Repeated Convolution for Three Summands

Why Three-Summand Convolution Is Iterated

When three independent random quantities contribute to a total, the distribution of that total can be built by adding the variables in two stages. First find the distribution of a pair; then convolve that result with the distribution of the remaining variable. The method works for discrete variables, continuous variables, and sums that combine both types.

The order of the two stages can matter for the amount of work, even though it cannot change the answer. A useful choice is usually the one that produces a simple intermediate distribution, or has a smaller support or simpler density.

Definition: Let \(X\), \(Y\), and \(Z\) be mutually independent random variables, and let \(T=X+Y+Z\). For discrete variables, the convolution of \(X\) and \(Y\) has probability mass function \(p_{X+Y}(s)=\sum_x p_X(x)p_Y(s-x)\). For independent continuous variables with densities, it has density \(f_{X+Y}(s)=\int_{-\infty}^{\infty} f_X(x)f_Y(s-x)\,dx\). To obtain the distribution of \(T\), apply the relevant convolution again to the intermediate sum and the third variable.

Independence is essential: it lets us multiply the component probabilities or densities in these formulas. The formulas also require summing over all possible values, or integrating over the full real line, with zero probability or density outside each variable’s support. In practice, support information gives the finite sum or integral bounds.

Associativity and the Choice of Order

Addition of random variables is associative: \((X+Y)+Z=X+(Y+Z)\) for every outcome. For independent variables, the corresponding convolution calculations are associative as well. Thus either pair can be combined first, provided the variables being combined are independent.

Theorem: If \(X\), \(Y\), and \(Z\) are mutually independent and have densities, the density of their sum can be calculated as either \(f_{(X+Y)+Z}\) or \(f_{X+(Y+Z)}\); both give \(f_T(t)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f_X(x)f_Y(y)f_Z(t-x-y)\,dy\,dx\). For discrete variables, the analogous double sum gives the same probability mass function whichever pair is combined first.

The continuous result follows by writing the convolution in two stages. First form the density of \(X+Y\), then add \(Z\):

f_{(X+Y)+Z}(t) =\int_{-\infty}^{\infty} \left[\int_{-\infty}^{\infty} f_X(x)f_Y(u-x)\,dx\right] f_Z(t-u)\,du.

Set \(y=u-x\), so \(u=x+y\). Reordering the integrations gives

f_{(X+Y)+Z}(t) =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f_X(x)f_Y(y)f_Z(t-x-y)\,dy\,dx.

The integrand is nonnegative, so the order of integration may be interchanged. Starting instead with \(Y+Z\) gives the same double integral. For discrete variables, the same reasoning is a rearrangement of sums. This establishes why the order does not affect the answer; it does not imply that every order is equally convenient to calculate.

List the component supports.
Identify which values can occur and where each probability mass or density is zero.

Choose a first pair.
Look for a pair whose convolution has a compact support or a familiar, manageable form.

Convolve the intermediate sum with the third variable.
Use the intermediate distribution as one input to the second convolution.

Check the result.
Confirm that the support is plausible and that probabilities total one, or that the density integrates to one. If convenient, compare means.

Three Discrete Summands

For finite discrete supports, the first convolution creates a table of possible pair totals and their probabilities. The second convolution combines each intermediate total with each possible value of the third variable. Keeping zero-probability totals out of the table helps avoid unnecessary arithmetic.

Worked Example: Combining Three Discrete Contributions

Model and method. Three independent hypothetical process counts are \(X\), \(Y\), and \(Z\). Their distributions are \(P(X=0)=0.4\), \(P(X=1)=0.6\); \(P(Y=0)=0.5\), \(P(Y=1)=0.5\); and \(P(Z=0)=0.25\), \(P(Z=2)=0.75\). Find the probability mass function of \(T=X+Y+Z\). The method is two successive discrete convolutions.

First convolution: \(U=X+Y\). By independence, each pair probability is the product of its component probabilities: \(P(U=0)=0.4(0.5)=0.20\), \(P(U=1)=0.4(0.5)+0.6(0.5)=0.20+0.30=0.50\), and \(P(U=2)=0.6(0.5)=0.30\). The intermediate probabilities sum to \(0.20+0.50+0.30=1.00\).

