Decomposition

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

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

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 $t_{mi} -\Delta t_i \leq t_i \leq t_{mi} + \Delta t_i$:

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

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,

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

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

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

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:

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

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