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Consider a renewal process with the lifetimes $X_1,X_2,\ldots$ having the continuous uniform distribution $\mathrm{Unif(1.5,4)}$. Determine the asymptotic expression for the expected number of renewals up to time $t$:

My attempt:
Using the formula $M(t) = \frac{t}{\mu} + \frac{\sigma^2-\mu^2}{2\mu^2}+o(1)$.

Calculating for $\mu$ and $\sigma^2$ since it follows $\mathrm{Unif(1.5,4)}$, $\mu = 2.75$ and $\sigma^2 = \frac{6.25}{12}$ and so the equation yields $M(t) = \frac{t}{2.75} + \frac{\frac{6.25}{12}-2.75^2}{2\cdot2.75^2}+o(1)$

But this was marked as incorrect, most notably $\mu$. I do not know how what else I can do for this, I would appreciate any help in solving this.

waterr
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  • Your values of $\mu=2.75, \sigma^2=6.25/12$ look correct. – Michael Mar 18 '24 at 22:28
  • @Michael Yeah, I'm just confused since 2.75 was marked as wrong in the field where $\mu$ goes, it is probably an error in the coding of the question so I just wanted to ask just in case if it was an error on my part – waterr Mar 18 '24 at 22:32
  • @Michael In a similar question to this one, I used $\mu$ and it was marked as correct. – waterr Mar 18 '24 at 22:34
  • You can ask the grader. Is your renewal equation correct? All I remember is it should be $t/\mu$ plus or minus a constant that is independent of $t$. – Michael Mar 18 '24 at 22:40
  • @Michael This is the equation that is shown in our textbook, I am thinking that if $\mu$ is wrongly marked as incorrect, it must mean that $M(t)$ is also wrongly marked as incorrect. I will try sending an email about this question to my professor. – waterr Mar 18 '24 at 22:43

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