Gradients & Hessians

Inverse GaussianGradient VectorHessian MatrixRight-CensoringGaussianGradient VectorHessian MatrixRight CensoringLogNormalGradient VectorHessian MatrixRight Censoring

Inverse Gaussian

$$\mu(H_{u_{i}},\theta_t)=\theta_0+\sum_{j=1}^{p}\theta_ju_{i-j+1}$$

$$p(u_i|k,\theta_t) = \left[\frac{k}{2\pi\cdot{u_i}^3}\right]^{1/2}e^{-\frac{1}{2}\frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}}$$

$$\mathcal{L}(u_{t-n:t}|k,\mathbf{\eta}_t,\theta_t) = -\sum_{i=1}^{n}\eta_i\log p({u}_i|k,\theta_t)\\$$

Gradient Vector

$$\nabla\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t)= \begin{bmatrix} \frac{\part}{\part k}\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t)\\ \frac{\part}{\part \theta_0}\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t)\\ \vdots\\ \frac{\part}{\part \theta_{p}}\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t) \end{bmatrix}$$

For $k$ we have:

$$\frac{\part}{\part k}\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part k}\left(-\sum_{i=0}^{m-1}\eta_i\log p(u_i|k,\theta_t)\right) = \\ \frac{\part}{\part k} \left( - \sum_{i=0}^{m-1} \eta_i \log \left( \left[\frac{k}{2\pi\cdot{u_i}^3}\right]^{1/2}e^{-\frac{1}{2}\frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}} \right) \right)=\\ \frac{\part}{\part k} \left( \sum_{i=0}^{m-1} - \frac{1}{2} \eta_i \log \left( \left[\frac{k}{2\pi\cdot{u_i}^3}\right] \right) +\frac{1}{2} \eta_i \frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i} \right)=\\ \frac{\part}{\part k} \left( \sum_{i=0}^{m-1} - \frac{1}{2} \eta_i \log \left( k \right) +\frac{1}{2}\eta_i\log \left(2\pi\cdot{u_i}^3\right) +\frac{1}{2} \eta_i \frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i} \right)=\\ \color{blue} { \sum_{i=0}^{m-1} \frac{\eta_i}{2} \left(-\frac{1}{k}+\frac{\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}\right) }$$

For $\theta_j$ we have:

$$\frac{\part}{\part \theta_j}\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part \theta_j}\left(-\sum_{i=0}^{m-1}\eta_i\log p(u_i|k,\theta_t)\right) = \\ \frac{\part}{\part \theta_j} \left( - \sum_{i=0}^{m-1} \eta_i \log \left( \left[\frac{k}{2\pi\cdot{u_i}^3}\right]^{1/2}e^{-\frac{1}{2}\frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}} \right) \right)=\\ \frac{\part}{\part \theta_j} \left( \sum_{i=0}^{m-1} - \frac{1}{2} \eta_i \log \left( \left[\frac{k}{2\pi\cdot{u_i}^3}\right] \right) +\frac{1}{2} \eta_i \frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i} \right)=\\ \frac{\part}{\part \theta_j} \left( \sum_{i=0}^{m-1} - \frac{1}{2} \eta_i \log \left( k \right) +\frac{1}{2}\eta_i\log \left(2\pi\cdot{u_i}^3\right) +\frac{1}{2} \eta_i \frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i} \right)=\\ \frac{\part}{\part \theta_j} \left( \sum_{i=0}^{m-1} \frac{\eta_ik}{2u_i} \frac { \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2 } { \mu(H_{u_{i-1}},\theta_t)^2 } \right)=\\ \sum_{i=0}^{m-1} \frac{\eta_ik}{2u_i} \cdot \frac{ 2\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(-u_{i-j})\cdot\mu(H_{u_{i-1}},\theta_t)^2 - 2\mu(H_{u_{i-1}},\theta_t)(u_{i-j})\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2 } {\mu(H_{u_{i-1}},\theta_t)^4}=\\ \sum_{i=0}^{m-1} \frac{\eta_ik}{u_i} \cdot \frac{ -\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(u_{i-j})\cdot\mu(H_{u_{i-1}},\theta_t) - (u_{i-j})\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2 } {\mu(H_{u_{i-1}},\theta_t)^3}=\\ \sum_{i=0}^{m-1} \frac{\eta_ik}{u_i} \cdot \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(u_{i-j}) \left( -\mu(H_{u_{i-1}},\theta_t) -u_i +\mu(H_{u_{i-1}},\theta_t) \right) } {\mu(H_{u_{i-1}},\theta_t)^3}=\\ -\sum_{i=0}^{m-1} \frac{\eta_ik}{u_i} \cdot \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(u_{i-j}) \left( u_i \right) } {\mu(H_{u_{i-1}},\theta_t)^3}=\\ \color{blue} { -\sum_{i=0}^{m-1} {\eta_ik} \cdot \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(u_{i-j}) } {\mu(H_{u_{i-1}},\theta_t)^3} }$$

