Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$ \frac{\int_{t+h}^{\infty} \lambda e^{-\lambda x} dx}{\int_{t}^{\infty} \lambda e^{-\lambda x} dx} = \frac{\int_{h}^{\infty} \lambda e^{-\lambda x} dx}{\int_{0}^{\infty} \lambda e^{-\lambda x} dx} $$ This is known as the memoryless property of the exponential distribution. From the plots we can see the shape of the curve looks the same wherever you start plotting. Is there a concept describing this?

enter image description here

enter image description here

share|cite|improve this question
I have seen it called scale invariance. – Ross Millikan Dec 4 '12 at 14:44
Thanks. How is it exactly defined, and how do you judge whether a curve is scale-invariant? – qed Dec 4 '12 at 14:49
It is defined precisely by being an exponential or geometric distribution. The motivation is just what you have seen-the curve looks the same at all scales. – Ross Millikan Dec 4 '12 at 15:53

$$ \frac{\int_{t+h}^{\infty} e^{-\lambda x} dx}{\int_{t}^{\infty} e^{-\lambda x} dx} =\frac{[-\frac{1}{\lambda}e^{-\lambda x}]_{t+h}^{\infty}}{[-\frac{1}{\lambda}e^{-\lambda x}]_{t}^{\infty}} =\frac{e^{-\lambda (t+h)}}{e^{-\lambda t}} =e^{-\lambda h} = \frac{\int_{h}^{\infty} e^{-\lambda x} dx}{\int_{0}^{\infty} e^{-\lambda x} dx} $$

share|cite|improve this answer
The question was asking about a term for this concept, not a proof of this equation. So how is this here an answer? – MvG Dec 7 '12 at 10:28
it demonstrates the scale invariance... – draks ... Dec 7 '12 at 10:48
@MartinSleziak thanks for pointing that out. I added the link to the tag. Sorry for the inconvience. – draks ... Dec 10 '12 at 13:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.