Ten successive exponential components

Topic: Functions of random variables and limit theorems · more questions on this topic
‹ A sieve for blueberries

A component has an exponential lifetime with expectation 1/2 year. On failure it is replaced at once by an identical new component. Ten components are used this way, and T is the time at which the tenth fails.

  1. Find \mathbb{E} T and \operatorname{Var}(T).

  2. Approximate \mathbb{P}(T>6) with the central limit theorem.

  3. Name the exact distribution of T.

  1. T=X_1+\cdots+X_{10} with the X_i independent \operatorname{Exp}(2), each with mean 1/2 and variance 1/4. So \mathbb{E} T=5 years and \operatorname{Var}(T)=10/4=2.5.

  2. By the central limit theorem,

\mathbb{P}(T>6)\approx1-\Phi\Bigl(\frac{6-5}{\sqrt{2.5}}\Bigr)=1-\Phi(0.632)=0.264.

  1. T is exactly \operatorname{Gamma}(10,2), a sum of ten independent exponentials with a common rate. The exact probability is \mathbb{P}(T>6)=\mathbb{P}(\operatorname{Poisson}(12)\le9)=0.242; the normal approximation is a little high because the gamma law is skewed to the right.