Since $\mu(H_{u_{i-1}},\theta_t)=\theta_0+\sum_{j=1}^{n-1}\theta_ju_{i-j}$ we note that in case $\color{black}{j=0}$, we must use $1$ instead of $u_{i-j} = u_i$

(Note: In code this case is handled by the fact the u[i] = 1)

Hessian Matrix

$$\nabla^2\mathcal{L}(u_{t-n:t}|k,\mathbf{\eta}_t,\theta_t)= \begin{bmatrix} \frac{\part^2}{\part^2 k}\mathcal{L}(\cdot) & \frac{\part}{\part k\part\theta_0}\mathcal{L}(\cdot) & \cdots & \frac{\part}{\part k\part\theta_{p}}\mathcal{L}(\cdot)\\ \\ \cdot & \frac{\part^2}{\part^2 \theta_0}\mathcal{L}(\cdot) & \cdots & \frac{\part}{\part \theta_0\theta_{p}}\mathcal{L}(\cdot) \\ \cdot & \cdot & \ddots & \vdots \\ \cdot & \cdot & \cdots & \frac{\part^2}{\part^2 \theta_{p}}\mathcal{L}(\cdot) \end{bmatrix}$$

$$\frac{\part^2}{\part^2 k}\mathcal{L}(u_{t-n:t}|k,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part k} \left( { \sum_{i=1}^{n} \frac{\eta_i}{2} \left(-\frac{1}{k}+\frac{\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}\right) } \right)=\\ \frac{\part}{\part k} \left( - \sum_{i=1}^{n} \frac{\eta_i}{2}k^{-1} \right)=\\ \color{blue}{ \sum_{i=1}^{n} \frac{\eta_i}{2}k^{-2} }$$

$$\frac{\part}{\part k\part\theta_{j}}\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part \theta_j} \left( { \sum_{i=1}^{n} \frac{\eta_i}{2} \left(-\frac{1}{k}+\frac{\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}\right) } \right)=\\ \frac{\part}{\part \theta_j} \left( { \sum_{i=1}^{n} \frac{\eta_i}{2u_i} \frac{\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2} } \right)=\\ \cdots\\ \color{blue} { -\sum_{i=1}^{n} {\eta_i} \cdot \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(u_{i-j}) } {\mu(H_{u_{i-1}},\theta_t)^3} }$$

$$\frac{\part}{\part k\part\theta_{j}}\mathcal{L}(u_{t-n:t}|k,\mathbf{\eta}_t,\theta_t) = \color{blue}{ \begin{cases} -\sum_{i=1}^{n} {\eta_i} \cdot \frac{ \left(u_i-\mu_i\right)(u_{i-j}) } {\mu_i^3} \ \ \ \ \ \ \ \ j\neq0\\ -\sum_{i=1}^{n} {\eta_i} \cdot \frac{ \left(u_i-\mu_i\right)(1) } {\mu_i^3} \ \ \ \ \ \ \ \ \ \ \ j=0 \end{cases}}$$

$$\frac{\part}{\part \theta_j\part k}\mathcal{L}(u_{t-m:t}|k,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part k} \left( -\sum_{i=1}^{n} {\eta_ik} \cdot \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(u_{i-j}) } {\mu(H_{u_{i-1}},\theta_t)^3} \right) =\\ \color{purple} { -\sum_{i=1}^{n} {\eta_i} \cdot \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)(u_{i-j}) } {\mu(H_{u_{i-1}},\theta_t)^3} }$$

We have that $\frac{\part}{\part \theta_j\part k}\mathcal{L}(\cdot) = \frac{\part}{\part k\part \theta_j}\mathcal{L}(\cdot)$ as expected.

$$\frac{\part}{\part \theta_j \part \theta_q}\mathcal{L}(u_{t-n:t}|k,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part \theta_q} \left( -\sum_{i=1}^{n}\eta_i{k}\cdot \left( \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)\cdot(u_{i-j}) } {\mu(H_{u_{i-1}},\theta_t)^3} \right) \right)=\\ -\sum_{i=1}^{n}\eta_i{k} (u_{i-j})\cdot \frac{\part}{\part \theta_q} \left( \frac{ \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)\ } {\mu(H_{u_{i-1}},\theta_t)^3} \right)=\\ -\sum_{i=1}^{n}\eta_i{k} (u_{i-j})\cdot \left( \frac{ (-u_{i-q})\mu(H_{u_{i-1}},\theta_t)^3 - 3\mu(H_{u_{i-1}},\theta_t)^2\left(u_{i-q}\right) \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)\ } {\mu(H_{u_{i-1}},\theta_t)^6} \right)=\\ -\sum_{i=1}^{n}\eta_i{k} (u_{i-j})\cdot \left( \frac{ (-u_{i-q})\mu(H_{u_{i-1}},\theta_t) - 3\left(u_{i-q}\right) \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)\ } {\mu(H_{u_{i-1}},\theta_t)^4} \right)=\\ \sum_{i=1}^{n}\eta_i{k} (u_{i-j})\cdot \left( \frac{ (u_{i-q})\mu(H_{u_{i-1}},\theta_t) + 3\left(u_{i-q}\right) \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)\ } {\mu(H_{u_{i-1}},\theta_t)^4} \right)=\\ \sum_{i=1}^{n}\eta_i{k} (u_{i-j}) \left(u_{i-q}\right) \cdot \left( \frac{ \mu(H_{u_{i-1}},\theta_t) + 3 \left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)\ } {\mu(H_{u_{i-1}},\theta_t)^4} \right)=\\ \color{blue} { \sum_{i=1}^{n}\eta_i{k}(u_{i-j})(u_{i-q})\cdot \left( \frac{ 3\cdot u_i - 2 \cdot \mu(H_{u_{i-1}},\theta_t) } {\mu(H_{u_{i-1}},\theta_t)^4} \right) }$$

