Consider $k$ functions $T_i(x_1,\dots,x_n)$ on $\mathbb{R}^n$, Let $p(x_1,\dots,x_n)$ be a pdf of random variable $X$, then what is the pdf of $$T(x)=(T_1,\dots,T_k)?$$

Some books state that $$p_T(t_1,\dots,t_k)=\int_S p(x_1,\dots,x_n)\left[\displaystyle\sum_{i_1<\cdots<i_k}\left|\frac{\partial(T_1,\dots,T_k)}{\partial(x_{i_1},\dots,x_{x_k})}\right|^2\right]^{-\frac{1}{2}}\,\mathrm{d}S,$$ where $\frac{\partial(T_1,\dots,T_k)}{\partial(x_{i_1},\dots,x_{x_k})}$ is Jacobian determinant, and the $(n-k)$-dimensional surface $S$ is definited by equations $$T_i(x_1,\dots,x_n)=t_i,\ 1\leq i\leq k.$$

Why? What is the relation between surface element $\mathrm{d}S$ and differential form $\sum f_{i_1,\dots,i_k}\mathrm dx_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}$?

In addition, I know that if $S$ has parameterization (with some abuse of notations) $$x_i=x_i(u_1,\dots,u_k),\ 1\leq i\leq n.$$ then $$\begin{align*} &\quad\;\sum a_{i_1,\dots,i_k}\mathrm{d}x_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}\\ &=\left(\sum a_{i_1,\dots,i_k}\frac{\partial(x_{i_1},\dots,x_{i_k})}{\partial(u_1,\dots,u_k)}\right)\mathrm{d}u_1\wedge\cdots\wedge\mathrm{d}u_k. \end{align*}$$ But is there exists an inversion of the formula, i.e., can we express $$a\,\mathrm{d}u_1\wedge\cdots\wedge\mathrm{d}u_k,$$ in form of $\mathrm{d}x_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}$?

  • $\begingroup$ Which books? I actually have a question about this topic that no one answered a while ago. $\endgroup$
    – William M.
    Mar 1, 2022 at 18:26
  • $\begingroup$ A book about mathematical statistics in chinese. $\endgroup$ Mar 1, 2022 at 18:39
  • $\begingroup$ Unfortunately, the change of variables formula for non-invertible transformations is quite hard to track down. The most I can tell is that if $T:\mathbf{R}^n \to \mathbf{R}^k,$ then you can hope to find $T':\mathbf{R}^n \to \mathbf{R}^{n-k}$ such that $(T, T')$ is a bijection and use the usual change of variables formula. This, however, requires $T$ to be injective. A many-to-one function is largely mysterious to me and all I have seen are statements of theorems but never proofs or proper references (e.g. what you just did). $\endgroup$
    – William M.
    Mar 1, 2022 at 19:01
  • $\begingroup$ This looks a lot like the Gram determinant, which gives the surface area element for a parametric submanifold. However, here it's a level set, defined implicitly. Have you tried to work out low-dimensional examples to make this more understandable? $\endgroup$ Mar 1, 2022 at 19:08

2 Answers 2


This is the coarea formula from geometric measure theory. If we assume that $T$ has constant rank $k$ (which is implicit in the assumption that the level sets $T_i=t_i$, $i=1,\dots,k$, are smooth), then this is just integration over the fiber.

For example, if $k=1$, then we want to write $dV = dx_1\wedge\dots\wedge dx_n= F(x)\,dS\wedge dt$, where $dS$ is the surface area element on the hypersurface $T=t$. We have the classic formula $dS = \iota_{X}dV = dV(X,\cdot)$, where $X$ is the unit normal vector $\nabla T/\|\nabla T\|$ of the level hypersurface. Since $dt(\nabla T)=dT(\nabla T) = \|\nabla T\|^2$, it follows that (up to sign) $dV = \frac1{\|\nabla T\|}dS\wedge dt$. Integration over the fiber gives $$p_T(t) = \int_{\{T=t\}} p(x_1,\dots,x_n) \frac1{\|\nabla T\|}dS.$$ More generally, if $k>1$, then the hypersurfaces $T_i=t_i$ will not necessarily be orthogonal, and we will have (up to sign) $$dS = \frac{dV\left(\tfrac{\nabla T_1}{\|\nabla T_1\|},\dots,\tfrac{\nabla T_k}{\|\nabla T_k\|},\cdot\right)}{\|dT_1\wedge\dots\wedge dT_k\|},$$ since $(dT_1\wedge\dots\wedge dT_k)\left(\tfrac{\nabla T_1}{\|\nabla T_1\|},\dots,\tfrac{\nabla T_k}{\|\nabla T_k\|}\right)$ is the conorm of $dT_1\wedge \dots\wedge dT_k$. This, in turn, is the expression in your formula.


One simple example if permitted. I did this mainly to better understand the matter for myself. For \begin{align} T(x_1,x_2)=\sqrt{x_1^2+x_2^2}\,,\quad X_1,X_2\sim N(0,1)\,,\text{ independent } \end{align} it is easy to see that \begin{align} \mathbb P\big(T(X_1,X_2)\le t\big)&=\int_{T(x_1,x_2)\le t}\frac{1}{2\pi}e^{-\frac{x_1^2+x_2^2}{2}}\,dx_1\,dx_2 & =\int_0^t\underbrace{\int_0^{2\pi}\frac{1}{2\pi}e^{-\frac{r^2}{2}}\,r\,d\varphi}_{(*)}\,dr\\ &=1-e^{-t^2/2}\,,\\ \end{align} which has density \begin{align} p_T(t)&=te^{-t^2/2}\,. \end{align} Since, \begin{align} \nabla T&=\frac{1}{\sqrt{x_1^2+x_2^2}}\left(\begin{matrix}x_1\\x_2\end{matrix}\right)\,,\quad\frac{1}{||\nabla T||}=1 \end{align} and \begin{align} p(x_1,x_2)=\frac{1}{2\pi}e^{-\frac{x_1^2+x_2^2}{2}} \end{align} we can write this in the form given by Ted Shifrin: \begin{align} p_T(t)&=\int_{\{T=t\}}p(x_1,x_2)\frac{1}{||\nabla T||}\,dS =\underbrace{\int_0^{2\pi}\frac{1}{2\pi}e^{-\frac{t^2}{2}}\,t\,d\varphi}_{(*)}\,. \end{align}

  • $\begingroup$ This should be $p_T(t)$ :) $\endgroup$ Mar 2, 2022 at 18:42
  • $\begingroup$ You are abusing function notation. The input is a single variable, not a pair. In fact, that’s the whole point of this construction. $\endgroup$ Mar 3, 2022 at 5:09
  • $\begingroup$ No, that is correct for a bivariate distribution function. You continue to miss the point. $p_T$ is a function on the set of level surfaces if $T$, not on the original space. $\endgroup$ Mar 3, 2022 at 5:28
  • $\begingroup$ @TedShifrin . Now I see it :) . The notion of $t_1$ and $t_2$ doesn't even exist. Thanks once more. You are impressive. $\endgroup$
    – Kurt G.
    Mar 3, 2022 at 6:16

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