Hypoexponential Laws from Unequal-Rate Exponential Sums

From Exponential Waiting Times to a New Sum Distribution

A single exponential random variable models a waiting time with a constant rate. When a process must pass through several independent stages, with a different rate at each stage, the total waiting time is a sum of exponential variables with unequal rates. It is generally not exponential or gamma, but its distribution has an explicit form obtained by partial fractions.

The resulting law is called a hypoexponential distribution. Its density is a finite sum of exponential terms. Some coefficients in that sum can be negative, so the expression may not look like a density term by term. The full expression is nevertheless nonnegative: a transform argument below identifies it with the probability law of the sum.

Definition: Let \(X_1,\ldots,X_m\) be independent exponential random variables, where \(X_i\) has rate \(\lambda_i>0\). If the rates are pairwise distinct, the sum \(S=X_1+\cdots+X_m\) has a hypoexponential distribution. An exponential variable with rate \(\lambda_i\) has density \(f_i(x)=\lambda_i e^{-\lambda_i x}\) for \(x\geq 0\), and zero otherwise.

Partial Fractions Give the Density

For \(s\geq 0\), the Laplace transform of \(X_i\) is \(E[e^{-sX_i}]=\lambda_i/(\lambda_i+s)\). Independence makes the transform of the sum a product:

E[e^{-sS}]=\prod_{i=1}^{m}\frac{\lambda_i}{s+\lambda_i}.

Because the rates are distinct, the denominator has distinct linear factors. Its partial-fraction decomposition has one term for each rate. Define

w_i=\prod_{\substack{1\leq j\leq m\\j\neq i}} \frac{\lambda_j}{\lambda_j-\lambda_i}, \qquad i=1,\ldots,m.

Then the transform can be written as

\prod_{i=1}^{m}\frac{\lambda_i}{s+\lambda_i} =\sum_{i=1}^{m}w_i\frac{\lambda_i}{s+\lambda_i}.

To verify the coefficient \(w_i\), multiply the decomposition by \(s+\lambda_i\) and set \(s=-\lambda_i\). On the product side, the result is \(\lambda_i\prod_{j\neq i}\lambda_j/(\lambda_j-\lambda_i)\), while the corresponding term on the sum side is \(w_i\lambda_i\). Equating these gives the stated formula.

Theorem: For pairwise distinct positive rates \(\lambda_1,\ldots,\lambda_m\), the sum \(S=X_1+\cdots+X_m\) has density and survival function

f_S(t)=\sum_{i=1}^{m}w_i\lambda_i e^{-\lambda_i t}, \qquad P(S>t)=\sum_{i=1}^{m}w_i e^{-\lambda_i t}, \qquad t\geq 0,

where \(w_i=\prod_{j\neq i}\lambda_j/(\lambda_j-\lambda_i)\).

A justification needs care: the partial-fraction expression is not automatically a probability density, because individual \(w_i\) can be negative. Let \(g(t)=\sum_i w_i\lambda_i e^{-\lambda_i t}\) for \(t\geq0\). Since this is a finite sum of integrable exponential functions, the signed measure \(g(t)\,dt\) has finite total variation. Its Laplace transform is the partial-fraction expression above, which equals the transform of the probability law of \(S\).

For completeness, Laplace transforms determine finite signed measures on \([0,\infty)\). If two such measures have the same transform for all \(s\geq0\), push their difference forward under \(t\mapsto e^{-t}\), obtaining a finite signed measure on \([0,1]\) with no mass at zero. The transform equalities at integer \(s=k\geq0\) say that the integral of \(y^k\) against this difference is zero for every \(k\). Thus its integral against every polynomial is zero. Polynomials are dense in the continuous functions on \([0,1]\), so the integral against every continuous function is zero; uniqueness of finite signed Borel measures then makes the difference measure zero. The map \(t\mapsto e^{-t}\) is one-to-one on \([0,\infty)\), so the original measures are equal.

Applying this uniqueness result identifies \(g(t)\,dt\) with the law of \(S\). In particular, \(g\geq0\) almost everywhere and integrates to one. Integrating its terms from \(t\) to infinity gives the survival formula, since \(\int_t^\infty \lambda_i e^{-\lambda_i u}\,du=e^{-\lambda_i t}\). This proves the theorem. \(\square\)

Worked Forms and Calculations

Worked Example: Two Stages with Rates 1 and 2

Let \(X_1\sim\operatorname{Exp}(1)\) and \(X_2\sim\operatorname{Exp}(2)\) be independent. For the rate 1 term, \(w_1=2/(2-1)=2\). For the rate 2 term, \(w_2=1/(1-2)=-1\). Therefore

f_S(t)=2e^{-t}-2e^{-2t}, \qquad P(S>t)=2e^{-t}-e^{-2t}, \qquad t\geq0.

The negative coefficient in \(w_2\) does not make the density negative. In fact, \(f_S(t)=2e^{-t}(1-e^{-t})\), which is nonnegative for \(t\geq0\). At \(t=0\), the density is zero, as expected for the sum of two continuously distributed positive waiting times. The survival function is one at zero because \(2-1=1\), and it tends to zero as \(t\) tends to infinity.

Worked Example: Three Stages with Rates 1, 2, and 4

Let the three independent rates be \(1,2,4\). Compute each coefficient directly:

w_1=\frac{2}{2-1}\frac{4}{4-1}=\frac{8}{3}, \qquad w_2=\frac{1}{1-2}\frac{4}{4-2}=-2, \qquad w_3=\frac{1}{1-4}\frac{2}{2-4}=\frac{1}{3}.