$$\frac{\part}{\part \theta_j \part \theta_q}\mathcal{L}(u_{t-n:t}|k,\mathbf{\eta}_t,\theta_t) = \color{blue}{ \begin{cases} \sum_{i=1}^{n}\eta_i{k}(u_{i-j})(u_{i-q})\cdot \left( \frac{ 3\cdot u_i - 2 \cdot \mu_i } {\mu_i^4} \right) \ \ \ \ \ \ j\neq0,q\neq0 \\ \sum_{i=1}^{n}\eta_i{k}(1)(u_{i-q})\cdot \left( \frac{ 3\cdot u_i - 2 \cdot \mu_i } {\mu_i^4} \right) \ \ \ \ \ \ \ \ \ \ \ j=0,q\neq0 \\ \sum_{i=1}^{n}\eta_i{k}(u_{i-j})(1)\cdot \left( \frac{ 3\cdot u_i - 2 \cdot \mu_i } {\mu_i^4} \right) \ \ \ \ \ \ \ \ \ \ \ j\neq0,q=0 \\ \sum_{i=1}^{n}\eta_i{k}(1)(1)\cdot \left( \frac{ 3\cdot u_i - 2 \cdot \mu_i } {\mu_i^4} \right) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ j=0,q=0 \end{cases} }$$

 

Right-Censoring

Some mathematical trick to treat the pdf and cdf of the Inverse Gaussian:

$$p(u_i|k,\theta_t) = \\ \left[\frac{k}{2\pi\cdot{u_i}^3}\right]^{1/2}e^{-\frac{1}{2}\frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}}=\\ \text{exp}\left(\log\left(\left[\frac{k}{2\pi\cdot{u_i}^3}\right]^{1/2}e^{-\frac{1}{2}\frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}}\right)\right) = \\ \text{exp}\left(\log\left(\left[\frac{k}{2\pi\cdot{u_i}^3}\right]^{1/2}\right)+\log\left(e^{-\frac{1}{2}\frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i}}\right)\right) = \\ \text{exp}\left( \frac{1}{2}\log\left(\frac{k}{2\pi\cdot{u_i}^3}\right) - \frac{1}{2}\frac{k\left(u_i-\mu(H_{u_{i-1}},\theta_t)\right)^2}{\mu(H_{u_{i-1}},\theta_t)^2\cdot u_i})\right) = \\$$

$$P(u_i|k,\theta_t) )= \\ \Phi\left(\sqrt{\frac{k}{u_i}}\left(\frac{u_i}{\mu(H_{u_{i-1}},\theta_t)}-1\right)\right)+e^{\frac{2k}{\mu(H_{u_{i-1}},\theta_t)}}\Phi\left(-\sqrt{\frac{k}{u_i}}\left(\frac{u_i}{\mu(H_{u_{i-1}},\theta_t)}+1\right)\right) = \\ \Phi\left(\sqrt{\frac{k}{u_i}}\left(\frac{u_i}{\mu(H_{u_{i-1}},\theta_t)}-1\right)\right) + e^{\frac{2k}{\mu(H_{u_{i-1}},\theta_t)}+\log\left[\Phi\left(-\sqrt{\frac{k}{u_i}}\left(\frac{u_i}{\mu(H_{u_{i-1}},\theta_t)}+1\right)\right)\right]}$$

Where $\Phi$ is the standard normal distribution cdf.

