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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 $\frac{1}{\mathbb E [V]}=\frac{\mathbb E[R_i]}{\mathbb E [X_i]}$

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  • $\begingroup$ 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
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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 $E(V)=40/3$

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  • $\begingroup$ 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
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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 $\frac{40}{3}$.

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