The course consists of an in-depth study of multivariate statistical methods (in particular, the binary logistic regression) and the impact on these methods of problems encourtered on real data as well as the possible solutions. The course will focus on the following themes
The Newton-Raphson method is a numerical technique to solve the equation $f(x)=0$. It is an iterative procedure illustrated in Fig. 7.1. We assume that the first approximate solution is $x_{1}$, where the function is $f\left(x_{1}\right)$, and the first derivative $f^{\prime}\left(x_{1}\right)$. Let the zero crossing of $f(x)$ be at $x_{1}+h$, then $f\left(x_{1}+h\right)=0$. From Fig. $7.1$ we see that if $h$ is small
$$
\begin{array}{r}
f\left(x_{1}+h\right) \approx f\left(x_{1}\right)+f^{\prime}\left(x_{1}\right) h \
\therefore f\left(x_{1}\right)+f^{\prime}\left(x_{1}\right) h \approx 0 \
\therefore h \approx-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}
\end{array}
$$
This allows us to write down a second approximation to the zero crossing:
$$
x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}
$$
The process can be repeated to obtain successively closer approximations. If after $s$ iterations the approximate solution is $x_{s}$ then the next iteration is $x_{s+1}$, and these quantities are related via the relation:
$$
x_{s+1}=x_{s}-\frac{f\left(x_{s}\right)}{f^{\prime}\left(x_{s}\right)} .
$$
STAT0007 COURSE NOTES ：
In the previous section we saw that in the final stages of a MarquardtLevenberg fitting procedure the curvature matrix is evaluated at the best-fit parameters. When fitting a straight line to a data set the two-dimensional error surface has the same curvature with respect to the slope and intercept for any values of these two parameters. Hence, in this special case, the curvature matrix can be calculated quite simply, without even having to perform the leastsquares fit.
For the special case of fitting to $y=m x+c$ the four elements of the curvature matrix are given by (see Exercise (7.7) for a derivation)
$$
\begin{aligned}
A_{c c} &=\sum_{i} \frac{1}{\alpha_{i}^{2}}, \
A_{c m}=A_{m c} &=\sum_{i} \frac{x_{i}}{\alpha_{i}^{2}}, \
A_{m m} &=\sum_{i} \frac{x_{i}^{2}}{\alpha_{i}^{2}} .
\end{aligned}
$$
The error matrix can thus be found simply by inverting this $2 \times 2$ matrix. $^{6}$