Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Jensen's inequality applied to geometric growth

  1. Jul 12, 2012 #1
    When population growth is geometric (e.g. Nt/N0 = λ^t, where t = time in years, lambda is annual finite rate of increase, Nt is population size at time t and N0 is initial population size), population growth is an accelerating function of lambda (e.g. f(λ) (i.e. growth) = λ^t, such that for a given t, larger lambda gives greater than an additive increase growth). According to Jensen's inequality, because the relationship between lambda and total population growth is concave up, the average of the function with variable lambda across years should be greater than the function of the mean lambda across years. This means that growth should be enhanced by variability in lambda.


    Given an average growth rate of 1.1, and t = 2.

    The first population has constant lambda equal to 1.1 in both years.

    growth = 1.1 * 1.1 = 1.1^2 = 1.21

    The second population has variable lambda but still with mean equal to 1.1. According to my reasoning above, average population growth should be increased as a consequence of variability, but:

    growth = 1.0 * 1.2 = 1.2.

    1.2 < 1.21, so growth is reduced.

    a more extreme example:

    growth = 0.6 * 1.6 = 0.96. More variability reduces the growth rate even further.

    Any idea where I am going wrong?
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Threads - Jensen's inequality applied Date
Inequality of functions Jan 15, 2017
Proof sin x < x for all x>0. Oct 11, 2015
Help proving triangle inequality for metric spaces Sep 12, 2015
Jensen's inequality Dec 12, 2008
Using Jensen's Inequality Apr 4, 2006