---

The additional term we add to our cost function when we apply right censoring is:

$$- \color{green}{\eta_n\log\left(1-P(t-u_{n})\right)}$$

$$t_n = t-u_n\\ P(t) = \int_{0}^{t}p(v)dv\\ \mathcal{L}(t)=-\eta_n\log\left(1-P(t)\right)\\ \frac{\part}{\part k} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \frac{\part \left(-\eta_n\log\left(1-P(t_n)\right)\right)}{\part \left(1-P(t_n)\right)} \cdot \frac{\part \left(1-P(t_n)\right)}{\part P(t_n)} \cdot \frac{\part P(t_n)}{\part k}=\\ -\frac{\eta_n}{1-P(t_n)}\cdot(-1)\cdot\frac{\part P(t_n)}{\part k}=\\ \frac{\eta_n}{1-P(t_n)}\cdot\frac{\part P(t_n)}{\part k}$$

$$P(t) =Pr(X< t)= \Phi\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)+e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)$$

Where $\Phi$ is the standard normal distribution cdf.

$$\frac{\part}{\part k}P(t)=\\ \Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right) \color{blue}{\frac{\part}{\part k}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)} + \\ \color{green}{\frac{\part}{\part k}\left(\frac{2k}{\mu}\right)}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) + \\ \Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) \color{purple}{\frac{\part}{\part k}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)} e^{\frac{2k}{\mu}} \\= \\ \Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right) \color{blue}{\left(\frac{1}{2t\sqrt{\frac{k}{t}}}\left(\frac{t}{\mu}-1\right)\right)} +\\ \color{green}{\frac{2}{\mu}}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) + \\ \Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) \color{purple}{ \left(-\frac{1}{2t \sqrt{\frac{k}{t}}}\left(\frac{t}{\mu}+1\right)\right)} e^{\frac{2k}{\mu}}$$

Which can be rewritten as

$$\frac{\part}{\part k}P(t)= \\ \Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right) \color{blue}{\left(\frac{1}{2t\sqrt{\frac{k}{t}}}\left(\frac{t}{\mu}-1\right)\right)} +\\ \color{green}{\frac{2}{\mu}}e^{\frac{2k}{\mu}+\log\left(\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)\right)} + \\ \color{purple}{ \left(-\frac{1}{2t \sqrt{\frac{k}{t}}}\left(\frac{t}{\mu}+1\right)\right)} e^{\frac{2k}{\mu} + \log\left( \Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)\right)}$$

We prefer this formulation since computing exp(2k/mu) alone would be impracticable since the values for $k$ may be greater than $10^3$, moreover there exists valid approximations of exact formulations for the logcdf and logpdf of the normal distribution.

$$\frac{\part}{\part \mu}P(t)=\\ \Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right) \color{blue}{\frac{\part}{\part \mu}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right)} + \\ \color{green}{\frac{\part}{\part \mu}\left(\frac{2k}{\mu}\right)}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) + \\ \Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) \color{purple}{\frac{\part}{\part \mu}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)} e^{\frac{2k}{\mu}} \\ = \\ \Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right) \color{blue}{\left(-\sqrt{\frac{k}{t}}\cdot \frac{t}{\mu^2}\right)} +\\ \color{green}{\left(-\frac{2k}{\mu^2}\right)}e^{\frac{2k}{\mu}}\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) +\\ \Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right) \color{purple}{\left(\sqrt{\frac{k}{t}}\cdot\frac{t}{\mu^2}\right)} e^{\frac{2k}{\mu}}$$

Which can be rewritten as

$$\frac{\part}{\part \mu}P(t)= \\ \Phi^{'}\left(\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}-1\right)\right) \color{blue}{\left(-\sqrt{\frac{k}{t}}\cdot \frac{t}{\mu^2}\right)} +\\ \color{green}{\left(-\frac{2k}{\mu^2}\right)}e^{\frac{2k}{\mu}+\log\left(\Phi\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)\right)} +\\ \color{purple}{\left(\sqrt{\frac{k}{t}}\frac{t}{\mu^2}\right)} e^{\frac{2k}{\mu}+\log\left(\Phi^{'}\left(-\sqrt{\frac{k}{t}}\left(\frac{t}{\mu}+1\right)\right)\right)}$$

Then, we can add to the gradient the following terms

$$\frac{\part}{\part k} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \frac{\eta_n}{1-P(t_n)}\cdot\frac{\part P(t_n)}{\part k} \\ \frac{\part}{\part \theta_j} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \frac{\eta_n}{1-P(t_n)}\cdot\frac{\part P(t_n)}{\part \mu}\cdot\frac{\part \mu}{\part \theta_j}$$

(Remember that $\mu(H_{u_{i}},\theta_t)=\theta_0+\sum_{j=1}^{p}\theta_ju_{i-j+1}$)

Gaussian

$$\mu(H_{u_{i}},\theta_t)=\theta_0+\sum_{j=1}^{p}\theta_ju_{i-j+1}$$

$$p(u_i|\sigma,\theta_t) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}}$$

$$\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) = -\sum_{i=1}^{n}\eta_i\log p({u}_i|\sigma,\theta_t)\\$$

Gradient Vector

$$\nabla\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t)= \begin{bmatrix} \frac{\part}{\part \sigma}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t)\\ \frac{\part}{\part \theta_0}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t)\\ \vdots\\ \frac{\part}{\part \theta_{p}}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t) \end{bmatrix}$$

