Stationary Probabilities for Number in System

Reading the Stationary Distribution

An M/M/1 queue has one server, Poisson arrivals at rate $\lambda$, and independent exponential service times with rate $\mu$. Its number in the system includes both any customer in service and all customers waiting. When the arrival rate is below the service rate, the queue has a stationary distribution, and its probabilities form a geometric sequence. That compact formula makes it possible to calculate empty-system probabilities, exact counts, and tail probabilities directly.

Let $(X_t)_{t\ge0}$ be the continuous-time queue-length process on the probability space $(\Omega,\mathcal{F},\mathbb{P})$, with state space $S=\{0,1,2,\ldots\}$. The state $X_t=n$ means there are $n$ customers in the system at time $t$. Assume $\lambda>0$, $\mu>0$, and define the traffic intensity $\rho=\lambda/\mu$. In this tutorial we assume $0<\rho<1$, the stability condition for a stationary probability distribution.

Definition: A stationary distribution for the queue-length process is a probability vector $\pi=(\pi_0,\pi_1,\ldots)$ such that if $X_0$ has distribution $\pi$, then $X_t$ has distribution $\pi$ for every $t\ge0$. Here $\pi_n$ is the stationary probability of $n$ customers in the system.

The earlier material on local balance and the stationary distribution gives the adjacent-state equations $\lambda\pi_n=\mu\pi_{n+1}$, for $n\ge0$. The stability and existence results establish that the stationary distribution exists uniquely when $0<\lambda<\mu$. The calculation below turns those equations and the requirement that probabilities sum to one into the explicit formula used throughout this tutorial.

Theorem: For an M/M/1 queue with $\lambda>0$, $\mu>0$, and $\rho=\lambda/\mu<1$, the stationary probabilities are $$ \pi_n=(1-\rho)\rho^n,\qquad n=0,1,2,\ldots. $$

Proof. Local balance gives $\pi_{n+1}=(\lambda/\mu)\pi_n=\rho\pi_n$ for every $n\ge0$. Iterating this equality yields $\pi_n=\rho^n\pi_0$. Since $\pi$ is a probability distribution, its entries sum to one. Because $0<\rho<1$, the geometric series converges and $$ 1=\sum_{n=0}^{\infty}\pi_n =\pi_0\sum_{n=0}^{\infty}\rho^n =\frac{\pi_0}{1-\rho}. $$ Therefore $\pi_0=1-\rho$, and substituting this value into $\pi_n=\rho^n\pi_0$ gives $\pi_n=(1-\rho)\rho^n$ for every $n\ge0$. The entries are nonnegative and their sum is one, so they define a probability distribution. $\square$

The index $n$ is a count of customers, so it is always a nonnegative integer. The expression $\pi_n$ is the probability of exactly $n$ customers, not the probability of at least $n$. In particular, the factor $1-\rho$ is essential: it is the probability that the system is empty, and it normalizes the entire geometric sequence.

Exact Counts and Cumulative Probabilities

Once $\pi_n$ is known, an exact-count probability is obtained by substituting the desired integer into the formula. Cumulative and tail probabilities require adding a range of these exact-count probabilities. For an integer $n\ge0$, “at most $n$” means $X_t\le n$; “more than $n$” means $X_t>n$, which starts at $n+1$; and “at least $n$” means $X_t\ge n$.

Theorem: In stationarity, for every integer $n\ge0$, $$ \mathbb{P}(X_t=0)=1-\rho,\qquad \mathbb{P}(X_t=n)=(1-\rho)\rho^n, $$ $$ \mathbb{P}(X_t\le n)=1-\rho^{n+1},\qquad \mathbb{P}(X_t>n)=\rho^{n+1},\qquad \mathbb{P}(X_t\ge n)=\rho^n. $$

