Erlang Waiting Times and Gamma Distributions

Why the Time to Several Events Is Not Exponential

An exponential model describes the waiting time until the next event in a constant-rate process. Many questions instead ask how long it takes for several events to occur: for example, the time until a service desk receives its fifth request or a sensor records its third signal. These waiting times have a related model, the Erlang distribution, which is a gamma distribution with a positive-integer shape parameter.

The key is to connect event times with event counts. If \(T_k\) is the time of the \(k\)th event and \(N(t)\) is the number of events observed by time \(t\), then \(T_k\leq t\) exactly when at least \(k\) events have occurred by \(t\). This equivalence makes probabilities for \(T_k\) accessible through the Poisson distribution.

Definition: Suppose events follow a homogeneous Poisson process with rate \(\lambda>0\) per unit time. This means the process has independent increments and the number of events in an interval of length \(t\) has a Poisson distribution with mean \(\lambda t\). The waiting time \(T_k\) from time zero to the \(k\)th event, for a positive integer \(k\), has an Erlang distribution with shape \(k\) and rate \(\lambda\). Equivalently, it has a gamma distribution with shape \(k\) and scale \(1/\lambda\).

The rate and time units must match. If \(\lambda\) is measured in events per hour, then \(t\) in \(\lambda t\) must be measured in hours. The product \(\lambda t\) is dimensionless and is the mean number of events accumulated by time \(t\).

From Poisson Counts to the Erlang Distribution

For a homogeneous Poisson process, \(N(t)\sim\operatorname{Poisson}(\lambda t)\). The event-time/count relationship gives the cumulative distribution function (CDF) of \(T_k\):

F_{T_k}(t)=P(T_k\leq t)=P(N(t)\geq k) =1-\sum_{j=0}^{k-1}e^{-\lambda t}\frac{(\lambda t)^j}{j!},\qquad t\geq 0.

For \(t<0\), \(F_{T_k}(t)=0\). The complement in the formula is the probability of observing fewer than \(k\) events by time \(t\). In particular, this formula involves a sum only through \(k-1\); omitting one of those terms changes the event being calculated.

Differentiating the CDF gives the density. To see the cancellation, write \(x=\lambda t\) and differentiate \(1-e^{-x}\sum_{j=0}^{k-1}x^j/j!\) with respect to \(t\). The derivative of the finite sum is \(\lambda\sum_{j=0}^{k-2}x^j/j!\), while differentiating \(e^{-x}\) contributes a factor \(-\lambda e^{-x}\). The terms from \(0\) through \(k-2\) cancel, leaving:

f_{T_k}(t)=\frac{\lambda^k t^{k-1}e^{-\lambda t}}{(k-1)!},\qquad t>0.

This is the gamma density with shape \(k\) and rate \(\lambda\). In shape-scale notation, the scale is \(1/\lambda\), not \(\lambda\). The two parameterizations therefore give \(E[T_k]=k/\lambda\) and \(\operatorname{Var}(T_k)=k/\lambda^2\).

Theorem: Under a homogeneous Poisson process with rate \(\lambda\), \(T_k\) is the sum of \(k\) independent exponential waiting times, each with rate \(\lambda\). Consequently, \(T_k\) has an Erlang (gamma) distribution with shape \(k\) and rate \(\lambda\). The result requires a constant rate and the Poisson-process assumptions; it does not apply automatically when the rate varies over time or events are dependent.

The sum-of-waiting-times description offers another interpretation. If \(W_j\) is the time between event \(j-1\) and event \(j\), then \(T_k=W_1+\cdots+W_k\), with each \(W_j\) exponential at rate \(\lambda\). The intervals are independent in a homogeneous Poisson process. This explains why \(E[T_k]=k/\lambda\) and why the variances of the \(k\) intervals add to \(k/\lambda^2\).

