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One can write a logistic function in the following form:

$y(t) = \frac{K}{\left(1 + Q \exp(-bt)\right)^{1/v}}$

where $v>0$ is considered a parameter, and $b$ is called the growth rate parameter. I am slightly confused as to the nature of this growth rate parameter, as this equation implies that it is a constant. But, when I read the phrase "growth rate", I read it as $dy/dt$, which is not just $b$. Can someone please explain the nature of $b$?

Further, say you had a bunch of y-values, say $(100,300,350,400,450,600,690,691,692)$. Could you estimate what $r$ was by looking at these values, assuming that $t=(1,2,3,4,5,6,7,8)$?

Thanks.

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  • $\begingroup$ When your model is just exponential growth you can put your data into Excel and ask it for the best exponential fit, and get values for $c$ and $r$. That's essentially linear regression on the logarithm. If it's for a logistic model, I'm sure there are analogous procedures to find the best constants. Now that the vocabulary is clearer I suggest you post the the technical part of the question at stats.stackexchange.com. I note that your model has four parameters, and doesn't look to me like the usual logistic equation. But statisticians will know. I see that you've asked there before. $\endgroup$ – Ethan Bolker Jun 9 '18 at 14:14
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For ordinary exponential growth governed by $y(t) = c e^{rt}$ we call $r$ the growth rate even though $dy/dt = ry$ is not constant. In this case as in yours the growth rate is essentially a proportion. Its units are $1/\text{time}$.

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  • $\begingroup$ Hi. Thanks for your answer. So, if I had a bunch of y-values, say $(100, 300, 350, 400, 450, 600, 690, 691, 692)$. Could you estimate what $r$ was by looking at these values, assuming that $t = (1,2,3,4,5,6,7,8)$? $\endgroup$ – Thomas Moore Jun 9 '18 at 14:06
  • $\begingroup$ See my comment on your edited question. $\endgroup$ – Ethan Bolker Jun 9 '18 at 14:15

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