Sums of Exponentials with Equal Rates

Why Equal-Rate Exponential Sums Have a Useful Form

An exponential random variable often represents a waiting time until an event. A total time made up of several successive stages is therefore a sum of waiting times. If the stage times are independent and have the same exponential rate, their sum has a particularly convenient distribution: the Erlang distribution. Its density and event probabilities can be calculated directly, without setting up a new convolution integral for every problem.

The conditions matter. The number of stages must be a fixed positive integer, the stage times must be independent, and each must have the same rate. If the rates differ, or the stages are dependent, the formulas below do not apply as stated.

Definition: Let \(X_1,\ldots,X_n\) be independent exponential random variables, each with rate \(\lambda>0\), so that \(f_{X_j}(x)=\lambda e^{-\lambda x}\) for \(x\geq 0\). The sum \(T=X_1+\cdots+X_n\) has an Erlang distribution with shape \(n\) and rate \(\lambda\). Its support is \(t\geq 0\).

The rate \(\lambda\) is the exponential rate, not the mean. Each stage has mean \(1/\lambda\), so the total has mean \(n/\lambda\). Units must match: if time is measured in hours, \(\lambda\) is expressed per hour and \(\lambda t\) is dimensionless.

Deriving the Density by Convolution

For \(n=1\), the sum is just \(X_1\), with density \(\lambda e^{-\lambda t}\). To see how the pattern develops, first add two independent stages. For \(t\geq 0\), convolution gives

f_{X_1+X_2}(t) =\int_0^t \lambda e^{-\lambda u}\lambda e^{-\lambda(t-u)}\,du =\lambda^2 e^{-\lambda t}\int_0^t du =\lambda^2t e^{-\lambda t}.

The common rate makes the exponential factor constant with respect to the integration variable: \(e^{-\lambda u}e^{-\lambda(t-u)}=e^{-\lambda t}\). The remaining integral contributes a power of \(t\). Repeating this argument gives the general density.

Theorem: For a fixed positive integer \(n\), the sum \(T\) of \(n\) independent exponential random variables with common rate \(\lambda\) has density

f_T(t)= \begin{cases} \dfrac{\lambda^n t^{n-1}e^{-\lambda t}}{(n-1)!},&t\geq 0,\\[6pt] 0,&t<0. \end{cases}

Here is the induction step that justifies the formula. Suppose the sum of \(n-1\) stages has density \(\lambda^{n-1}u^{n-2}e^{-\lambda u}/(n-2)!\). Convolving it with one more independent exponential stage gives

\begin{aligned} f_T(t) &=\int_0^t \frac{\lambda^{n-1}u^{n-2}e^{-\lambda u}}{(n-2)!} \lambda e^{-\lambda(t-u)}\,du\\ &=\frac{\lambda^n e^{-\lambda t}}{(n-2)!} \int_0^t u^{n-2}\,du\\ &=\frac{\lambda^n t^{n-1}e^{-\lambda t}}{(n-1)!}. \end{aligned}

This proves the result from the \(n=1\) base case by induction. The lower integration limit is zero because each stage time is nonnegative, and the upper limit is \(t\) because the first \(n-1\) stages cannot total more than \(t\) if the full sum is \(t\).

Worked Example: Deriving a Three-Stage Density and Threshold Probability

A hypothetical process has three independent stages. Each stage duration is exponential with rate \(2\) per hour. Let \(T\) be the total duration. Find its density and calculate the probability that the process finishes within one hour.

Method. The fixed number of stages is \(n=3\), and the common rate is \(\lambda=2\) per hour. Independence and the common rate allow the Erlang formulas.

Substitute \(n=3\) and \(\lambda=2\) in the density:

f_T(t)=\frac{2^3t^{3-1}e^{-2t}}{(3-1)!} =\frac{8t^2e^{-2t}}{2} =4t^2e^{-2t},\qquad t\geq 0.

For the cumulative distribution function, use \(F_T(t)=1-e^{-\lambda t}\sum_{k=0}^{n-1}(\lambda t)^k/k!\). At \(t=1\) hour, \(\lambda t=2\), so

\begin{aligned} P(T\leq 1) &=1-e^{-2}\left(1+2+\frac{2^2}{2!}\right)\\ &=1-5e^{-2}\\ &\approx 1-5(0.1353353)\\ &\approx 0.3233. \end{aligned}

Thus the probability of finishing within one hour is \(0.3233\), to four decimal places. As a check, the complementary probability of taking more than one hour is \(5e^{-2}\approx0.6767\), and the two probabilities sum to \(1.0000\). This calculation avoids the common mistake of using the density value at one hour as though it were a probability.

Cumulative Distribution and Survival Formulas

Integrating the density gives the cumulative distribution function (cdf). For a nonnegative sum, \(F_T(t)=P(T\leq t)\). The Erlang cdf and its complement are

F_T(t)= \begin{cases} 0,&t<0,\\[2pt] 1-e^{-\lambda t}\displaystyle\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!},&t\geq 0, \end{cases} \qquad P(T>t)=e^{-\lambda t}\displaystyle\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!} \quad(t\geq 0).

One way to verify the cdf formula is to differentiate its proposed survival expression. Set \(Q(t)=e^{-\lambda t}\sum_{k=0}^{n-1}(\lambda t)^k/k!\). Differentiation makes all but the final term cancel:

Q'(t) =-\lambda e^{-\lambda t}\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!} +\lambda e^{-\lambda t}\sum_{k=0}^{n-2}\frac{(\lambda t)^k}{k!} =-\frac{\lambda^n t^{n-1}e^{-\lambda t}}{(n-1)!} =-f_T(t).

