For an M/M/1 queueing system, distribution of the time ($t_e$) between two consecutive events (either arrival or departure) can be derived as follows with the independent assumption,

$$F(t_e\ge t)=F(t_a\ge t)\cdot F(t_d\ge t)=e^{-\lambda t}e^{-\mu t}=e^{-(\lambda+\mu)t}$$

where $t_a$ and $t_d$ are the time between two consecutive arrivals and departures respectively.

From the complementary CDF (CCDF) derived above, we can say that $t_e$ is also exponentially distributed (with mean $1/(\lambda+\mu)$).

However, if we derive the result using CDF rather than CCDF, the outcome is,

$$F(t_e\le t)=F(t_a\le t)\cdot F(t_d\le t)\\=(1-e^{-\lambda t})(1-e^{-\mu t})=1-e^{-\lambda t}-e^{-\mu t}+e^{-(\lambda+\mu)t}\\<1-e^{-(\lambda+\mu)t}=1-F(t_e\ge t)$$

From this point of view, it seems we cannot say $t_e$ is exponentially distributed.

I am quite puzzled about this, could anyone shed some light upon this? Thanks.


Your first attempt uses the claim that $$ \{ t_e \ge t \} = \{ t_a \ge t \} \cap \{ t_d \ge t \}. \tag{$\dagger$} $$ (Here the "$\{...\}$" notation just means "the event that [...] happens".)

In your second attempt, you look at $\{t_e \le t\}$. Since everything is with exponential times, "$\le$" or "<" doesn't matter; let's look at $\{t_e < t\}$. You then claim that $$ \{ t_e < t \} = \{ t_a < t \} \cap \{ t_d < t \}. \tag{$\ddagger$}$$

However, these two displays, $(\dagger, \ddagger)$, are incompatible. Note that $\{t_e \ge t\} = \{t_e < t\}^c$. (Here $\{...\}^c$ means the complementary event.) But, for general sets/events $A$ and $B$, $$ \text{if}\quad C = A \cap B \quad\text{then}\quad C^c = A^c \cup B^c. $$ Hence, applying this in your situation, assuming that $(\dagger)$ is the correct statement, we get $$ \{t_e < t\} = \{t_e \ge t\}^c = \bigl( \{t_a \ge t\} \cap \{t_d \ge t\} \bigr)^c = \{t_a < t\} \cup \{t_d < t\}, $$ which you then need to deal with -- basically, you deal with this by doing your first computation for the different parts, and using the independence of $\{t_a \ge t\}$ and $\{t_d \ge t\}$ (which you assumed in your first calculation).

  • $\begingroup$ Thanks, it points out where I went wrong. Since the r.v. $t_e$ is the time till the next event, therefore $t_e\ge t$ means $t_a\ge t$ as well as $t_d\ge t$. However, for $t_e\le t$, at least one of $t_a\le t$ and $t_d\le t$ happens will guarantee $t_e\le t$. The probability, therefore, is equal to $1-P(t_a\ge t, t_d\ge t)=1-F(t_e\ge t)$. I see you applied de Morgan's law, but $C^c$ is not equal to $(A\cup B)\backslash(A\cap B)$ if $(A\cup B)$ is not the universal set. It's supposed to be $S\backslash(A\cap B)$. $\endgroup$ – Guoyang Qin Feb 12 at 19:14
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    $\begingroup$ Oops, thank you for pointing out the mistake! Let me just correct that... $\endgroup$ – Sam T Feb 12 at 20:52

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