5

Hoping you can help me understand the probability density function for the exponential distribution.

Given that the distribution's PDF is described as follows when x > 0:

$$\lambda e^{-\lambda x}$$

Why do we multiply by λ? Sure I'm missing something basic here, but seems that if we're modeling time to event, using e and the exponents would be enough. Is it because we might expect the event to happen multiple times in the time interval?

Thank you in advance.

T_T
  • 153
  • 2

1 Answers1

5

So that when you integrate the pdf between 0 and infinity the probability density sums to 1 as required under the axioms of probability.

user135784
  • 266
  • 1
  • 3
  • 11
  • Thank you -- definitely understand that this works, but feel I'm still missing a basic understanding of *why* this particular way of manipulating the formula makes sense. Have followed proofs through the CDF to the PDF and still feel I'm missing that intuitive understanding. – T_T Apr 21 '14 at 11:59
  • 6
    A PDF by definition has to satisfy that its integral over all possible x is 1. The intuition is that of all possible events, something has to happen. Now, when you integrate e^-lambda (without the lambda multiplied beforehand) over 0 to infinity (which are all possible outcomes), you will easily find that it evaluates to 1/lambda. If you then multiply by lambda, you get the desired area (=probability) of 1. – Julian Schuessler Apr 21 '14 at 12:05
  • Yes Julian is correct. You also may want to look at this question if you want a deeper understanding of where this distribution comes from: http://stats.stackexchange.com/questions/2092/relationship-between-poisson-and-exponential-distribution :) – user135784 Apr 21 '14 at 12:12
  • 1
    Aha, got it! Was missing the integral = 1/lambda. Thank you both. – T_T Apr 21 '14 at 13:31