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If I'm calculating a consistent rate, it's easy to model that as exponential decay. For example, if I lose 10% of customers each year, starting with A customers.

$$ y = A 0.9^x $$

However, what if that churn is less with each year. Rather than $ A * 0.9 * 0.9 * 0.9 ... $ I need $ A * 0.9 * 0.88 * 0.86 ... $ or something along those lines.

Is there a standard way to model something like that?

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You can create a function like that, e.g., $$y=A*(0.9-0.02x)^x \ ,$$ for $x=1,2,\ldots$. But it leaves 2 questions:

  • Does this equation makes sense in terms of your model? Is there data available that fits in this model? And, in this specific case, what should we do for $x>45$?

  • Is the expression in the equation both accurate in therms of modelling but still easy enough so that you can obtain some other properties of the model that you may be interested in?

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Adding on to Ertxiem's answer here, maybe the rate of decrease isn't necessarily linear. Then the base would no longer be $r-kx$ raised to some exponent $cx$. We can derive this by assuming a few starting points: $$f(0)=1$$ $$f(1)=0.9$$ $$f(2)=0.9\cdot(0.9*0.98)=0.9\cdot0.882=0.9^2\cdot0.98$$ $$f(3)=0.9\cdot(0.9*0.98)\cdot(0.9*0.98^2)=0.9\cdot0.882\cdot0.86436=0.9^3\cdot0.98^3$$ We can see a pattern emerging in the exponents if we keep going. The exponent for $0.9$ will be increasing by 1, but the exponent for $0.98$ will actually be the triangular numbers! Essentially $f(x) = (0.9)^x\cdot (0.98)^{x(x-1)/2}$

Just some more food for thought.

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    $\begingroup$ Nice equation, it has no problem for $x>45$ like mine. Another way to look at it is: $f(x)=\left(0.9 \cdot 0.98^{(x-1)/2}\right)^x$, that may help to understand the initial 0.9 part and the new part with the 0.98. $\endgroup$ – Ertxiem Mar 29 at 0:56

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