Find the expectation of the arrival times of a queuing system?

Question

Suppose ${X_1, X_2, \dots}$ are independent identically distributed random variables defined by the density $f(x)=\lambda e^{-\lambda x}$. The renewal process $N={N(t): t>=0}$ is defined by the sequence of arrival times, for $n>=2$: $$W_0=0, W_1=X_1, \dots, W_n=\sum_{k=1}^{n}X_k$$

1. Find the expectation of $W_n$.
2. Find the expectation of $W_{N(t)+1}$ by evaluating $[W_{N(t)+1}|N(t)=n]$.
3. Verify that $E[W_{N(t)+1}]=(E[X_1])[M(t)+1]$ where $M(t)=E[N(t)]$.

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Since each $X$ is Poisson, we have a sum of Poissons, correct? So just to get started, I think that $E[W_n]=n\lambda$. Is that right?

Answer 1

Since $\mathbb E[X_1]=\frac1\lambda$, we compute $$\mathbb E[W_n] = \mathbb E\left[\sum_{k=1}^n X_k\right]=\sum_{k=1}^n\mathbb E[X_k] = n\mathbb E[X_1] = \frac n\lambda. $$ To compute $\mathbb E[W_{N(T)+1}]$, use the law of total expectation: \begin{align} \mathbb E[W_{N(T)+1}] &= \sum_{n=0}^\infty \mathbb E[W_{N(T)+1}\mid N(T)=n]\mathbb P(N(T)=n)\\ &=\sum_{n=0}^\infty \mathbb E[W_{N(T)}+X_{N(T)+1}\mid N(T)=n]\mathbb P(N(T)=n)\\ &=\sum_{n=0}^\infty \mathbb (E[W_n] + \mathbb E[X_1])\left(\frac{(\lambda T)^n}{n!}e^{-\lambda T}\right)\\ &=\sum_{n=0}^\infty \left(\frac{n+1}\lambda \right)\left(\frac{(\lambda T)^n}{n!}e^{-\lambda T}\right)\\ &=\frac1\lambda\sum_{n=0}(n+1)\left(\frac{(\lambda T)^n}{n!}e^{-\lambda T}\right)\\ &=\frac1\lambda\mathbb E[N(T)+1]\\ &=\frac1\lambda(\mathbb E[N(T)]+1). \end{align}