Second convolution: \(T=U+Z\). The variable \(Z\) is zero with probability \(0.25\) and two with probability \(0.75\). For each total, add the contributions from the possible values of \(Z\):

\begin{aligned} P(T=0)&=0.20(0.25)=0.05,\\ P(T=1)&=0.50(0.25)=0.125,\\ P(T=2)&=0.30(0.25)+0.20(0.75)=0.075+0.150=0.225,\\ P(T=3)&=0.50(0.75)=0.375,\\ P(T=4)&=0.30(0.75)=0.225. \end{aligned}

The probabilities total \(0.05+0.125+0.225+0.375+0.225=1.000\). As a further check, the component means are \(E[X]=0.6\), \(E[Y]=0.5\), and \(E[Z]=1.5\), so \(E[T]=2.6\). The computed distribution gives \(0(0.05)+1(0.125)+2(0.225)+3(0.375)+4(0.225)=2.6\), as required.

Order and mistake avoided. One could first combine \(Y\) and \(Z\), then add \(X\); associativity guarantees the same distribution. A common mistake is to treat the two ways to get \(T=2\) as mutually exclusive without adding both contributions. The correct mass is \(0.30(0.25)+0.20(0.75)=0.225\), not either term alone.

Three Continuous Summands

For continuous variables, the intermediate density is itself an input to the next convolution. A useful example is the sum of three independent uniform variables, where the first pair gives a triangular density. The final density is piecewise because the integration limits change as the target total moves across the support.

Worked Example: Sum of Three Independent Uniform Variables

Model and method. Let \(X\), \(Y\), and \(Z\) be mutually independent, each with distribution \(\operatorname{Uniform}(0,1)\), so each density is 1 on \([0,1]\) and 0 elsewhere. Find the density of \(T=X+Y+Z\), then calculate \(P(T\leq 1.5)\). Convolve two variables first, then convolve their sum with the third.

First convolution. For \(V=X+Y\), the density at \(v\) is the length of the overlap of \(0\leq x\leq1\) with \(0\leq v-x\leq1\). Therefore,

f_V(v)= \begin{cases} v, & 0\leq v\leq1,\\ 2-v, & 1<v\leq2,\\ 0, & \text{otherwise}. \end{cases}

Second convolution. Since \(Z\) has density 1 on \([0,1]\), \(f_T(t)=\int_{\max(0,t-1)}^{\min(2,t)} f_V(v)\,dv\) when the lower bound does not exceed the upper bound. For \(0\leq t\leq1\), this is \(\int_0^t v\,dv=t^2/2\). For \(1<t\leq2\), split at \(v=1\):

\begin{aligned} f_T(t) &=\int_{t-1}^{1}v\,dv+\int_{1}^{t}(2-v)\,dv\\ &=\frac{1-(t-1)^2}{2} +\left(2t-\frac{t^2}{2}-\frac{3}{2}\right)\\ &=-t^2+3t-\frac{3}{2}. \end{aligned}

For \(2<t\leq3\), the integration interval lies in the descending part of \(f_V\), from \(t-1\) to 2. Thus \(\int_{t-1}^{2}(2-v)\,dv=(3-t)^2/2\). The complete density is

f_T(t)= \begin{cases} t^2/2, & 0\leq t\leq1,\\ -t^2+3t-\frac{3}{2}, & 1<t\leq2,\\ (3-t)^2/2, & 2<t\leq3,\\ 0, & \text{otherwise}. \end{cases}

To calculate \(P(T\leq1.5)\), integrate the first piece from 0 to 1 and the second from 1 to 1.5:

\begin{aligned} P(T\leq1.5) &=\int_0^1\frac{t^2}{2}\,dt +\int_1^{1.5}\left(-t^2+3t-\frac{3}{2}\right)\,dt\\ &=\frac{1}{6} +\left[-\frac{t^3}{3}+\frac{3t^2}{2}-\frac{3t}{2}\right]_{1}^{1.5}\\ &=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}. \end{aligned}

The density is symmetric about \(1.5\), which is consistent with the probability \(1/2\). As an integration check, the lower and upper pieces each have area \(1/6\), and the middle piece has area \(2/3\), giving total area 1.

Order and mistake avoided. The three variables have the same distribution, so any pair is equally simple to convolve first. A common bounds error is to integrate the middle piece from 0 to \(t\), ignoring that \(f_V(v)\) changes form at 1 and that \(v\geq t-1\) is required. Splitting the integral at 1 and using the actual overlap bounds gives the stated density.

A Mixed Sum: Choose the Simple Pair First

When a sum contains both discrete and continuous variables, a useful order is often to combine the continuous variables first. Adding a discrete variable to a continuous sum then produces a probability-weighted mixture of shifted densities. The mixture weights are the discrete probabilities; they are not additional density values to be added without weighting.

Worked Example: A Discrete Shift of a Continuous Pair

Model and method. Let \(X\) be independent of \(Y\) and \(Z\), with \(P(X=0)=0.75\) and \(P(X=2)=0.25\). Let \(Y\) and \(Z\) be independent \(\operatorname{Uniform}(0,1)\) variables, also independent of each other. Find the density of \(T=X+Y+Z\), then calculate \(P(T<2.5)\). First convolve the two uniforms; then mix the resulting density over the two values of \(X\).