For $\sigma$ we have:

$$\frac{\part}{\part \sigma}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part \sigma}\left(-\sum_{i=0}^{m-1}\eta_i\log p(u_i|\sigma,\theta_t)\right) = \\ \frac{\part}{\part \sigma}\left(-\sum_{i=0}^{m-1}\eta_i\log \left(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i\left(\frac{\part}{\part\sigma}\log \left(\frac{1}{\sigma\sqrt{2\pi}}\right)+\frac{\part}{\part\sigma}\log\left(e^{-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i\left(\frac{\part}{\part\sigma}\log \left(\frac{1}{\sigma\sqrt{2\pi}}\right)+\frac{\part}{\part\sigma}\left(-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i \left(-\sigma\sqrt{2\pi}\frac{1}{2\pi\sigma^2} + \frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma^2}\right) = \\ \color{blue}{ -\sum_{i=0}^{m-1}\eta_i \left( -\frac{1}{\sigma} + \frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma^2} \right)}$$

For $\theta_j$ we have:

$$\frac{\part}{\part \theta_j}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part \theta_j}\left(-\sum_{i=0}^{m-1}\eta_i\log p(u_i|\sigma,\theta_t)\right) = \\ \frac{\part}{\part \theta_j}\left(-\sum_{i=0}^{m-1}\eta_i\log \left(\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i\left(\frac{\part}{\part\theta_j}\log\left(\frac{1}{\sigma\sqrt{2\pi}}\right)+\frac{\part}{\part\theta_j}\log\left(e^{-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i\left(\frac{\part}{\part\theta_j}\log\left(\frac{1}{\sigma\sqrt{2\pi}}\right)+\frac{\part}{\part\theta_j}\left(-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i\left(\frac{\part}{\part\theta_j}\left(-\frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i\left( -\frac{1}{2\sigma}\frac{\part}{\part\theta_j}\left(u_i-\mu_i\right)^2\right) = \\ \color{blue}{ -\sum_{i=0}^{m-1}\eta_i\left( \frac{1}{\sigma}\left(u_i-\mu_i\right)u_{i-j}\right)}$$

Since $\mu(H_{u_{i-1}},\theta_t)=\theta_0+\sum_{j=1}^{n-1}\theta_ju_{i-j}$ we note that in case $\color{black}{j=0}$, instead of $u_{i-j} = u_i$ we have to use $1$

Hessian Matrix

$$\nabla^2\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t)= \begin{bmatrix} \frac{\part^2}{\part^2 \sigma}\mathcal{L}(\cdot) & \frac{\part}{\part k\part\theta_0}\mathcal{L}(\cdot) & \cdots & \frac{\part}{\part \sigma\part\theta_{p}}\mathcal{L}(\cdot)\\ \\ \cdot & \frac{\part^2}{\part^2 \theta_0}\mathcal{L}(\cdot) & \cdots & \frac{\part}{\part \theta_0\theta_{p}}\mathcal{L}(\cdot) \\ \cdot & \cdot & \ddots & \vdots \\ \cdot & \cdot & \cdots & \frac{\part^2}{\part^2 \theta_{p}}\mathcal{L}(\cdot) \end{bmatrix}$$

$$\frac{\part^2}{\part^2 \sigma}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\sigma} \left( -\sum_{i=0}^{m-1}\eta_i \left(-\frac{1}{\sigma} + \frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma^2}\right) \right)=\\ -\sum_{i=0}^{m-1}\eta_i\left( \frac{\part}{\part\sigma} \left(-\frac{1}{\sigma}\right) + \frac{1}{2}\left(u_i-\mu_i\right)^2 \frac{\part}{\part\sigma}\frac{1}{\sigma^2} \right) =\\ \color{purple}{ -\sum_{i=0}^{m-1}\eta_i \left(\frac{1}{\sigma^2} - \left(u_i-\mu_i\right)^2 \frac{1}{\sigma^3} \right) }$$

$$\frac{\part^2}{\part\sigma\part\theta_j}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\theta_j} \left( -\sum_{i=0}^{m-1}\eta_i \left(-\frac{1}{\sigma} + \frac{1}{2}\frac{\left(u_i-\mu_i\right)^2}{\sigma^2}\right) \right)=\\ -\sum_{i=0}^{m-1}\eta_i\left( \frac{1}{2\sigma^2}\frac{\part}{\part\theta_j}\left(u_i-\mu_i\right)^2 \right) =\\ -\sum_{i=0}^{m-1}\eta_i\left( -\frac{1}{\sigma^2}\left(u_i-\mu_i\right)u_{i-j} \right) =\\ \color{purple}{ \sum_{i=0}^{m-1}\eta_i\left( \frac{1}{\sigma^2}\left(u_i-\mu_i\right)u_{i-j} \right) }$$

Consistency check:

$$\frac{\part^2}{\part\theta_j\part\sigma}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\sigma} \left( -\sum_{i=0}^{m-1}\eta_i\left( \frac{1}{\sigma}\left(u_i-\mu_i\right)u_{i-j}\right) \right) = \\ \color{purple}{ \sum_{i=0}^{m-1}\eta_i\left( \frac{1}{\sigma^2}\left(u_i-\mu_i\right)u_{i-j} \right) }$$

