Find E(V), where V is the first time a bulb working longer than 5 days is replaced, where bulb life $X_i$ is uniformly distributed over [0,20]. So $V= \min\{T_n:n\ge1, X_n>5\}$

I know that $$ \lim_{t\to \infty} \frac{C(t)}{t}=\frac{\mathbb E[R_i]}{\mathbb E [X_i]}$$ A solution (not mine) states that $$\lim_{t\to \infty} \frac{N(t)\in[0,t]}{t}=\frac{1}{\mathbb E [V]}=\frac{\mathbb E[R_i]}{\mathbb E [X_i]}=\frac{0.75}{10}\Rightarrow \mathbb E[V]=40/3$$

I don't get how $\frac{1}{\mathbb E [V]}=\frac{\mathbb E[R_i]}{\mathbb E [X_i]}$

  • $\begingroup$ What is $T_n$? You haven't defined it anywhere. $\qquad$ $\endgroup$ – Michael Hardy Apr 13 '16 at 5:26
  • $\begingroup$ $T_n$ is the time that the $n^{th}$ bulb is replaced $\endgroup$ – ak87 Apr 13 '16 at 5:27
  • $\begingroup$ What is $R_i$? And what is $C(t)$? $\qquad$ $\endgroup$ – Michael Hardy Apr 13 '16 at 5:30
  • $\begingroup$ sorry, I have incorrectly assumed that the notation was universally used with renewal processes. $R_i$ is the reward associated with the renewal event. $C(t)$ is the sum of the values of all $R_i$ until time t $\endgroup$ – ak87 Apr 13 '16 at 5:34
  • $\begingroup$ ok, I believe I've seen all this before, but the notation conventionally accompanying renewal processes is something I haven't seen in a long time and I didn't master it the way I have some other things I learned around that same time. $\qquad$ $\endgroup$ – Michael Hardy Apr 13 '16 at 5:36

I didn't totally follow the solution you presented, but I would go about it as follows:

If the first light bulb lasts more than 5 days, the expected lifetime is 12.5 days. The probability of this occurring is .75.

There is a .25 probability that bulb #1 does not last more than 5 days. In this scenario, the expected lifetime of bulb #1 is 2.5. The expected time from the death of bulb #1 until $V$ is the same as the original $E(V)$ since we are back to square one at that point. Hence, the expected value of $V$ given that bulb #1 didn't last more than 5 days is $2.5+E(V)$.

This gives us the equation $E(V)=.75*12.5+.25*(2.5+E(V))$

Solving this equation gives you $E(V)=40/3$

  • $\begingroup$ just a question: how are (both of) you calculating E(V)=12.5 given that the bulb lasted more than 5 days? $\endgroup$ – ak87 Apr 13 '16 at 6:13
  • $\begingroup$ Given that the life is greater than $5$, the life is uniformly distributed in the interval $[5,20]$, so has mean $\frac{5+20}{2}$. $\endgroup$ – André Nicolas Apr 13 '16 at 6:17

We give a different solution, conditioning on the life of the first bulb.

If the life of the first bulb is $\le 5$ (probability $\frac{1}{4}$), then the conditional mean life is $\frac{5}{2}$, and the conditional expectation of $V$ is $\frac{5}{2}+E(V)$. If the life of the first bulb is $\gt 5$ (probability $\frac{3}{4}$), then the conditional expectation of $V$ is $\frac{25}{2}$. Thus $$E(V)=\frac{1}{4}\left(\frac{5}{2}+E(V)\right)+\frac{3}{4}\left(\frac{25}{2}\right).$$ Solve for $E(V)$. We get $\frac{40}{3}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.