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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

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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.

enter image description here

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  • $\begingroup$ 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 $x=\frac{i}{11}$. $\endgroup$
    – Korr4K
    Commented Aug 23, 2019 at 17:57
  • $\begingroup$ Have you mentioned that in your question? $\endgroup$
    – callculus42
    Commented Aug 23, 2019 at 17:58
  • $\begingroup$ 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"" $\endgroup$
    – Korr4K
    Commented Aug 23, 2019 at 18:01
  • $\begingroup$ You start at t=0 and then after ten steps ($t=10$) it is updated. $\endgroup$
    – callculus42
    Commented Aug 23, 2019 at 18:02
  • $\begingroup$ 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) $\endgroup$
    – Korr4K
    Commented Aug 23, 2019 at 18:09

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