# Can't understand the proof of the Time-Rescaling theorem

I was reading the following paper: The time-rescaling theorem and its application to neural spike train data analysis and I have some difficulties understanding the proof of the time-rescaling-theorem. Here is the first part of the demonstration, my main concern is about the form of the Jacobian of the transformation between $$u_k$$ and $$\tau_k$$, so it's not particularly important to understand every bit of information provided below, however, I report part of the paper to add a little bit of context:

Time-Rescaling Theorem.

Let $$0 be a realization from a point process with a conditional intensity function $$\lambda\left(t \mid H_{t}\right)$$ satisfying $$0<\lambda\left(t \mid H_{t}\right)$$ for all $$t \in(0, T] .$$ Define the transformation $$\Lambda\left(u_{k}\right)=\int_{0}^{u_{k}} \lambda\left(u \mid H_{u}\right) d u$$ for $$k=1, \ldots, n$$, and assume $$\Lambda(t)<\infty$$ with probability one for all $$t \in(0, T]$$. Then the $$\Lambda\left(u_{k}\right)$$ 's are a Poisson process with unit rate.

Proof.

Let $$\tau_{k}=\Lambda\left(u_{k}\right)-\Lambda\left(u_{k-1}\right)$$ for $$k=1, \ldots, n$$ and set $$\tau_{T} = \int_{u_n}^{T} \lambda\left(u \mid H_{u}\right) d u$$. To establish the result, it suffices to show that the $$\tau_{k}$$s are independent and identically distributed exponential random variables with mean one. Because the $$\tau_{k}$$ transformation is one-to-one and $$\tau_{n+1}>\tau_{T}$$ if and only if $$u_{n+1}>T$$, the joint probability density of the $$\tau_{k}$$ 's is $$\begin{array}{l} f\left(\tau_{1}, \tau_{2}, \ldots, \tau_{n} \cap \tau_{n+1}>\tau_{T}\right) =f\left(\tau_{1}, \ldots, \tau_{n}\right) \operatorname{Pr}\left(\tau_{n+1}>\tau_{T} \mid \tau_{1}, \ldots, \tau_{n}\right) \end{array}$$ The following two events are equivalent $$\left\{\tau_{n+1}>\tau_{T} \mid \tau_{1}, \ldots, \tau_{n}\right\}=\left\{u_{n+1}>T \mid u_{1}, u_{2}, \ldots, u_{n}\right\}$$ Hence \color{black}{ \begin{aligned} \operatorname{Pr}\left(\tau_{n+1}>\tau_{T} \mid \tau_{1}, \tau_{2}, \ldots, \tau_{n}\right) &=\operatorname{Pr}\left(u_{n+1}>T \mid u_{1}, u_{2}, \ldots, u_{n}\right) \\ &=\exp \left\{-\int_{u_n}^{T} \lambda\left(u \mid H_{u_{n}}\right) d u\right\} \\ &=\exp\left\{-\tau_T\right\} \end{aligned} } where the last equality follows from the definition of $$\tau_T$$.

By the multivariate change-of-variable formula (Port, 1994) we have that $$f\left(\tau_{1}, \tau_{2}, \ldots, \tau_{n}\right)=|J| f\left(u_{1}, u_{2}, \ldots, u_{n} \cap N\left(u_{n}\right)=n\right)$$ where $$|J|$$ is the determinant of the Jacobian matrix $$J$$ of the transformation between $$u_{j}, j=1, \ldots, n$$ and $$\tau_{k}, k=1, \ldots, n$$.

Now, since we're talking about the transformation from $$u_{j}, j=1, \ldots, n$$ to $$\tau_{k}, k=1, \ldots, n$$ I'd expect the Jacobian to look like:

$$J = \begin{bmatrix} \frac{\partial \tau_1}{\partial u_1} & \frac{\partial \tau_1}{\partial u_2} & \dots &\frac{\partial \tau_1}{\partial u_n} \\ \frac{\partial \tau_2}{\partial u_1} & \frac{\partial \tau_2}{\partial u_2} & \dots &\frac{\partial \tau_2}{\partial u_n} \\ \vdots & \vdots & \ddots &\vdots \\ \frac{\partial \tau_n}{\partial u_1} & \frac{\partial \tau_n}{\partial u_2} & \dots &\frac{\partial \tau_n}{\partial u_n} \end{bmatrix}$$ And, given the fact that $$\tau_{k}=\Lambda\left(u_{k}\right)-\Lambda\left(u_{k-1}\right) = \int_{0}^{u_{k}} \lambda\left(u \mid H_{u}\right) du - \int_{0}^{u_{k-1}} \lambda\left(u \mid H_{u}\right) du = \int_{u_{k-1}}^{u_{k}} \lambda\left(u \mid H_{u}\right) du$$

From the first fundamental theorem of calculus we know that $$\frac{d}{d x} \int_{a}^{x} f(t) d t=f(x)$$, so $$\frac{\partial \tau_k}{\partial u_k} = \frac{\partial}{\partial u_k}\int_{u_{k-1}}^{u_{k}} \lambda\left(u \mid H_{u}\right) du = \lambda\left(u_k \mid H_{u_k}\right)$$

Which would translate to

$$J = \begin{bmatrix} \frac{\partial \tau_1}{\partial u_1} & 0 & \dots & 0 \\ 0 & \frac{\partial \tau_2}{\partial u_2} & \dots & 0 \\ \vdots & \vdots & \ddots &\vdots \\ 0 & 0 & \dots &\frac{\partial \tau_n}{\partial u_n} \end{bmatrix}$$

$$J = \begin{bmatrix} \lambda\left(u_1 \mid H_{u_1}\right) & 0 & \dots & 0 \\ 0 & \lambda\left(u_2 \mid H_{u_2}\right) & \dots & 0 \\ \vdots & \vdots & \ddots &\vdots \\ 0 & 0 & \dots & \lambda\left(u_n \mid H_{u_n}\right) \end{bmatrix}$$

However, the paper continues as:

Because $$\tau_{k}$$ is a function of $$\color{black}{u_{1}, \ldots, u_{k}}, J$$ is a lower triangular matrix, and its determinant is the product of its diagonal elements defined as $$|J|=\left|\prod_{k=1}^{n} J_{k k}\right| .$$

By assumption $$0<\lambda\left(t \mid H_{t}\right)$$ and the definition of $$\tau_{k}$$, the mapping of $$u$$ into $$\tau$$ is one-to-one. Therefore, by the inverse differentiation theorem (Protter & Morrey, 1991), the diagonal elements of $$J$$ are $$J_{k k}=\frac{\partial u_{k}}{\partial \tau_{k}}=\lambda\left(u_{k} \mid H_{u_{k}}\right)^{-1}$$

So apparently the correct Jacobian is

$$J = \begin{bmatrix} \frac{\partial u_1}{\partial \tau_1} & \frac{\partial u_1}{\partial\tau_2} & \dots &\frac{\partial u_1}{\partial\tau_n} \\ \frac{\partial u_2}{\partial \tau_1} & \frac{\partial u_2}{\partial\tau_2} & \dots &\frac{\partial u_2}{\partial\tau_n} \\ \vdots & \vdots & \ddots &\vdots \\ \frac{\partial u_n}{\partial \tau_1} & \frac{\partial u_n}{\partial\tau_2} & \dots &\frac{\partial u_n}{\partial\tau_n} \end{bmatrix}$$

and the matrix is lower-triangular.

What am I doing wrong? Which elements of the Jacobian are indeed non-zeros?

Thanks