Proof. The empty-system formula is the stationary formula at $n=0$, since $(1-\rho)\rho^0=1-\rho$. The exact-count formula is the theorem already proved. For the cumulative probability, sum the entries from $0$ through $n$: $$ \mathbb{P}(X_t\le n) =\sum_{k=0}^{n}(1-\rho)\rho^k =(1-\rho)\frac{1-\rho^{n+1}}{1-\rho} =1-\rho^{n+1}. $$ The strict tail is the complementary event to $X_t\le n$, so $\mathbb{P}(X_t>n)=1-\mathbb{P}(X_t\le n)=\rho^{n+1}$. For the inclusive tail, sum from $k=n$ onward: $$ \mathbb{P}(X_t\ge n) =\sum_{k=n}^{\infty}(1-\rho)\rho^k =(1-\rho)\frac{\rho^n}{1-\rho} =\rho^n. $$ All these sums converge because $0<\rho<1$. $\square$

The exponents record the event boundary. The strict tail $X_t>n$ begins at $n+1$, hence its exponent is $n+1$. The inclusive tail $X_t\ge n$ begins at $n$, hence its exponent is $n$. At $n=0$, these formulas give $\mathbb{P}(X_t>0)=\rho$ and $\mathbb{P}(X_t\ge0)=1$, as required.

Worked Example: Empty, Exactly Three, and More Than Two

Suppose arrivals occur at rate $\lambda=4$ customers per hour and service occurs at rate $\mu=10$ customers per hour. Find the stationary probabilities that the system is empty, contains exactly three customers, and contains more than two customers.

First check stability and compute the traffic intensity: $$ \rho=\frac{\lambda}{\mu}=\frac{4}{10}=\frac25. $$ Since $0<\frac25<1$, the stationary formulas apply. The probability of an empty system is $$ \pi_0=1-\rho=1-\frac25=\frac35. $$ The probability of exactly three customers is $$ \pi_3=(1-\rho)\rho^3 =\frac35\left(\frac25\right)^3 =\frac35\cdot\frac8{125} =\frac{24}{625}. $$ For more than two customers, the strict event is $X_t>2$, so the tail formula gives $$ \mathbb{P}(X_t>2)=\rho^{2+1} =\left(\frac25\right)^3 =\frac8{125}. $$

Check: The probability of at most two customers is $1-\rho^3=1-\frac8{125}=\frac{117}{125}$. The individual probabilities are $\pi_0=\frac35=\frac{75}{125}$, $\pi_1=\frac35\cdot\frac25=\frac6{25}=\frac{30}{125}$, and $\pi_2=\frac35\cdot\frac4{25}=\frac{12}{125}$. Their sum is $\frac{75+30+12}{125}=\frac{117}{125}$; adding $\mathbb{P}(X_t>2)=\frac8{125}$ gives $1$.

Worked Example: Probabilities Through Two Customers

Let $\lambda=3$ and $\mu=5$, in the same time units. Compute the probabilities of zero, one, and two customers, and then calculate the probability of more than two customers.

The rates give $\rho=\frac35$, which is strictly less than one. Thus $1-\rho=\frac25$. Substitution gives $$ \pi_0=\frac25=\frac{50}{125},\qquad \pi_1=\frac25\cdot\frac35=\frac6{25}=\frac{30}{125}, $$ $$ \pi_2=\frac25\left(\frac35\right)^2 =\frac25\cdot\frac9{25} =\frac{18}{125}. $$ Since “more than two” means at least three, its probability is $$ \mathbb{P}(X_t>2)=\left(\frac35\right)^3=\frac{27}{125}. $$

Check: Adding the exact probabilities through two gives $\frac{50+30+18}{125}=\frac{98}{125}$. The complement is $1-\frac{98}{125}=\frac{27}{125}$, matching the strict-tail calculation. The four disjoint events $X_t=0$, $X_t=1$, $X_t=2$, and $X_t>2$ therefore have probabilities summing to one.

Worked Example: At Most Three Versus Exactly Four

Suppose $\lambda=8$ and $\mu=12$. Find the probability of at most three customers, the probability of more than three, and the probability of exactly four.

