We greatly thank the reviewer for your constructive feedback.
> For the log-normalization, could you kindly provide the detailed derivation for obtaining $\hat \lambda_{\text{ASM,Log}}$ and $\hat \lambda _{\text{ASM,Log Interval}}$?
Sure, we are more than delighted to discuss this with you.
For log-normalization on timestamps, since the transformed variable $x_n$ has conditional pdf,
$
p(x_n|\mathcal F_{x_{n-1}})=\lambda \exp\left[-\lambda \exp(x_n)\right]\exp(x_n).
$
The conditional score and its partial derivative w.r.t. $x_n$ will be,
$
\psi(x_n|\mathcal F_{x_{n-1}}) = -\lambda \exp(x_n) + 1,
$
$
\frac{\partial}{\partial x_n}\psi_{\theta}(x_n|\mathcal F_{x_{n-1}})=-\lambda \exp(x_n).
$
Now we perform ASM on the transformed variable $x_n$. Denote $\mathcal X = (x_1,\ldots, x_{N_T})^T$ as the vector of transformed timestamps, the implicit ASM objective (Eq. 13) for $\mathcal X$ will be,
$
\mathcal J_\text{ASM}(\theta)=\mathbb E_{p(\mathcal X)}[\sum_{n=1}^{N_T}\frac{1}{2}\psi^2_\theta(x_n|\mathcal F_{x_{n-1}})+\frac{\partial}{\partial x_n}\psi_\theta(x_n|\mathcal F_{x_{n-1}})].
$
We plug in the conditional score we just computed into the above equation and get,
$
\mathcal J_{\text{ASM}}(\theta) =\mathbb E_{p(\mathcal X)}[\sum_{n=1}^{N_T}\frac{1}{2}\exp(2x_n)\lambda^2-\sum_{n=1}^{N_T}2\exp(x_n)\lambda ]+C,
$
where $C$ is a constant not containing $\theta$.
The empirical objective based on samples $t_1^{(m)},\ldots t_{N_m}^{(m)}, m=1,\ldots M$ will be,
$\hat J_\text{ASM}(\theta)=\frac{1}{M}\sum_{m=1}^M\sum_{n=1}^{N_m}[\frac{1}{2}\exp(2x_n^{(m)})\lambda^2-2\exp(x_n^{(m)})\lambda].
$
This is quadratic with respect to $\lambda$, and its minimal $\hat \lambda_{\text{ASM,Log}}$ can be computed analytically as
$
\hat \lambda_{\text{ASM,Log}} = 2\frac{\sum_{m=1}^M\sum_{n=1}^{N_m}\exp(x_n^{(m)})}{\sum_{m=1}^M\sum_{n=1}^{N_m}\exp(2x_n^{(m)})}.
$
For log transformation on time interval $y_n$, since the conditional pdf of $\tau_n = t_n-t_{n-1}$ is,
$
p(\tau_n|\mathcal F_{\tau_{n-1}})=\lambda \exp(-\lambda \tau_n),
$
thus the conditional pdf of $y_n = \log \tau_n$ will be,
$
p(y_n|\mathcal F_{y_{n-1}})=\lambda \exp\left[-\lambda \exp(y_n)\right]\exp(y_n),
$
it has the same analytical expression as that of $x_n$, except that $y_n$ and $x_n$ have different support. Therefore, following exactly the same steps as those of $x_n$, we can get the analytical expression of $\hat \lambda_{\text{ASM,Log, Interval}}$ in our rebuttal (with minor modification for adding superscript $(m)$).
> Additionally, is it possible to express $\hat \lambda_{\text{AWSM,Log}}$ and $\hat \lambda_{\text{AWSM, Log Interval}}$ in closed form?
Yes, it is possible. We take $\hat \lambda_{\text{AWSM, Log}}$ as an example. The derivation of $\hat \lambda_{\text{AWSM, Log Interval}}$ is basically the same.
First, the implicit AWSM objective as Eq. 17 in our paper for $\mathcal X$ will be,
$
\mathcal J_\text{AWSM}(\theta)=\mathbb E_{p(\mathcal X)}[\sum_{n=1}^{N_T}\frac{1}{2}\psi^2_\theta(x_n|\mathcal F_{x_{n-1}})h_n(\mathcal X)+\frac{\partial}{\partial x_n}\psi_\theta(x_n|\mathcal F_{x_{n-1}})h_n(\mathcal X)+\psi_\theta(x_n|\mathcal F_{x_{n-1}})\frac{\partial}{\partial x_n}h_n(\mathcal X)].
$
Plug in $\psi_\theta(x_n|\mathcal F_{x_{n-1}})$ and $\frac{\partial}{\partial x_n}\psi_\theta(x_n|\mathcal F_{x_{n-1}})$ into the empirical version of the above equation and get,
$
\hat J_\text{AWSM}(\theta)=\frac{1}{M}\sum_{m=1}^M\sum_{n=1}^{N_m}[\frac{1}{2}\exp(2x_n^{(m)})h_n(\mathcal X^{(m)})\lambda^2-2\exp(x_n^{(m)})h_n(\mathcal X^{(m)})\lambda - \exp(x_n^{(m)})\frac{\partial}{\partial x_n}h_n(\mathcal X^{(m)})\lambda].
$
This is still quadratic w.r.t $\lambda$, and its minimal will be,
$
\hat \lambda_{\text{AWSM, Log}}=\frac{\sum_{m=1}^M\sum_{n=1}^{N_m}[2\exp(x_n^{(m)})h_n(\mathcal X^{(m)})+\exp(x_n^{(m)})\frac{\partial}{\partial x_n}h_n(\mathcal X^{(m)})]}{\sum_{m=1}^M\sum_{n=1}^{N_m}\exp(2x_n^{(m)})h_n(\mathcal X^{(m)})}.
$
Since we choose weight $h_n(\mathcal X)=(x_n-x_{n-1})(\log T-x_n)$ with derivative $\frac{\partial}{\partial x_n}h_n(\mathcal X)=\log T - 2x_n+x_{n-1}$. Plug them in the above equation, then we get a closed form of $\hat \lambda_{\text{AWSM, Log}} $.