We have that $\frac{\part}{\part \theta_j\part \sigma}\mathcal{L}(\cdot) = \frac{\part}{\part \sigma\part \theta_j}\mathcal{L}(\cdot)$ as expected.

$$\frac{\part^2}{\part\theta_j\theta_k}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\theta_k} \left( -\sum_{i=0}^{m-1}\eta_i\left( \frac{1}{\sigma}\left(u_i-\mu_i\right)u_{i-j}\right) \right)= \\ \frac{\part}{\part\theta_k} \left( \sum_{i=0}^{m-1}\eta_i\left( \frac{1}{\sigma}u_{i-j}\mu_i\right) \right)= \\ \color{purple}{ \sum_{i=0}^{m-1}\eta_i\left( \frac{1}{\sigma}u_{i-j}u_{i-k} \right) }$$

 

Right Censoring

The additional term we add to our cost function when we apply right censoring is:

$$- \color{green}{\eta_n\log\left(1-P(t-u_{n})\right)}$$

$$t_n = t-u_n\\ P(t) = \int_{0}^{t}p(v)dv\\ \mathcal{L}(t)=-\eta_n\log\left(1-P(t)\right)$$

$$\frac{\part}{\part \sigma} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \frac{\part \left(-\eta_n\log\left(1-P(t_n)\right)\right)}{\part \left(1-P(t_n)\right)} \cdot \frac{\part \left(1-P(t_n)\right)}{\part P(t_n)} \cdot \frac{\part P(t_n)}{\part\sigma}=\\ -\frac{\eta_n}{1-P(t_n)}\cdot(-1)\cdot\frac{\part P(t_n)}{\part \sigma}=\\ \frac{\eta_n}{1-P(t_n)}\cdot\frac{\part P(t_n)}{\part \sigma}$$

$$\frac{\part}{\part \theta_j} \left(-\eta_n\log\left(1-P(t_n)\right)\right)= \frac{\eta_n}{1-P(t_n)}\cdot\frac{\part P(t_n)}{\part \theta_j}$$

Given that $\phi$ is the standard normal distribution pdf, we know that:

$$\frac{\part P(t_n)}{\part \sigma} = -\left(\frac{t_n-\mu}{\sigma^2}\right)\phi\left(\frac{t_n-\mu}{\sigma}\right)$$

$$\frac{\part P(t_n)}{\part \mu} = -\frac{1}{\sigma}\phi\left(\frac{t_n-\mu}{\sigma}\right)$$

Which yields:

$$\frac{\part}{\part \sigma} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \color{green}{\frac{-\eta_n}{1-P(t_n)}\left(\frac{t_n-\mu}{\sigma^2}\right)\phi\left(\frac{t_n-\mu}{\sigma}\right)}$$

$$\frac{\part}{\part \mu} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \color{green}{\frac{-\eta_n}{1-P(t_n)}\frac{1}{\sigma}\phi\left(\frac{t_n-\mu}{\sigma}\right)}$$

So

$$\frac{\part}{\part \theta_j} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \color{green}{\frac{-\eta_n}{1-P(t_n)}\frac{1}{\sigma}\phi\left(\frac{t_n-\mu}{\sigma}\right)u_{i-j}}$$

The hessian is approximated.

We can use the gaussian distribution parameters as a starting point for the Inverse Gaussian parameters.

We can use $\theta_{\textit{InverseGaussian}} = \theta_{\textit{Gaussian}}$ and $k_{\textit{InverseGaussian}} = \frac{\mu_{\text{mean}}^3}{\sigma^2}$

LogNormal

$$\mu(H_{u_{i}},\theta_t)= \theta_0+\sum_{j=1}^{p}\theta_ju_{i-j+1}$$

$$p(u_i|\sigma,\theta_t) = \frac{1}{u_i\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{\left(\log(u_i)-\log(\mu_i)\right)^2}{\sigma^2}}$$

$$\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) = -\sum_{i=1}^{n}\eta_i\log p({u}_i|\sigma,\theta_t)\\$$

Gradient Vector

$$\nabla\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t)= \begin{bmatrix} \frac{\part}{\part \sigma}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t)\\ \frac{\part}{\part \theta_0}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t)\\ \vdots\\ \frac{\part}{\part \theta_{p}}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t) \end{bmatrix}$$

