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For the renewal process whose interarrival times are uniformly distributed over (0,1), determine the expected time from t = 1 until the next renewal.

I know that $\lim_{t \to\infty}\frac{S_{N(t)+1}}{N(t)}=\mu $ and for the uniform distribution, $\mu=\frac{1}{2}$. I know I want to find $E[S_{N(1)+1}]$. I'm just now sure how to put these ideas together. Or if I'm even headed the right direction.

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$$S_{N(t)+1} = \sum_{i=1}^{N(t)+1}X_i$$

where $X_i \sim \text{Uniform}(0,1)$ are iid. Since $N(t)+1$ is a stopping time and $\mathbb{E}[N(t)+1] = m(t)+ 1$ where $m(t) = \mathbb{E}[N(t)]$, if $m(t) < \infty$, we can use Wald's equation to conclude that,

$$\mathbb{E}[S_{N(t)+1}] = \mathbb{E}[X_1]\mathbb{E}[N(t)+1] = \frac{1}{2}(m(t)+1)$$

The only thing left is to compute $m(t)$ and simultaneously show that $m(t) < \infty$. It is easy to show that,

$$m(t) = \sum_{n=1}^{\infty}F_{n}(t)$$

where $F_{n}(t)$ is the $n$ convolutions of the cdf of uniform random variable. It follows from above,

$$m(t) = F(t) + (F*\sum_{k=1}^{\infty}F_{k})(t) = F(t) + (F*m)(t) = F(t) + \int_{0}^{t}m(t-x)dF(x)$$

Now, use $F(t)=t$ and $dF(x) = 1\cdot dx$ to show that,

$$m(t) = t + \int_{0}^{t}m(t-x)dx = t+\int_{0}^{t}m(x)dx \implies \frac{dm(t)}{dt} = 1+m(t) \implies m(t) = e^{t}-1$$

Therefore, $m(t) < \infty$ when $t < \infty$ and $\mathbb{E}[S_{N(t)+1}] = e^t/2$.

P.S. Please comment if something is not clear.

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  • $\begingroup$ Can you please explain the step in which $m(t-x)$ is changed to $m(x)$ (in the integration part). $\endgroup$ – Harry Jul 3 at 11:31
  • $\begingroup$ Use change of variables $u \leftarrow t-x$ and $du = -dx$. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Jul 3 at 15:43

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