# Correct interpretation of exponential decay and decay factor

my question is about exponential decay and its factor.

English isn't my native language and therefore I'm not sure about the precise definition in my particular case.

I'm reading a specific paper and here it is described, not so well, an algorithm. The part I'm not sure about is as follow: I have a x variable of value 1e-7, this algorithm has a loop and it is said that after every 10 iterations it is applied to x "an exponential decay with decay factor 0.95".

Do you think the correct interpretation would be to multiply the actual x value for $$e^{-0.95}$$ at each step that the update is required? As wikipedia states, the $$\lambda$$ is called constant and not factor

Another option could be for me to multiply by 0.95 and not $$e^{-0.95}$$

I'm sorry if my question is dumb but I can't verify the answer with brute force and I think this is the best place to find the most accurate one

If you have an exponential decay of $$5\%$$ after 10 steps then the equation is

$$e^{-\lambda x}=e^{-\lambda 10}=0.95\Rightarrow \lambda=\frac{-\ln(0.95)}{10}\approx 0.00512933$$

If we have an inital value of $$5$$ then after 10 steps we have $$y(10)=5\cdot e^{-0.00512933\cdot 10}=4.75=5\cdot 0.95$$

This is what we wanted. For a reference see here.

Here is a graph with the initial value $$5$$. You see that the decay is not linear.

• I don't need to update it at each step i, but only when $i=11,22,33...$, therefore I would consider $x=1$ when $i=11$, $x=2$ when $i=22$ or simply $x=\frac{i}{11}$. Commented Aug 23, 2019 at 17:57
• Have you mentioned that in your question? Commented Aug 23, 2019 at 17:58
• Hum, I reported the exact same statement as in the paper and this is my interpretation of "after every 10 iterations it is applied to x "an exponential decay with decay factor 0.95"" Commented Aug 23, 2019 at 18:01
• You start at t=0 and then after ten steps ($t=10$) it is updated. Commented Aug 23, 2019 at 18:02
• Yes, if I'm correct you are telling me that "to apply to x an exponential decay with decay factor 0.95" correspond to a simple $x=x*0.95$ to use only when the step is 11,21,31 (I was wrong when I said 11,22,33).. (or 10,20,30 if you start to count at 0) Commented Aug 23, 2019 at 18:09