For $\sigma$ we have:

$$\frac{\part}{\part \sigma}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part \sigma}\left(-\sum_{i=0}^{m-1}\eta_i\log p(u_i|\sigma,\theta_t)\right) = \\ \frac{\part}{\part \sigma} \left(-\sum_{i=0}^{m-1}\eta_i\log \left( \frac{1}{u_i\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{\left(\log(u_i)-\log(\mu_i)\right)^2}{\sigma^2}} \right) \right) = \\ -\sum_{i=0}^{m-1}\eta_i \left( \frac{\part}{\part\sigma} \log \left(\frac{1}{u_i\sigma\sqrt{2\pi}}\right) +\frac{\part}{\part\sigma}\log\left(e^{-\frac{1}{2}\frac{\left(\log\left(u_i\right)-\log(\mu_i)\right)^2}{\sigma^2}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i \left( \frac{\part}{\part\sigma} \log \left(\frac{1}{u_i\sigma\sqrt{2\pi}}\right) -\frac{\part}{\part\sigma}\left({\frac{1}{2}\frac{\left(\log\left(u_i\right)-\log(\mu_i)\right)^2}{\sigma^2}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i \left( -\frac{u_i\sigma\sqrt{2\pi}}{u_i\sigma^2\sqrt{2\pi}} -\frac{\left(\log\left(u_i\right)-\log(\mu_i)\right)^2}{2}\frac{\part}{\part\sigma}\left(\frac{1}{\sigma^2}\right)\right) = \\ \color{blue}{ -\sum_{i=0}^{m-1}\eta_i \left( -\frac{1}{\sigma} +\left(\log\left(u_i\right) - \log(\mu_i)\right)^2\left(\frac{1}{\sigma^3}\right)\right) }$$

For $\theta_j$ we have:

$$\frac{\part}{\part \theta_j}\mathcal{L}(u_{t-m:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part \theta_j}\left(-\sum_{i=0}^{m-1}\eta_i\log p(u_i|\sigma,\theta_t)\right) = \\ \frac{\part}{\part \theta_j} \left(-\sum_{i=0}^{m-1}\eta_i\log \left( \frac{1}{u_i\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\frac{\left(\log(u_i)-\log(\mu_i)\right)^2}{\sigma^2}} \right) \right) = \\ -\sum_{i=0}^{m-1}\eta_i \left( \frac{\part}{\part\theta_j} \log \left(\frac{1}{u_i\sigma\sqrt{2\pi}}\right) +\frac{\part}{\part\theta_j}\log\left(e^{-\frac{1}{2}\frac{\left(\log\left(u_i\right)-\log(\mu_i)\right)^2}{\sigma^2}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i \left( \frac{\part}{\part\theta_j} \log \left(\frac{1}{u_i\sigma\sqrt{2\pi}}\right) -\frac{\part}{\part\theta_j}\left({\frac{1}{2}\frac{\left(\log\left(u_i\right)-\log(\mu_i)\right)^2}{\sigma^2}}\right)\right) = \\ -\sum_{i=0}^{m-1}\eta_i \left( -\frac{1}{2\sigma^2}\frac{\part}{\part\theta_j}\left(\left(\log\left(u_i\right)-\log(\mu_i)\right)^2\right)\right) = \\ \color{blue}{ -\sum_{i=0}^{m-1}\eta_i \left( \frac{1}{\sigma^2}\left(\log\left(u_i\right)-\log(\mu_i)\right)\frac{1}{\mu_i}u_{i-j}\right) }$$

Since $\mu(H_{u_{i-1}},\theta_t)=\theta_0+\sum_{j=1}^{n-1}\theta_ju_{i-j}$ we note that in case $\color{black}{j=0}$, instead of $u_{i-j} = u_i$ we have to use $1$

Hessian Matrix

$$\nabla^2\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t)= \begin{bmatrix} \frac{\part^2}{\part^2 \sigma}\mathcal{L}(\cdot) & \frac{\part}{\part k\part\theta_0}\mathcal{L}(\cdot) & \cdots & \frac{\part}{\part \sigma\part\theta_{p}}\mathcal{L}(\cdot)\\ \\ \cdot & \frac{\part^2}{\part^2 \theta_0}\mathcal{L}(\cdot) & \cdots & \frac{\part}{\part \theta_0\theta_{p}}\mathcal{L}(\cdot) \\ \cdot & \cdot & \ddots & \vdots \\ \cdot & \cdot & \cdots & \frac{\part^2}{\part^2 \theta_{p}}\mathcal{L}(\cdot) \end{bmatrix}$$

$$\frac{\part^2}{\part^2 \sigma}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\sigma} \left( -\sum_{i=0}^{m-1}\eta_i \left( -\frac{1}{\sigma} +\left(\log\left(u_i\right) - \log(\mu_i)\right)^2\left(\frac{1}{\sigma^3}\right)\right) \right)=\\ \color{purple}{ -\sum_{i=0}^{m-1}\eta_i \left( \frac{1}{\sigma^2} -\frac{3\left(\log\left(u_i\right) - \log(\mu_i)\right)^2}{\sigma^4}\right) }$$

$$\frac{\part^2}{\part\sigma\part\theta_j}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\theta_j} \left( -\sum_{i=0}^{m-1}\eta_i \left( -\frac{1}{\sigma} +\left(\log\left(u_i\right) - \log(\mu_i)\right)^2\left(\frac{1}{\sigma^3}\right)\right) \right)=\\ \color{purple}{ \sum_{i=0}^{m-1}\eta_i \left( \frac{2\left(\log\left(u_i\right) - \log(\mu_i)\right)}{\sigma^3}\frac{1}{\mu_i}u_{i-j}\right)}$$

