$\begingroup$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.
$\endgroup$11 Answer
$\begingroup$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: $$ \det (J_{F^{-1}}) = \det\big((J_F)^{-1}\big) = \frac{1}{\det(J_F)} = n. $$
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