# Model churn when churn rate diminishes over time

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?

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 $$x>45$$?
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 $$f(x) = (0.9)^x\cdot (0.98)^{x(x-1)/2}$$
• 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. – Ertxiem Mar 29 at 0:56