4 ms·

It seems like the Lindy effect for a positive real number valued ramdom variable X, is the following identity E[X | X >= a] = 2 * a

• nonrandomstring 3y ago
  It's more likely described by the non-linear portion corresponding to the first stage of the Bathtub Curve [0] we use in engineering. [0] https://en.wikipedia.org/wiki/Bathtub_curve https://en.wikipedia.org/wiki/Bathtub_curve Most complex but faulted things fail very early.
• sieste 3y ago
  > E[X | X >= a] = 2 * a Is it possible to derive the maximum entropy probability distribution subject to that constraint?
  • golol 3y ago
    Let f: R -> R be the probability density function of X. It fulfills int f(x) dx = 1 and we furthermore assume that f is differentiable. We can write the identity as int_a^oo x f(x) dx / int_a^oo f(x) dx = 2 a. We rewrite this as int_a^oo x f(x) dx = 2 a int_a^oo f(x) dx. Taking the derivative with respect to a on both sides yields - a f(a) = 2 int_a^oo f(x) dx + 2 a (- f(a)). Taking another derivative yields - f(a) - a f'(a) = - 2 f(a) - 2 f(a) - 2 a f'(a). We rewrite this as f'(a) = - 3 f(a) / a, and furthermore f'(a) / f(a) = - 3 / a. We can write this as (ln ○ f)'(a) = - 3 / a. As far as I can tell, we can not find a solution on all of [0, oo). Let's assume that failure is impossible up to an early time t. This means that we assume f(x) = 0 for all x < t, and that possibly f(t) > 0, meaning we have a jump discontinuity in the probability density at x = t. We integrate from t to x int_t^x (ln ○ f)'(a) da = int_t^x - 3 / a da. This evaluates to ln(f(x)) = ln(f(t)) - int_t^x 3 / a da = ln(f(t)) - 3 ln(x) + 3 ln(t) = ln(f(t)) - ln(x^3) + ln(t^3) = ln(f(t) t^3 / x^3). In particular f(x) = f(t) t^3 / x^3. Right now f(t) is still a free parameter, but recall that 1 = int f(x) dx = int_t^oo f(t) t^3 / x^3 dx = f(t) t^3 [- 1 / (2 oo^2) + 1 / (2 t^2)] = f(t) t / 2 has to hold, so we find that f(t) = 2 / t and f(x) = 2 t^2 / x^3. So these are the probability densities that fulfill the "modified" Lindy effect, where failure is impossible before time t > 0. I haven't double checked this so maybe there is a mistake, but I'm quite convinced that this line of reasoning leads to a unique distribution for each t > 0 and impossibility when t = 0.
    • sieste 3y ago
      So the Pareto distribution with alpha=2 produces the Lindy effect. Nice result! It looks like this generalises to E(X|X>a)=ka (with alpha depending on k).
      • golol 3y ago
        I didn't notice that is a Pareto distribution! Well, that is quite nice indeed.