 '18 at 23:35
drerDdrerD
1559
asked Dec 9 '18 at 23:35
drerDdrerD
1559
asked Dec 9 '18 at 23:35
drerDdrerD
1559
1559
$begingroup$
$$F_Z(z) = iint_{xyle z} f_{X,Y}(x,y),mathrm dxmathrm dy = int_{0}^z int_0^infty y^{-1}, f_{X,Y}(y^{-1}z,y),mathrm dymathrm dz .$$
$endgroup$
– Kanu Kim
Dec 10 '18 at 0:36
add a comment |
$begingroup$
$$F_Z(z) = iint_{xyle z} f_{X,Y}(x,y),mathrm dxmathrm dy = int_{0}^z int_0^infty y^{-1}, f_{X,Y}(y^{-1}z,y),mathrm dymathrm dz .$$
$endgroup$
– Kanu Kim
Dec 10 '18 at 0:36
$begingroup$
$$F_Z(z) = iint_{xyle z} f_{X,Y}(x,y),mathrm dxmathrm dy = int_{0}^z int_0^infty y^{-1}, f_{X,Y}(y^{-1}z,y),mathrm dymathrm dz .$$
$endgroup$
– Kanu Kim
Dec 10 '18 at 0:36
$begingroup$
$$F_Z(z) = iint_{xyle z} f_{X,Y}(x,y),mathrm dxmathrm dy = int_{0}^z int_0^infty y^{-1}, f_{X,Y}(y^{-1}z,y),mathrm dymathrm dz .$$
$endgroup$
– Kanu Kim
Dec 10 '18 at 0:36
add a comment |
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$begingroup$
$$F_Z(z) = iint_{xyle z} f_{X,Y}(x,y),mathrm dxmathrm dy = int_{0}^z int_0^infty y^{-1}, f_{X,Y}(y^{-1}z,y),mathrm dymathrm dz .$$
$endgroup$
– Kanu Kim
Dec 10 '18 at 0:36