# PDF of transformed random variable and Surface Element in High Dimension.

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 $$p_T(t_1,\dots,t_k)=\int_S p(x_1,\dots,x_n)\left[\displaystyle\sum_{i_1<\cdots where $$\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 $$\mathrm{d}x_{i_1}\wedge\cdots\wedge\mathrm{d}x_{i_k}$$?

• Which books? I actually have a question about this topic that no one answered a while ago. Mar 1, 2022 at 18:26
• A book about mathematical statistics in chinese. Mar 1, 2022 at 18:39
• 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). Mar 1, 2022 at 19:01
• 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? Mar 1, 2022 at 19:08

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}
• This should be $p_T(t)$ :) Mar 2, 2022 at 18:42
• 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. Mar 3, 2022 at 5:28
• @TedShifrin . Now I see it :) . The notion of $t_1$ and $t_2$ doesn't even exist. Thanks once more. You are impressive. Mar 3, 2022 at 6:16