Question: Lamps are of a type with an average lifetime of 1200 hours. Assume that we can model the probability of failure of these bulbs by an exponential density function with mean μ = 1200.
If one bulb burns out and is replaced by a bulb of the same type, find the probability that the two bulbs fail within a total of 1200 hours.
I tried solving it by
$$\int_0^{1200}\int_0^{1200-x}\frac{1}{1440000}e^{-x/1200}e^{-y/1200}dy\, \, dx = 1-e^{-1}$$
but this seems not to be the correct answer, as I know that the value must be lower than this. Can anyone show where I got wrong?
Best Answer
Since you are modeling $T_1 + T_2 = 1200$ where T ~ exp($\frac{1}{1200}$), you need to perform the convolution of two iid exponential variables. You may want to read about the Erlang distribution, or Gamma distribution which is the analytical solution to this problem in the time domain.
Another way to look at it is to use the connection to the Poisson(1) and ask what is the probability that N $\geq$ 2. This is not as strict a formulation, as we are not forcing the time of the second failure to be at 1200 hours.
As for your approach (integrating over the joint distribution):
$\int\limits_{0}^{1200} \int\limits_0^{1200-x}(\frac{1}{1200})^2 e^{\frac{-x}{1200}} e^{\frac{-y}{1200}} dydx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}} \int\limits_0^{1200-x}(\frac{1}{1200}) e^{\frac{-y}{1200}} dydx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}} [-e^{\frac{-y}{1200}}]^{1200-x}_0 dydx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}}[1-e^{-(1-\frac{x}{1200})}]dx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}}-\frac{1}{1200}e^{-1}dx = [1-e^{-1}]-e^{-1} =1-2e^{-1} = Poisson(N\geq2, \lambda = 1)$