Find E(V), where V is the first time a bulb working longer than 5 days is replaced, where bulb life is uniformly distributed 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 $\frac{1}{\mathbb E [V]}=\frac{\mathbb E[R_i]}{\mathbb E [X_i]}$
 A: 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 $\frac{40}{3}$.
A: 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 $E(V)=40/3$
