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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 $0<u_{1}<u_{2}<, \ldots,<u_{n}<T$ 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

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