Resampling

Let's consider the input spectrum $R$. We assume that the intensity $T_R$ at channel $i$ is a random variable which follows the normal distribution $N(\mu_R,{\sigma_R}^2)$. Its probability density function is:

$$\ensuremath{\mathrm{pdf}}_R(x) = \frac{1}{\sigma_R \sqrt{2 ... ...thrm{exp}} \left( - \frac{(x-\mu_R)^2}{2 {\sigma_R}^2} \right)$$ (8)

where $\mu_R$ is the mean and $\sigma_R$ the standard deviation. The variance of such a distribution is:

$$\ensuremath{\mathrm{var}}(T_R(i)) = {\sigma_R}(i)^2$$ (9)

Choosing a weight $w_R(i)$ of the following form:

$$w_R(i) = \frac{1}{{\sigma_R}(i)^2}$$ (10)

ensure that the weighted mean is the maximum likelihood estimator of the mean, under the assumption of independent (i.e. uncorrelated at first order) and normally distributed channel intensities with the same mean.

Let's call $R'$ the resampled version of $R$ such as it is aligned with the desired output spectrum $S$. The intensity of $R'$ at channel $i$ can be written (see fig. ):

where:

• $j_{\ensuremath{\mathrm{min}}}$ and $j_{\ensuremath{\mathrm{max}}}$ are the channel range of input spectrum $R$ which are overlapped at least partially by the output channel number $i$,
• $f(j)$ is the used fraction of the different input channels. It quantifies their contribution in the selected range:
  

  $f(j) = 0$	if the input channel is out of the range,
  $f(j) = 1$	if the output channel fully overlaps the input channel,
  $0 < f(j) < 1$	otherwise (proportionally to the overlap).

We can also define the normalization factor $\beta$ and the weight $\alpha(j)$:

$$\beta = \sum_{j=j_{\ensuremath{\mathrm{min}}}}^{j_{\ensuremath{\mathrm{max}}}} f_R(j) \times w_R(j)$$ (11)

$$\alpha_R(j) = \frac{f_R(j) \times w_R(j)}{\beta}$$ (12)

The normalization factor $\beta$ ensures that the total integrated intensity is preserved during the resampling process. With these, $T_{R'}(i)$ can be written:

$$T_{R'}(i) = \sum_{j=j_{\ensuremath{\mathrm{min}}}}^{j_{\ensuremath{\mathrm{max}}}} \alpha_R(j) \times T_R(j)$$ (13)