Batteries A and B have lifetimes that are independent and
exponentially distributed with a common mean of m years. The
probability that battery B outlasts battery A by more than one year is
0.33. Calculate m.
I am pretty lost. I know that A, B must be ~ exp($\lambda = 1/m$). Are we calculating P(B > A+1)?
(Finan exam p 39.29)
Best Answer
Mean $m$ years implies that $m = 1/\lambda \implies \lambda = 1/m$. This is because you need an actual distribution parameter $\lambda$ rather than given mean for calculations below (although your exam might have meant $m$ is a parameter, in which case calculations below still apply, but the final step will be different).
You have to do some re-writing. First, $$\mathbb{P}[B > A + 1] = \mathbb{P}[\cup_{x}(B > x + 1) \cap (A = x)] = \int_{0}^{\infty}\int_{x+1}^{\infty}\lambda*e^{-\lambda x}\lambda*e^{-\lambda y}dy dx = \frac{e^{-\lambda}}{2}$$
Now, since you know the probability, you can write $$\frac{e^{-\lambda}}{2} = 0.33 \implies e^{-\lambda} = 0.66 \implies -\lambda = \ln0.66 \implies \lambda = -\ln0.66$$ If you original task meant $\lambda = 1/m$, then just express $m$ through $\lambda$, if it meant $\lambda = m$, you have the answer.