Let \(V=Y+Z\). Its density is \(v\) for \(0\leq v\leq1\), \(2-v\) for \(1<v\leq2\), and zero otherwise. Conditional on \(X=0\), the total has the density of \(V\); conditional on \(X=2\), the density is shifted two units to the right. Weighting by the probabilities of those cases gives

f_T(t)= \begin{cases} 0.75t, & 0\leq t\leq1,\\ 0.75(2-t), & 1<t\leq2,\\ 0.25(t-2), & 2<t\leq3,\\ 0.25(4-t), & 3<t\leq4,\\ 0, & \text{otherwise}. \end{cases}

For the requested probability, the \(X=0\) case always has \(T=V\leq2<2.5\), so it contributes \(0.75\). In the \(X=2\) case, \(T<2.5\) means \(V<0.5\). On that interval \(f_V(v)=v\), hence

\begin{aligned} P(T<2.5) &=0.75+0.25P(V<0.5\mid X=2)\\ &=0.75+0.25\int_0^{0.5}v\,dv\\ &=0.75+0.25(0.125)=0.78125. \end{aligned}

The density integrates to \(0.75(1)+0.25(1)=1\), since each shifted triangular density has total area 1 before weighting. Also, \(E[X]=2(0.25)=0.5\), and \(E[Y+Z]=1\), so \(E[T]=1.5\). This is consistent with a distribution whose largest density contributions lie below 2.

Order and mistake avoided. Combining \(Y\) and \(Z\) first gives one triangular density, followed by two shifts. Combining a discrete variable with a continuous one first is also possible, but it creates a mixture before the final continuous convolution. A common error is to count the \(X=2\) case as impossible whenever \(T<2.5\); it remains possible when \(Y+Z<0.5\), which accounts for the additional \(0.03125\).

Common Checks and Exam Traps

Repeated convolution is a sequence of calculations, so an error in the first stage can be carried into the second. Before presenting a result, check that the intermediate support is the sum of the two component supports and that the final support extends from the sum of the three lower endpoints to the sum of the three upper endpoints. For discrete distributions, verify that the masses add to one. For continuous distributions, verify that the density is nonnegative on its support and integrates to one.

Dropping one route to a total. In discrete convolution, several combinations can produce the same total. Add all their probability contributions.
Using the wrong support bounds. Include only values for which all component probabilities or densities are nonzero.
Forgetting mixture weights. When a discrete summand shifts a continuous density, multiply each shifted density by the probability of that discrete value.
Assuming the easiest order changes the answer. Associativity changes the workload, not the distribution. The variables must still be independent for the stated product-convolution formulas.

This method is useful whenever a total is built from three independent components and the component distributions do not reduce immediately to a single familiar distribution. It also gives a practical way to control complexity: calculate an intermediate distribution that is easy to describe, then use it as one input to the final convolution. The subject is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, use successive convolution and check that the final probabilities or density are consistent with the stated support.

Independent Bernoulli variables \(A\), \(B\), and \(C\) have success probabilities \(0.5\), \(0.25\), and \(0.5\), respectively. Find the probability mass function of \(A+B+C\). (Answer: \(P(0)=0.1875,\ P(1)=0.4375,\ P(2)=0.3125,\ P(3)=0.0625\).)
Independent discrete variables have distributions \(P(X=0)=0.6,\ P(X=2)=0.4\); \(P(Y=1)=0.25,\ P(Y=3)=0.75\); and \(P(Z=0)=P(Z=1)=0.5\). Find the probability mass function of \(X+Y+Z\). (Answer: \(P(1)=0.075,\ P(2)=0.075,\ P(3)=0.275,\ P(4)=0.275,\ P(5)=0.15,\ P(6)=0.15\).)
Three independent \(\operatorname{Uniform}(0,1)\) variables are added. Find the probability that their sum is greater than 2. (Answer: \(1/6\), or \(0.1667\) to four decimals.)
\(X\) is independent of \(Y,Z\), with \(P(X=0)=0.8\) and \(P(X=3)=0.2\). The independent variables \(Y,Z\) are each \(\operatorname{Uniform}(0,2)\). Find \(P(X+Y+Z<3)\). (Answer: \(0.7000\).)
Three independent \(\operatorname{Uniform}(0,1)\) variables are added. Using the middle piece of the sum density, find \(P(1<T<2)\). (Answer: \(2/3\), or \(0.6667\) to four decimals.)

What You'll Learn