Consistency check:

$$\frac{\part^2}{\part\theta_j\part\sigma}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\sigma} \left( -\sum_{i=0}^{m-1}\eta_i \left( \frac{1}{\sigma^2}\left(\log\left(u_i\right)-\log(\mu_i)\right)\frac{\part\mu_i}{\part\theta_j}\right) \right) = \\ \color{purple}{ \sum_{i=0}^{m-1}\eta_i \left( \frac{2\left(\log\left(u_i\right) - \log(\mu_i)\right)}{\sigma^3}\frac{1}{\mu_i}u_{i-j}\right)}$$

We have that $\frac{\part}{\part \theta_j\part \sigma}\mathcal{L}(\cdot) = \frac{\part}{\part \sigma\part \theta_j}\mathcal{L}(\cdot)$ as expected.

$$\frac{\part}{\part\theta_j\theta_k}\mathcal{L}(u_{t-n:t}|\sigma,\mathbf{\eta}_t,\theta_t) =\\ \frac{\part}{\part\mu} \left( -\sum_{i=0}^{m-1}\eta_i \left( \frac{1}{\sigma^2}\left(\log\left(u_i\right)-\log(\mu_i)\right)\frac{1}{\mu_i}u_{i-j}\right) \right) \frac{\part \mu}{\part \theta_k} = \\ -\sum_{i=0}^{m-1} \eta_i \frac{u_{i-j}}{\sigma^2} \frac{\part }{\part \mu} \left( \frac{\log(u_i)-\log(\mu_i)}{\mu_i} \right) \frac{\part \mu}{\part \theta_k} = \\ -\sum_{i=0}^{m-1} \eta_i \frac{u_{i-j}}{\sigma^2} \left( \frac{ - \frac{1}{\mu_i}\mu_i - (\log(u_i) -\log(\mu_i)) }{\mu_i^2} \right) \frac{\part \mu}{\part \theta_k} = \\ \color{purple}{ \sum_{i=0}^{m-1} \eta_i \left( \frac{ 1 + (\log(u_i) -\log(\mu_i)) }{\mu_i^2\sigma^2} \right) u_{i-j} u_{i-k} }$$

 

 

Right Censoring

The additional term we add to our cost function when we apply right censoring is:

$$- \color{green}{\eta_n\log\left(1-P(t-u_{n})\right)}$$

$$t_n = t-u_n\\ P(t) = \int_{0}^{t}p(v)dv\\ \mathcal{L}(t)=-\eta_n\log\left(1-P(t)\right)$$

$$\frac{\part}{\part \sigma} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \frac{\part \left(-\eta_n\log\left(1-P(t_n)\right)\right)}{\part \left(1-P(t_n)\right)} \cdot \frac{\part \left(1-P(t_n)\right)}{\part P(t_n)} \cdot \frac{\part P(t_n)}{\part\sigma}=\\ -\frac{\eta_n}{1-P(t_n)}\cdot(-1)\cdot\frac{\part P(t_n)}{\part \sigma}=\\ \frac{\eta_n}{1-P(t_n)}\cdot\frac{\part P(t_n)}{\part \sigma}$$

$$\frac{\part}{\part \theta_j} \left(-\eta_n\log\left(1-P(t_n)\right)\right)= \frac{\eta_n}{1-P(t_n)}\cdot\frac{\part P(t_n)}{\part \theta_j}$$

Given that $\phi$ is the standard normal distribution pdf, we know that:

$$\frac{\part P(t_n)}{\part \sigma} = \phi\left(\frac{\log t_n-\log(\mu)}{\sigma}\right)\frac{\part}{\part \sigma}\left(\frac{\log t_n-\log(\mu)}{\sigma}\right) = \\ -\phi\left(\frac{\log t_n-\log(\mu)}{\sigma}\right)\left(\frac{\log t_n-\log(\mu)}{\sigma^2}\right)$$

$$\frac{\part P(t_n)}{\part \mu} = \phi\left(\frac{\log t_n-\log(\mu)}{\sigma}\right)\frac{\part}{\part \mu}\left(\frac{\log t_n-\log(\mu)}{\sigma}\right) = \\ -\phi\left(\frac{\log t_n-\log(\mu)}{\sigma}\right)\left(\frac{1}{\mu\sigma}\right)$$

Which yields:

$$\frac{\part}{\part \sigma} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \color{green}{ -\frac{\eta_n}{1-P(t_n)} \phi\left(\frac{\log t_n-\log(\mu)}{\sigma}\right)\left(\frac{\log t_n-\log(\mu)}{\sigma^2}\right)}$$

$$\frac{\part}{\part \mu} \left(-\eta_n\log\left(1-P(t_n)\right)\right) = \color{green}{ -\frac{\eta_n}{1-P(t_n)} \phi\left(\frac{\log t_n-\log(\mu)}{\sigma}\right)\left(\frac{1}{\mu\sigma}\right) }$$

The hessian is approximated.