Worked Probability Calculations

Worked Example: The Third Event by Day One or Day Two

Hypothetical scenario. A monitoring system receives signals according to a homogeneous Poisson process at a rate of \(1.5\) signals per day. Find (a) the probability that the third signal arrives by the end of day 1 and (b) the probability that it arrives after day 1 but by the end of day 2. The rate is \(1.5\) per day, so use \(t\) in days. No interest or mortality assumption is involved.

Method and distribution. The time \(T_3\) has an Erlang distribution with shape \(k=3\) and rate \(\lambda=1.5\) per day. From the CDF formula,

F_{T_3}(t)=1-e^{-1.5t}\left(1+1.5t+\frac{(1.5t)^2}{2}\right).

(a) At \(t=1\) day, \(1.5t=1.5\), so

F_{T_3}(1)=1-e^{-1.5}\left(1+1.5+\frac{1.5^2}{2}\right) =1-3.625e^{-1.5} =1-3.625(0.22313016) \approx 0.19115317.

Thus the probability is \(0.1912\) to four decimal places.

(b) The interval probability is a difference of CDF values. At \(t=2\), \(1.5t=3\), giving

F_{T_3}(2)=1-e^{-3}\left(1+3+\frac{3^2}{2}\right) =1-8.5(0.04978707) \approx 0.57680992.

Therefore,

P(1<T_3\leq 2)=F_{T_3}(2)-F_{T_3}(1) \approx 0.57680992-0.19115317 =0.38565675 \approx 0.3857.

Check and common mistake avoided. The interval probability is positive and less than \(F_{T_3}(2)\), as it must be. Rounding is done only at the end. Using the exact CDF expression at day 1 and then subtracting the unrounded values avoids carrying forward an inaccurate rounded probability.

Worked Example: Convert Minutes Before Finding an Interval Probability

Hypothetical scenario. A dispatch channel receives messages according to a homogeneous Poisson process at a rate of \(4\) messages per hour. Find the probability that the second message arrives between 15 and 45 minutes after monitoring begins. No interest or mortality assumption is involved.

Method and distribution. The second-event time \(T_2\) is Erlang with shape \(2\) and rate \(4\) per hour. Convert the bounds to hours: \(15\) minutes is \(15/60=0.25\) hours, and \(45\) minutes is \(45/60=0.75\) hours. For shape \(2\),

F_{T_2}(t)=1-e^{-4t}(1+4t).

At the lower bound, \(4t=4(0.25)=1\), so

F_{T_2}(0.25)=1-e^{-1}(1+1) =1-\frac{2}{e} \approx 1-0.73575888 =0.26424112.

At the upper bound, \(4t=4(0.75)=3\), so

F_{T_2}(0.75)=1-e^{-3}(1+3) =1-4(0.04978707) \approx 0.80085173.

Subtract the lower-tail probability from the upper-tail probability:

P(0.25<T_2\leq 0.75) =0.80085173-0.26424112 =0.53661061 \approx 0.5366.

Check and common mistake avoided. The result is the probability of the second event time falling in the interval, not the probability of exactly two messages in a 45-minute interval. Converting minutes to hours before multiplying by the hourly rate prevents a factor-of-60 error.

Worked Example: Probability the Fifth Event Has Not Yet Occurred

Hypothetical scenario. A test instrument records events in a homogeneous Poisson process at a rate of \(0.8\) events per hour. Find the probability that the fifth event has not occurred by 4 hours. No interest or mortality assumption is involved.

Method and distribution. \(T_5\) has an Erlang distribution with shape \(5\) and rate \(0.8\) per hour. “Has not occurred by 4 hours” means \(T_5>4\), equivalently that at most four events have occurred by that time. Here \(\lambda t=0.8(4)=3.2\), so

P(T_5>4)=e^{-3.2}\sum_{j=0}^{4}\frac{3.2^j}{j!}.

Evaluate each term of the sum:

1+\frac{3.2^1}{1!}+\frac{3.2^2}{2!}+\frac{3.2^3}{3!}+\frac{3.2^4}{4!} =1+3.2+5.12+5.46133333+4.36906667 =19.1504.

