Statistics for Maintenance Modelling: Proportional Hazards and Accelerated Life Models

# Exercise 1¤

Let t denote the time to failure of an electrical insulator subject to constant voltage y, and suppose for some and S0(t) = e-t.

1. Prove that both proportional hazards and accelerated life models can produce the same S(t; y) = .
2. Prove that t is exponentially distributed with a scale parameter g(y).
3. What is the mean of t?

## Answer 1

The survival function of the exponential distribution is usually of the form of (from The Reliability Function for the Weibull Distribution screen in the Continuous Lifetime Distributions 1 session) with the hazard function of h0(t) = . In this case, since S0(t) = e-t, = 1. If we use a PH model with = 1, then we have h(t; y) = h0(t)g(y) = and
S(t; y) = = (from screen 9). Using an accelerated life model we have S(t; y) = S0{t • g(y))} = S(t). Substituting t for the parameter in the baseline survival function we get S0(t) = , so the result is the same.

## Answer 2

If S(t; y) = , then the pdf (from The Reliability Function screen in the Continuous Lifetime Distributions 1 session) is , so t is exponentially distributed with scale parameter .

## Answer 3

If f(t) is in the form , then the mean is 1/. Since we've just proved that =g(y), the mean is 1/g(y).

If f(x) = aebx then f'(x) = abebx.