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Decomposition

Now consider the set of $ n$ logistic components $ C = \{c_1, \ldots,c_n\}$, where

$\displaystyle c_i = \bold{N_i}(x_i,\bold P_i) $

where $ x_i$ is an arbitrary subspace of $ t$. (The components $ c_i$ are analogous to $ N_i(t)$ as discussed in equation (7).) A typical choice for $ x_i$ is the interval over which $ c_i$ grows from 10% to 90% of its saturation level, namely $ (t_{mi} -\Delta t_i, t_{mi} + \Delta t_i)$.

Restricting the domain of $ c_i$ gives us a criterion for decomposing the associated data set $ D$ into subsets $ D_j$ ( $ 1 \leq j \leq n$) corresponding to each component logistic $ c_i$. A subset $ D_j$ contains a point $ (t_i,d_i)$ if $ t_{mi} -\Delta t_i \leq t_i \leq t_{mi} + \Delta t_i$:

$\displaystyle D_j = \{(t_i,d^\ast_i)\;\vert\; t_i \in x_i\}. $

where $ d^\ast_i$ is the adjusted value of $ d_i$. The adjustment is subtracting out the ``effects'' from other components, leaving us with the (approximate) contribution of component $ c_i$ to this data point. In other words,

$\displaystyle d^\ast_i = d_i - \sum_{j \neq i} \frac{\bold P_{j2}}{1 + \text{exp} \left[ {-\frac{\ln(81)}{\bold P_{j1}}}(t - \bold P_{j3}) \right] } $

In Figure 5C, the hypothetical data set plotted in Figure 5A is decomposed into subsets $ D_1$ (circles) and $ D_2$ (crosses); similarly, the fitted curve is also decomposed into its component logistics. Note that the subsets $ D_j$ are not necessarily mutually exclusive in the $ t$ domain; in this example, $ D_1$ and $ D_2$ share points with common $ t$-values around $ t = 40$. At these times, we can see that there are two concurrent growth processes; in addition, we can also quantify how much of the growth can be attributed to each process.

As we saw in Figure 3, we can apply the Fisher-Pry transform to each component and its corresponding data subset. This is useful because it normalizes each component on a semi-logarithmic scale, allowing for easy comparison when plotted on the same graph. Moreover, it allows the fitting of logistics using linear least squares. The Fisher-Pry transform of components $ \bold c_i$ is

$\displaystyle FP(c_i) = \frac{\displaystyle \frac {\bold c_i}{\kappa_i}}{1 - \displaystyle\frac{\bold c_i}{\kappa_i}} $

where plotting $ \bold x_i$ vs. $ \log(FP(c_i)$ produces a straight line. The component data subsets $ D_j$ are transformed in a similar manner:

$\displaystyle FP(d^\ast_i) = \frac{\displaystyle\frac{d^\ast_i}{\kappa_i}} {1 - \displaystyle\frac{d^ast_i}{\kappa_i}} $

Figure 5D shows the Fisher-Pry decomposition of the hypothetical data.

next up previous contents
Next: Growth rates and the Up: The Mathematics of Loglet Previous: Definitions and notation   Contents
Jason Yung 2004-01-28