How The Jacobian of the transformation can be shown to not depend on $X_i$ or $\bar X $ and is equal to the constant $n$ Transform the random variables, $X_i$, $i=1,2,\ldots,n$ to
$$
\begin{align}
Y_1 & =\bar X \\
Y_2 & =X_2-\bar X \\
Y_3 & = X_3-\bar X \\
& {}\  \vdots \\
Y_n & =X_n-\bar X
\end{align}
$$
Find the Jacobian of transformation.
 A: I am assuming that 
$$\bar{X} = \frac{1}{n}\sum^n_{i=1} X_i.$$
Then Jacobi matrix for the transformation $F$ that maps $(X_1,\ldots,X_n)$ to $(Y_1,\ldots,Y_n)$ is
$$
J_F=\begin{pmatrix} 
\dfrac{\partial Y_1}{\partial X_1} & \cdots & \dfrac{\partial Y_1}{\partial X_n} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial Y_n}{\partial X_1} & \cdots & \dfrac{\partial Y_n}{\partial X_n}
\end{pmatrix} = \begin{pmatrix} 
\dfrac{1}{n} & \dfrac{1}{n} &\cdots & \dfrac{1}{n} & \dfrac{1}{n}\\
-\dfrac{1}{n}& 1-\dfrac{1}{n} &\cdots &-\dfrac{1}{n} & -\dfrac{1}{n} \\ 
\vdots &\vdots &\ddots &\vdots &\vdots \\ 
-\dfrac{1}{n} &-\dfrac{1}{n}  &\cdots & -\dfrac{1}{n} & 1 -\dfrac{1}{n}
\end{pmatrix}.
$$
For computing $\det J$, notice that we can add the first row to every other row (this doesn't change the determinant) to get $J_1$:
$$
J_1 = \begin{pmatrix} 
\dfrac{1}{n} & \dfrac{1}{n} &\cdots & \dfrac{1}{n} & \dfrac{1}{n}\\
0& 1 &\cdots &0& 0\\ 
\vdots &\vdots &\ddots &\vdots &\vdots \\ 
0 &0 &\cdots & 0 & 1 
\end{pmatrix}.
$$
Again multiply the second row till the last row by $-1/n$ and add to the first row, what is left is just a diagonal matrix $J_2$:
$$
J_2 = \mathrm{diag}\,\{1/n,1,\ldots,1\}.
$$
Hence:
$$
\det J_F = \det J_2 = \frac{1}{n}.
$$

If we want to compute the Jacobian of the inverse transform $F^{-1}$ which maps $(Y_1,\ldots,Y_n)$ back to $(X_1,\ldots,X_n)$:
$$
J_{F^{-1}}=\begin{pmatrix} 
\dfrac{\partial X_1}{\partial Y_1} & \cdots & \dfrac{\partial X_1}{\partial Y_n} \\ \vdots & \ddots & \vdots \\ \dfrac{\partial X_n}{\partial Y_1} & \cdots & \dfrac{\partial X_n}{\partial Y_n}
\end{pmatrix},
$$
by inverse function theorem, the inverse transform $F^{-1}$'s Jacobian matrix is actually the inverse matrix of the Jacobian of $F$:
$$
J_{F^{-1}} = (J_F)^{-1},
$$
evaluated at different variables of course, but here we have constants so you don't have to worry. Now the problem changes to computing the determinant for an inverse matrix:
$$
\det (J_{F^{-1}}) = \det\big((J_F)^{-1}\big) = \frac{1}{\det(J_F)} = n.
$$ 
