Mean residual time

The Endeavour 2013-06-17

If something has survived this far, how much longer is it expected to survive? That’s the question answered by mean residual time.

For a positive random variable X, the mean residual time for X is a function e_X(t) given by

$e_X(t) = E(X - t \mid X > t) = \int_t^\infty \frac{1 - F_X(x)}{1-F_X(t)} \, dx$

provided the expectation and integral converge. Here F(t) is the CDF, the probability that X is greater than t.

For an exponential distribution, the mean residual time is constant. For a Pareto (power law) distribution, the mean residual time is proportional to t. This has some interesting consequences, known as the Lindy effect.

Now let’s turn things around. Given function a function e(t), can we find a density function for a positive random variable with that mean residual time? Yes.

The equation above yields a differential equation for F, the CDF of the distribution.

If we differentiate both sides of

$e(t) (1 - F(t)) = \int_t^\infty 1 - F(x)\, dx$

with respect to t and rearrange, we get the first order differential equation

$F'(t) + g(t)\, F(t) = g(t)$

where

$g(t) = \frac{e'(t) + 1}{e(t)}$

The initial condition must be F(0) = 0 because we’re looking for the distribution of a positive random variable, i.e. the probability of X being less than zero must be 0. The solution is then

$F(t) = 1 - \frac{e(0)}{e(t)} \exp\left( -\int_0^t \frac{dx}{e(x)} \right)$

This means that for a desired mean residual time, you can use the equation above to create a CDF function to match. The derivative of the CDF function gives the PDF function, so differentiate both sides to get the density.