Also \(Q(0)=1\). Therefore \(1-Q(t)\) is zero at \(t=0\) and has derivative \(f_T(t)\), as a cdf must. The finite sum ends at \(n-1\), a detail that is particularly important in threshold calculations.

For interval probabilities, subtract cdf values: \(P(a<T\leq b)=F_T(b)-F_T(a)\) when \(0\leq a\leq b\). The distribution is continuous, so including or excluding a single endpoint does not change the probability. When the event is \(T>t\), the survival formula is often shorter than calculating \(1-F_T(t)\) from scratch.

Worked Example: Probability That a Three-Stage Total Falls in an Interval

A hypothetical inspection sequence has three independent durations, each exponential with rate \(0.5\) per hour. Find the probability that the total duration \(T\) is between 2 and 6 hours.

Method. Here \(n=3\) and \(\lambda=0.5\) per hour. Evaluate the Erlang cdf at both endpoints, using \(\lambda t\) as the dimensionless argument in the finite sum.

At 6 hours, \(\lambda t=0.5(6)=3\):

F_T(6)=1-e^{-3}\left(1+3+\frac{3^2}{2!}\right) =1-8.5e^{-3} \approx 0.5768.

At 2 hours, \(\lambda t=0.5(2)=1\):

F_T(2)=1-e^{-1}\left(1+1+\frac{1^2}{2!}\right) =1-2.5e^{-1} \approx 0.0803.

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

P(2<T\leq 6)=F_T(6)-F_T(2) \approx 0.5768-0.0803 =0.4965.

The probability is \(0.4965\), to four decimal places. It is between zero and one, and it is smaller than \(F_T(6)\), as an interval contained in \([0,6]\) must be. Using the survival probabilities instead gives the same result: \(P(T>2)-P(T>6)\).

Rate Units, Moments, and Practical Checks

The product \(\lambda t\) must use consistent units. For example, a rate of \(6\) per day must be paired with time in days, not hours. Convert either the rate or the time before substituting. The density also carries units: if \(t\) is measured in days, \(f_T(t)\) is measured per day, while a cdf or event probability is dimensionless.

The mean and variance can provide useful checks. An exponential stage with rate \(\lambda\) has mean \(1/\lambda\) and variance \(1/\lambda^2\). By linearity of expectation and independence of the stages,

E[T]=\frac{n}{\lambda}, \qquad \operatorname{Var}(T)=\frac{n}{\lambda^2}.

These summaries do not replace the cdf when a threshold probability is required. They can, however, flag a unit or parameter error: the mean total duration should be \(n\) times the mean duration of one stage.

Worked Example: Applying the Survival Formula With a Different Number of Stages

A hypothetical facility processes four independent items in sequence. Each processing time is exponential with rate \(6\) per day. Find the probability that the combined time exceeds half a day, and find the density at half a day.

Method. Use \(n=4\), \(\lambda=6\) per day, and \(t=0.5\) day. The dimensionless product is \(\lambda t=6(0.5)=3\).

The survival formula ends at \(n-1=3\):

\begin{aligned} P(T>0.5) &=e^{-3}\left(1+3+\frac{3^2}{2!}+\frac{3^3}{3!}\right)\\ &=e^{-3}(1+3+4.5+4.5)\\ &=13e^{-3}\\ &\approx 0.6472. \end{aligned}

For the density, substitute into the density formula:

f_T(0.5) =\frac{6^4(0.5)^3e^{-3}}{3!} =\frac{1296(0.125)}{6}e^{-3} =27e^{-3} \approx 1.3443\text{ per day}.

The probability that the total is at most half a day is \(1-0.6472=0.3528\), to four decimal places. The density value is not a probability and need not be less than one; its units are per day. The two results are consistent with the support and units of the model.

Common Exam Trap: The Final Index

A frequent error is to include one extra term in the survival sum. For a sum of \(n\) exponential stages, the terms run from \(k=0\) through \(k=n-1\), not through \(k=n\). For the three-stage example with rate \(2\) per hour at \(t=1\) hour, the correct calculation is

P(T>1)=e^{-2}\left(1+2+\frac{2^2}{2!}\right)=5e^{-2}\approx0.6767.

The incorrect extra-term calculation would be

e^{-2}\left(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}\right) =\frac{19}{3}e^{-2} \approx0.8571.

That value is not the survival probability for three stages; it adds a term belonging to a different shape. Check the indexing against the density: its power is \(t^{n-1}\), and the cdf and survival sum likewise stop at \(n-1\). Also check that a stated exponential parameter is a rate before using it as \(\lambda\); a mean waiting time must first be converted to its reciprocal rate.

Key takeaway: For a fixed number \(n\) of independent exponential stages with common rate \(\lambda\), use the Erlang density for pointwise density calculations and the finite-sum cdf or survival formula for event probabilities. Confirm both the common-rate assumption and the time units before substituting.

Check Your Understanding

For each problem, state or use the common rate in the time units given. Probabilities are rounded to four decimal places.

Two independent exponential durations each have rate \(3\) per hour. What is the probability their sum exceeds 1 hour? (Answer: \(4e^{-3}\approx0.1991\))
Three independent exponential durations each have rate \(1\) per minute. What is the probability their sum is at most 2 minutes? (Answer: \(1-5e^{-2}\approx0.3233\))
Four independent exponential durations each have rate \(2\) per hour. What is the probability their sum is less than 1 hour? (Answer: \(1-\frac{19}{3}e^{-2}\approx0.1429\))
Three independent exponential durations each have rate \(0.25\) per minute. Find the probability their sum is between 4 and 8 minutes. (Answer: \(2.5e^{-1}-5e^{-2}\approx0.2430\))
Five independent exponential durations each have rate \(10\) per hour. Find the probability their sum is at most half an hour. (Answer: \(1-65.375e^{-5}\approx0.5595\))

What You'll Learn