Here $\rho=\frac8{12}=\frac23$, so $1-\rho=\frac13$, and the stationary condition holds. The cumulative probability through three is $$ \mathbb{P}(X_t\le3)=1-\rho^4 =1-\left(\frac23\right)^4 =1-\frac{16}{81} =\frac{65}{81}. $$ The strict tail is $$ \mathbb{P}(X_t>3)=\rho^4 =\left(\frac23\right)^4 =\frac{16}{81}. $$ Exactly four customers is one point in that tail, not the whole tail: $$ \pi_4=(1-\rho)\rho^4 =\frac13\cdot\frac{16}{81} =\frac{16}{243}. $$

Check: The probabilities for zero through three are $\pi_0=\frac13=\frac{27}{81}$, $\pi_1=\frac29=\frac{18}{81}$, $\pi_2=\frac4{27}=\frac{12}{81}$, and $\pi_3=\frac8{81}$. Their sum is $\frac{27+18+12+8}{81}=\frac{65}{81}$. Adding the strict tail $\frac{16}{81}$ gives one. Also, $\pi_4=\frac{16}{243}$ is smaller than $\mathbb{P}(X_t>3)=\frac{16}{81}$, as it must be because the tail includes the event of exactly four and all larger counts.

Interpreting What Is Being Counted

These formulas concern the number in the system: a customer in service counts as one, just as a customer waiting does. The number waiting in queue is a different random quantity, because a system with one customer has nobody waiting while the server is occupied. Its distribution is a separate calculation. By PASTA, a Poisson arrival sees this same stationary distribution of the number in the system immediately before arrival. This statement concerns the system count, not the number waiting in queue.

The geometric form also gives a simple way to assess how much probability lies beyond a chosen threshold. If the threshold is $n$, the chance of strictly exceeding it is $\rho^{n+1}$. A larger $\rho$, still below one, makes this tail larger at every fixed threshold. As $\rho$ approaches one, the powers decay more slowly, so high customer counts become more likely. These are statements about stationary probabilities; they do not by themselves provide a waiting-time distribution or a mean queue length.

Common mistake: Do not use $\rho^n$ as the probability of exactly $n$ customers. For example, with $\rho=\frac25$, that incorrect calculation would give $\mathbb{P}(X_t=2)=\rho^2=\frac4{25}$. The correct point probability includes the normalizing factor $1-\rho$: $$ \mathbb{P}(X_t=2)=(1-\rho)\rho^2 =\frac35\cdot\frac4{25} =\frac{12}{125}. $$ The value $\rho^2=\frac4{25}=\frac{20}{125}$ is instead $\mathbb{P}(X_t\ge2)$, which includes every count from two upward.

A reliable calculation starts by identifying the event precisely. “Exactly $n$” calls for $(1-\rho)\rho^n$; “at least $n$” calls for $\rho^n$; and “more than $n$” calls for $\rho^{n+1}$. For “at most $n$,” either sum the point probabilities from zero to $n$ or use the complement $1-\rho^{n+1}$. Checking that a cumulative probability and its complementary tail add to one is an effective way to catch an indexing error.

Check Your Understanding

Assume throughout that the rates satisfy $0<\lambda<\mu$, and use the stationary number-in-system distribution.

If $\rho=\frac14$, calculate the probability that the system is empty and the probability that it contains exactly two customers.
For $\rho=\frac35$, calculate $\mathbb{P}(X_t>1)$. Explain why the exponent is two rather than one.
For $\rho=\frac23$, find the probability of at most two customers and verify it by adding $\pi_0,\pi_1,\pi_2$.
Explain the difference between the events $X_t\ge3$ and $X_t>3$, and write the corresponding formulas in terms of $\rho$.
With $\rho=\frac25$, identify the event whose probability is $\rho^2$, and calculate the probability of exactly two customers.

Stationary Probabilities for Number in System

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