Proof of reciprocal of a Jacobian

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So I was reading 2 different statistics books, I read about the PDF variable transformation and one book cited that you can do it by

If $Y_i=u_i(X_1,X_2,...,X_n)$ where $i=1,2,...,n$, then

$f_Y(y_1,y_2,...y_n)=f_X(x_1,x_2,...,x_n)|J_1|$

Where $f_X$ and $f_Y$ are the joint PDF of $X_i$'s and $Y_i$'s respectively and

$J_1=beginvmatrixfracpartial x_1partial y_1&fracpartial x_1partial y_2&cdots&fracpartial x_1partial y_n\fracpartial x_2partial y_1&fracpartial x_2partial y_2&cdots&fracpartial x_2partial y_n\vdots&vdots&ddots&vdots\fracpartial x_npartial y_1&fracpartial x_npartial y_2&cdots&fracpartial x_npartial y_nendvmatrix$

Now the second book I read instead gives

$f_Y(y_1,y_2,...,y_n)=f_X(x_1,x_2,...,x_n)|J_2|^-1$

Where

$J_2=beginvmatrixfracpartial y_1partial x_1&fracpartial y_1partial x_2&cdots&fracpartial y_1partial x_n\fracpartial y_2partial x_1&fracpartial y_2partial x_2&cdots&fracpartial y_2partial x_n\vdots&vdots&ddots&vdots\fracpartial y_npartial x_1&fracpartial y_npartial x_2&cdots&fracpartial y_npartial x_nendvmatrix$

Now for transformation of 1 variable I know that

$fracdxdy=(fracdydx)^-1$

But for 2 or more variable can someone give me a proof that

$J_1=J_2^-1$

asked Jul 22 at 13:15

Germaniac

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up vote
0
down vote

favorite

So I was reading 2 different statistics books, I read about the PDF variable transformation and one book cited that you can do it by

If $Y_i=u_i(X_1,X_2,...,X_n)$ where $i=1,2,...,n$, then

$f_Y(y_1,y_2,...y_n)=f_X(x_1,x_2,...,x_n)|J_1|$

Where $f_X$ and $f_Y$ are the joint PDF of $X_i$'s and $Y_i$'s respectively and

Now the second book I read instead gives

$f_Y(y_1,y_2,...,y_n)=f_X(x_1,x_2,...,x_n)|J_2|^-1$

Where

Now for transformation of 1 variable I know that

$fracdxdy=(fracdydx)^-1$

But for 2 or more variable can someone give me a proof that

$J_1=J_2^-1$

asked Jul 22 at 13:15

Germaniac

142

add a commentÂ |Â

up vote
0
down vote

favorite

So I was reading 2 different statistics books, I read about the PDF variable transformation and one book cited that you can do it by

If $Y_i=u_i(X_1,X_2,...,X_n)$ where $i=1,2,...,n$, then

$f_Y(y_1,y_2,...y_n)=f_X(x_1,x_2,...,x_n)|J_1|$

Where $f_X$ and $f_Y$ are the joint PDF of $X_i$'s and $Y_i$'s respectively and

Now the second book I read instead gives

$f_Y(y_1,y_2,...,y_n)=f_X(x_1,x_2,...,x_n)|J_2|^-1$

Where

Now for transformation of 1 variable I know that

$fracdxdy=(fracdydx)^-1$

But for 2 or more variable can someone give me a proof that

$J_1=J_2^-1$

asked Jul 22 at 13:15

Germaniac

142

So I was reading 2 different statistics books, I read about the PDF variable transformation and one book cited that you can do it by

If $Y_i=u_i(X_1,X_2,...,X_n)$ where $i=1,2,...,n$, then

$f_Y(y_1,y_2,...y_n)=f_X(x_1,x_2,...,x_n)|J_1|$

Where $f_X$ and $f_Y$ are the joint PDF of $X_i$'s and $Y_i$'s respectively and

Now the second book I read instead gives

$f_Y(y_1,y_2,...,y_n)=f_X(x_1,x_2,...,x_n)|J_2|^-1$

Where

Now for transformation of 1 variable I know that

$fracdxdy=(fracdydx)^-1$

But for 2 or more variable can someone give me a proof that

$J_1=J_2^-1$

asked Jul 22 at 13:15

Germaniac

142

asked Jul 22 at 13:15

Germaniac

142

asked Jul 22 at 13:15

Germaniac

142

asked Jul 22 at 13:15

Germaniac

142

add a commentÂ |Â

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