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I have $t \sim \varepsilon(\lambda)$. Then if I let $X$ represent the sequence of occurrence of $t$ like: $t_1+t_2+t_3+...+t_n$ does $X \sim \mathcal{G}(n, \lambda)$? Since I know that when $X \sim \mathcal{G}(1, \lambda)$ then $X \sim \varepsilon(\lambda)$?


Reason I ask this is because in an attempt to do question below (part (b ii) in particular)

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My first impression is its asking the time to the start of the nth service. But if the previous part seem to be more useful if its until the end of the nth service?

If $T_n$ is referring to time until the start it will look like:

  • $T_1 = \tau$
  • $T_2 = 2 \tau + t_1$
  • $T_n = n \tau + \underbrace{t_1 + ... + t_{n-1}}_{X}$

Versus if its referring to time until end of service

  • $T_1 = \tau + t_1$
  • $T_2 = 2 \tau + t_1 + t_2$
  • $T_n = n \tau + \underbrace{t_1 + ... + t_{n}}_{X}$

Either way if I let the end portion be the random variable $X$ then I can reuse my answer for part (b i)? If each of those $t_i$ follows the exponential distribution, can I say $X$ follows a gamma distribution? Or am I supposed to do something different?

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Gamma distribution satisfies the following property: $A_i \sim \mathcal{G}(\alpha_i, \beta), i =1,2, \cdots, n \Rightarrow \sum_{i=1}^{n}A_i \sim \mathcal{G}(\sum_{i=1}^{n}\alpha_i , \beta)$.

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  • $\begingroup$ Err ... so is the answer yes? That if I have $\sum_i t_i$ where each $t_i \sim \varepsilon(\lambda)$ then $\sum_i t_i \sim \mathcal{G}(n, \lambda)$. In this statement, 1st parameter of gamma distribution is $\mathbb{Z}^+$ but in your case, is it the same for $\alpha_i$? $\endgroup$ – Jiew Meng Nov 5 '13 at 23:19
  • $\begingroup$ @JiewMeng Yes, it is. We only need that these Gamma distributions in the sum are independent between each other. The parameters $\alpha_i$ can be any positive numbers $\endgroup$ – Petite Etincelle Nov 7 '13 at 8:32

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