I have two identical lightbulbs, each is an IID exponential random variables with mean 1000 hours.
- How long does it take for the first bulb to burn out, on average?
- For the second to burn out, on average?
- Instead of using both of them simultaneously, I replace the first one after it burns out with the second out. How long on average until the second bulb burn out?
My Solutions
- $P(X > 1000) = e^{-\lambda t} = 0.368$
- $P(X + Y > 1000) = e^{-\lambda t}(1+\lambda t)$
- I don't know how to do this one, I'm thinking about Poisson but can't think of a way to set it up
Best Answer
Say we call our two bulb lifetimes $X$ and $Y$ and assume they are independent and identically distributed.
That the distributions of the lifetimes of light bulbs are memoryless does not seem plausible to me, but I'll play along.
We have $\Pr(X>x) = e^{-x/1000} = \Pr(Y>x)$.
So \begin{align} & \Pr(\min\{X,Y\} > x) = \Pr(X>x\ \&\ Y>x) = \Pr(X>x)\cdot\Pr(Y>x) \\[10pt] = {} & \Big( e^{-x/1000}\Big)^2 = e^{-x/500}. \end{align} This is an exponential distribution with expected value $500$, so that's how long it is on average until the first bulb burns out.
The average time until they have both burned out is $\operatorname{E}(\max\{X,Y\})$. We have
\begin{align} & \Pr(\max < x) = \Pr(X<x\ \&\ Y<x) = \Pr(X<x) \cdot \Pr(Y<x) \\[10pt] = {} & (1-e^{-x/1000})^2 = 1 - 2e^{-x/1000} + e^{-x/500} \end{align} So the density function is $$ f_{\max} (x) = \frac d {dx} \left( 1 - 2e^{-x/1000} + e^{-x/500} \right) = 2\cdot\frac 1 {1000} e^{-x/1000} - \frac 1 {500} e^{-x/500} \text{ for } x>0. $$ Since this density is $2$ times the density of an exponential distribution with mean $1000$ minus an exponential distribution with mean $500$, we get \begin{align} & \int_0^\infty x f_{\max}(x)\,dx \\[10pt] = {} & 2\int_0^\infty x\cdot(\text{that first density})\,dx - \int_0^\infty x\cdot(\text{that second density}) \,dx \\[10pt] = {} & 2\cdot1000 - 500 = 1500. \end{align}
The answer to the third question is merely the sum of the two averages: $1000+1000=2000$.