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.