Problem. Beginning at time $t = O$. we start using bulbs. one at a time, to illuminate a room. Bulbs are replaced immediately upon failure. Each new bulb is selected independently by an equally likely choice between a type-A bulb and a type-B bulb. The lifetime, $X$, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF:
for type-A bulbs: $$f(x) = \begin{cases} e^{-x}, & if\ x \ge 0, \\ 0, & otherwise; \end{cases}$$ for type-B bulbs: $$f(x) = \begin{cases} 3e^{-3x}, & if\ x \ge 0, \\ 0, & otherwise; \end{cases}$$ Determine the PDF of the time until the 12th bulb failure.
Let $Z$ be a random variable denoting the time until the 12th bulb failure.
Possible solutions to calculate $PDF$ of $Z$,
- First calculate $MGF$ and then deduce $PDF$. In this example, $MGF$ can be computed as $$ \left[\frac{1}{2} \left(\frac{1}{1-s} + \frac{3}{3-s}\right)\right]^{12}$$ Deducing $PDF$ of $Z$ is a tedious task in this case.
- Let $X_{i}$ be time between $i^{th}$ and $i+1^{th}$ failures, then $PDF$ of $X_{i}$ is $\frac{1}{2}(e^{-x} + 3e^{-3x})$. Then, $$Z = \sum_{i=1}^{12} X_{i}$$.
If $X_{i}$ would have been purely exponential then $Z$ would have been easy to compute. But still using convolution $Z$ can be computed.
Confused, what is the correct line of thought? Any help is appretiated.