Proof of Theorem 8

We will prove that if the distribution of $K$ is in ${\cal T}^{(n)}$, the distribution of $T$ is in ${\cal T}^{(n)}$ for $n \geq 2$. When $\rho=0$, $B=K$; thus, the theorem follows immediately.

Below, we assume $0<\rho<1$. The first three moments of $T$ are

$$\begin{eqnarray*} \mbox{{\bf\sf E}}\left[ T \right] & = & \frac{\mbox{{\bf\sf E}... ...}}\left[ G^2 \right])^2}{(\mbox{{\bf\sf E}}\left[ G \right])^2}, \end{eqnarray*}$$

where $\rho$ is the load of the M/G/1 queue. Let $t_2 \equiv \frac{\mbox{{\bf\sf E}}\left[ T^2 \right]}{(\mbox{{\bf\sf E}}\left[ T \right])^2}$ and $t_3 \equiv \frac{\mbox{{\bf\sf E}}\left[ T^3 \right]}{(\mbox{{\bf\sf E}}\left[ T \right])^3}$. Observe that if $t_2 > \frac{n+1}{n}$ and $t_3 - \frac{n+2}{n+1} t_2^2 \geq 0$, then the distribution of $T$ is in ${\cal T}^{(n)}$.

Thus, it suffices to prove that $t_2 > \frac{n+1}{n}$ and $t_3 - \frac{n+2}{n+1} t_2^2 \geq 0$, given that the distribution of $K$ is in ${\cal T}^{(n)}$. First, $t_2 > \frac{n+1}{n}$ is easy to prove:

$$t_2 = k_2 + \frac{\rho}{1-\rho}g_2\frac{\mbox{{\bf\sf E}}\l... ...]}{\mbox{{\bf\sf E}}\left[ K \right]} > k_2 > \frac{n+1}{n},$$

where $k_2=\frac{\mbox{{\bf\sf E}}\left[ K^2 \right]}{(\mbox{{\bf\sf E}}\left[ K \right])^2}$ and $g_2=\frac{\mbox{{\bf\sf E}}\left[ G^2 \right]}{(\mbox{{\bf\sf E}}\left[ G \right])^2}$. Below, we prove that $t_3 - \frac{n+2}{n+1} t_2^2 \geq 0$. Note that

$$\begin{eqnarray*} t_3 & = & k_3 + \frac{3\rho}{1-\rho} g_2k_2\frac{\mbox{{\bf\s... ...}}\left[ G \right]}{\mbox{{\bf\sf E}}\left[ K \right]}\right)^2, \end{eqnarray*}$$

where $k_3=\frac{\mbox{{\bf\sf E}}\left[ K^3 \right]}{(\mbox{{\bf\sf E}}\left[ K \right])^3}$ and $g_3=\frac{\mbox{{\bf\sf E}}\left[ G^3 \right]}{(\mbox{{\bf\sf E}}\left[ G \right])^3}$. Thus,

$$\begin{eqnarray*} \lefteqn{t_3-\frac{n+2}{n+1}t_2^2}\\ & = & k_3 -\frac{n+2}{n+... ...K \right]}\right)^2\\ & > & k_3 -\frac{n+2}{n+1}k_2^2 \geq 0. \end{eqnarray*}$$

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