Thus,

P(T_5>4)=e^{-3.2}(19.1504) \approx 0.04076220(19.1504) =0.78061251 \approx 0.7806.

Check and common mistake avoided. The answer is a tail probability and is greater than one-half, which is plausible because the expected time to the fifth event is \(5/0.8=6.25\) hours, longer than 4 hours. The correct complement includes counts \(0,1,2,3,\) and \(4\); stopping at three would instead calculate the probability that fewer than four events occurred.

Choosing the Correct Event and Formula

For event-time questions, first translate the wording into an event involving \(T_k\). Then select the CDF, survival probability, or difference of CDFs. The following relationships are useful:

Question wording	Probability to calculate	Equivalent count statement
By time \(t\), no later than time \(t\)	\(P(T_k\leq t)=F_{T_k}(t)\)	\(N(t)\geq k\)
After time \(t\)	\(P(T_k>t)=1-F_{T_k}(t)\)	\(N(t)\leq k-1\)
Between \(a\) and \(b\), where \(0\leq a<b\)	\(P(a<T_k\leq b)=F_{T_k}(b)-F_{T_k}(a)\)	At least \(k\) events by \(b\), but fewer than \(k\) by \(a\)

Because \(T_k\) has a continuous distribution, including or excluding a single endpoint does not change these probabilities. For example, \(P(T_k=a)=0\). Nevertheless, translating the wording carefully matters: “the \(k\)th event by time \(t\)” means at least \(k\) events, while “exactly \(k\) events by time \(t\)” describes a Poisson count and is a different event.

Common Exam Traps and Model Limits

Using the exponential model for the kth event. Only \(T_1\), the time to the first event, is exponential. For \(k>1\), use the Erlang/gamma distribution or the equivalent Poisson-count sum.
Reversing the count condition. \(T_k\leq t\) means \(N(t)\geq k\), not \(N(t)=k\). Its complement is \(N(t)\leq k-1\).
Mixing rate and scale. A gamma shape-rate model uses rate \(\lambda\); the corresponding scale is \(1/\lambda\). Substituting \(\lambda\) as the scale reverses the time scale.
Mixing time units. Convert time into the unit used by \(\lambda\) before computing \(\lambda t\).
Applying the result to a nonconstant rate. If the event rate changes over time, the homogeneous Poisson-process formula does not directly apply. A constant-rate model must be justified by the scenario or stated as an assumption.

This model is useful when a question is naturally framed as “how long until the next several events?” It also provides a cross-check: a CDF value for the kth event can be computed either from the Erlang formula or by summing Poisson probabilities for counts \(k\) and above. The two approaches must agree.

Key takeaway: Under a constant-rate Poisson process, \(T_k\) is gamma with shape \(k\) and rate \(\lambda\), and \(P(T_k\leq t)=P(N(t)\geq k)\). Translate the event carefully, match the time units to the rate, and use a CDF difference for an interval.

Check Your Understanding

For each hypothetical process, assume a homogeneous Poisson process and use the stated rate units consistently.

A sensor records events at a rate of \(3\) per day. Find the probability that the second event occurs by half a day. (Answer: \(0.4422\))
A queue receives requests at a rate of \(6\) per hour. Find the probability that the third request takes more than 20 minutes to arrive. (Answer: \(0.6767\))
A device produces signals at a rate of \(1.2\) per hour. Find the probability that the second signal arrives between 30 and 90 minutes after observation begins. (Answer: \(0.4153\))
A hypothetical monitoring process has rate \(2\) events per day. Find the probability that the fourth event occurs after day 1 but by day 3. (Answer: \(0.7059\))
A system logs failures at a rate of \(0.5\) per hour. Using the appropriate Erlang model, find the mean and variance of the waiting time to the sixth failure, in hours and hours squared, respectively. (Answer: Mean \(=12\) hours; variance \(=24\) hours squared.)

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