The density and survival function are consequently

f_S(t)=\frac{8}{3}e^{-t}-4e^{-2t}+\frac{4}{3}e^{-4t}, \qquad P(S>t)=\frac{8}{3}e^{-t}-2e^{-2t}+\frac{1}{3}e^{-4t}.

A direct check of the density's sign is possible by setting \(y=e^t\), so \(y\geq1\):

f_S(t)=\frac{4}{3}e^{-4t}(2e^{3t}-3e^{2t}+1) =\frac{4}{3}e^{-4t}(e^t-1)^2(2e^t+1)\geq0.

Also \(w_1+w_2+w_3=8/3-2+1/3=1\), so the survival function equals one at \(t=0\). These checks agree with the probability interpretation established by the theorem.

When Unequal Rates Become Equal

The distinct-rate formula has denominators \(\lambda_j-\lambda_i\), so its individual coefficients become undefined as rates coincide. This is a feature of the partial-fraction representation, not a discontinuity in the distribution of the sum. When all rates approach a common positive rate, the sum approaches a gamma law with integer shape.

Theorem: Suppose, for each \(r\), \(X_{1,r},\ldots,X_{m,r}\) are independent exponentials with rates \(\lambda_{1,r},\ldots,\lambda_{m,r}\), and \(\lambda_{i,r}\to\lambda>0\) for every \(i\). Then \(S_r=\sum_iX_{i,r}\) converges in distribution to a gamma random variable with shape \(m\) and rate \(\lambda\), whose density is

f(t)=\frac{\lambda^m}{(m-1)!}t^{m-1}e^{-\lambda t}, \qquad t\geq0.

To prove this, take independent unit-rate exponential variables \(E_1,\ldots,E_m\) and represent \(X_{i,r}=E_i/\lambda_{i,r}\). Each \(X_{i,r}\) has the required exponential distribution, and for every outcome with finite \(E_i\), \(\sum_i E_i/\lambda_{i,r}\to\sum_i E_i/\lambda\). Thus the coupled sums converge almost surely, and hence in distribution.

The limiting sum is a sum of \(m\) independent exponentials with common rate \(\lambda\). Its density can be obtained by induction using convolution. For \(m=1\) it is \(\lambda e^{-\lambda t}\). If the formula holds for \(m\), convolution with another exponential gives, for \(t\geq0\),

\int_0^t \frac{\lambda^m}{(m-1)!}u^{m-1}e^{-\lambda u} \lambda e^{-\lambda(t-u)}\,du = \frac{\lambda^{m+1}e^{-\lambda t}}{(m-1)!} \int_0^t u^{m-1}\,du = \frac{\lambda^{m+1}}{m!}t^m e^{-\lambda t}.

This is the claimed density at shape \(m+1\), completing the induction. The equal-rate limit is therefore the Erlang distribution, the gamma distribution with positive integer shape. \(\square\)

Worked Example: Two Rates Coalescing at 3

For two distinct rates \(a\) and \(b\), the density formula reduces to

f_{a,b}(t)=\frac{ab}{b-a}\left(e^{-at}-e^{-bt}\right), \qquad t\geq0.

Take \(a=3\) and let \(b\to3\). Write \(b=3+h\). Then \(e^{-3t}-e^{-(3+h)t}=e^{-3t}(1-e^{-ht})\), and \((1-e^{-ht})/h\to t\). Also \(ab/(b-a)=3(3+h)/h\). Therefore, for each \(t\geq0\),

f_{3,3+h}(t) =3(3+h)e^{-3t}\frac{1-e^{-ht}}{h} \longrightarrow 9t e^{-3t}.

The limit is the gamma density with shape 2 and rate 3. It is normalized: \(\int_0^\infty 9t e^{-3t}\,dt=9(1/3)^2=1\). This calculation also shows how two terms whose coefficients separately diverge can combine to give a finite limit.

Interpretation and Common Pitfall

The hypoexponential law is useful when a total completion time is formed by passing through a fixed sequence of independent stages with unequal exponential rates. Its transform is simple, and the partial-fraction form gives explicit survival probabilities and densities. As checks on calculations, the survival expression must equal one at zero and tend to zero at infinity. Independence also gives \(E[S]=\sum_i1/\lambda_i\) and \(\operatorname{Var}(S)=\sum_i1/\lambda_i^2\), since an exponential variable of rate \(\lambda_i\) has mean \(1/\lambda_i\) and variance \(1/\lambda_i^2\).

Common mistake: A negative partial-fraction coefficient does not by itself invalidate the formula, nor does it mean the sum has a negative density. The terms form one combined expression; positivity follows because that expression represents the law of the sum. Conversely, merely matching a transform to a signed sum is not enough if uniqueness is stated only for probability measures. The correct justification uses uniqueness for finite signed measures, as above, and then concludes that the signed density expression is nonnegative almost everywhere.

A second pitfall is to substitute equal rates directly into the distinct-rate coefficients. The denominators then vanish, even though the distribution has a well-defined limit. Use a limiting argument for the combined distribution, or work directly with the common-rate gamma law.

Check Your Understanding

Use the formulas and arguments developed here to answer the following questions.

For independent exponentials with rates 2 and 5, calculate both partial-fraction coefficients and write the density of their sum.
Why does a negative coefficient in the partial-fraction representation not imply that the sum has a negative density?
State the survival function for a sum with three pairwise distinct positive rates, using the coefficient notation in this tutorial.
What gamma distribution is obtained when four rates all tend to 2?
Why is uniqueness for finite signed measures relevant when identifying the density from the partial-fraction expression?