Setting Convolution Bounds from Supports

Why the Bounds Matter

For a sum $Z=X+Y$, the value of $X$ determines the value that $Y$ must take: if $Z=z$, then $Y=z-X$. A candidate value of $X$ contributes to the density of $Z$ only when both $X$ and $z-X$ lie in their respective supports. The integration limits therefore come from intersecting two support restrictions, not from copying the support of either variable.

This is a practical skill in continuous convolution. The integrand may be straightforward, yet an incorrect interval can produce a density that is too small, too large, or positive where the sum is impossible. The method developed here is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Suppose $X$ and $Y$ are independent continuous random variables with densities $f_X$ and $f_Y$, and let $Z=X+Y$. For a fixed $z$, the density of $Z$ is $$ f_Z(z)=\int_{-\infty}^{\infty} f_X(x)f_Y(z-x)\,dx. $$ If $X$ is supported on $[a,b]$ and $Y$ is supported on $[c,d]$, the integrand can be nonzero only when $a\leq x\leq b$ and $c\leq z-x\leq d$. Thus the effective limits are $$ L(z)=\max(a,z-d),\qquad U(z)=\min(b,z-c). $$ When $L(z)<U(z)$, integrate from $L(z)$ to $U(z)$; otherwise $f_Z(z)=0$.

Deriving the Intersection Rule

The two inequalities that determine the integration interval have different origins. The first, $a\leq x\leq b$, says that $x$ must be a possible value of $X$. The second comes from requiring $Y=z-x$ to lie in $[c,d]$. Rearranging that condition gives $c\leq z-x\leq d$, or $z-d\leq x\leq z-c$. Both conditions must hold at once.

Write the first support constraint.
Since $X\in[a,b]$, the integration variable must satisfy $a\leq x\leq b$.

Translate the other support.
Since $Y=z-x\in[c,d]$, rearrangement gives $z-d\leq x\leq z-c$.

Intersect the intervals.
The lower limit is the larger lower bound, $\max(a,z-d)$, and the upper limit is the smaller upper bound, $\min(b,z-c)$.

Integrate only if the intersection has positive length.
If the lower limit is at least the upper limit, there are no compatible values and the density is zero at that $z$.

This also explains why the sum has support $[a+c,b+d]$: no sum can be below $a+c$ or above $b+d$. Within that range, the active lower or upper bound can change as $z$ changes. Those changes produce the pieces of the density. The possible change points are sums of support endpoints, $a+c$, $a+d$, $b+c$, and $b+d$; sort them and determine which bound is active on each interval.

Worked Examples

Worked Example: A Sum of Two Uniform Variables

Let $X\sim U[0,4]$ and $Y\sim U[2,5]$ be independent. Find the integration interval and $f_{X+Y}(5)$, then give the full density of $Z=X+Y$. Method: intersect the support restrictions and use the constant density product. Common mistake avoided: using either variable’s support by itself as the convolution interval.

The densities are $f_X(x)=1/4$ for $0\leq x\leq4$ and $f_Y(y)=1/3$ for $2\leq y\leq5$, and each is zero elsewhere. At $z=5$, the $X$ restriction gives $0\leq x\leq4$. The $Y$ restriction gives $$ 2\leq 5-x\leq5 \quad\Longrightarrow\quad 0\leq x\leq3. $$ Intersecting these intervals gives $0\leq x\leq3$. Therefore $$ f_Z(5)=\int_0^3 \frac14\cdot\frac13\,dx =\frac{1}{12}(3-0)=\frac14=0.2500. $$

For the full density, the support of $Z$ is $[0+2,4+5]=[2,9]$. The bounds are $$ L(z)=\max(0,z-5),\qquad U(z)=\min(4,z-2). $$ The upper bound changes at $z=6$, and the lower bound changes at $z=5$. Thus the interval length is $z-2$ for $2\leq z\leq5$, $3$ for $5\leq z\leq6$, and $9-z$ for $6\leq z\leq9$. Multiplying each length by the density product $1/12$ gives $$ f_Z(z)= \begin{cases} \dfrac{z-2}{12},&2\leq z\leq5,\\[4pt] \dfrac14,&5\leq z\leq6,\\[4pt] \dfrac{9-z}{12},&6\leq z\leq9,\\[4pt] 0,&\text{otherwise}. \end{cases} $$

As a check, the areas under the three nonzero pieces are $3/8$, $1/4$, and $3/8$, respectively. Their sum is $1$, as required for a density. At $z=5$, both adjacent formulas give $3/12=1/4$, agreeing with the direct calculation.

Worked Example: A Nonuniform Density with a Changing Bound

Let $X$ and $Y$ be independent, with $f_X(x)=3x^2$ on $0\leq x\leq1$, and $f_Y(y)=1/2$ on $0\leq y\leq2$; both densities are zero elsewhere. Find the bounds and $f_{X+Y}(2.5)$, and derive the density over the full support. Method: find the active support intersection, then integrate the nonconstant product. Common mistake avoided: treating the product of the densities as constant just because one summand is uniform.

The densities are valid: $\int_0^1 3x^2\,dx=1$ and $\int_0^2(1/2)\,dy=1$. At $z=2.5$, the restrictions are $0\leq x\leq1$ and $0\leq2.5-x\leq2$. The second gives $0.5\leq x\leq2.5$. Their intersection is $0.5\leq x\leq1$. On this interval the density product is $$ f_X(x)f_Y(2.5-x)=3x^2\cdot\frac12=1.5x^2. $$ Hence $$ f_Z(2.5)=\int_{0.5}^{1}1.5x^2\,dx =\left[0.5x^3\right]_{0.5}^{1} =0.5(1-0.125)=0.4375. $$

For general $z$, the bounds are $L(z)=\max(0,z-2)$ and $U(z)=\min(1,z)$, with support $[0,3]$. For $0\leq z\leq1$, the interval is $[0,z]$, so $f_Z(z)=\int_0^z1.5x^2\,dx=0.5z^3$. For $1\leq z\leq2$, it is $[0,1]$, giving $f_Z(z)=0.5$. For $2\leq z\leq3$, it is $[z-2,1]$, giving $$ f_Z(z)=\int_{z-2}^{1}1.5x^2\,dx =0.5\left(1-(z-2)^3\right). $$ Thus $$ f_Z(z)= \begin{cases} \dfrac12z^3,&0\leq z\leq1,\\[4pt] \dfrac12,&1\leq z\leq2,\\[4pt] \dfrac12\left(1-(z-2)^3\right),&2\leq z\leq3,\\[4pt] 0,&\text{otherwise}. \end{cases} $$

The pieces agree at $z=1$ and $z=2$. Their areas are $1/8$, $1/2$, and $3/8$, adding to $1$. The formula at $z=2.5$ gives $0.5(1-0.5^3)=0.4375$, matching the direct integral.

Worked Example: Both Densities Vary Across the Active Interval

Let $X$ and $Y$ be independent, with $f_X(x)=2x$ on $0\leq x\leq1$ and $f_Y(y)=2(y-1)$ on $1\leq y\leq2$, both zero elsewhere. Find the bounds and $f_{X+Y}(2.5)$. Method: intersect both supports before substituting into the density product. Common mistake avoided: overlooking that the lower support endpoint of $Y$ imposes an upper bound on $x$.

Both functions integrate to $1$: $\int_0^1 2x\,dx=1$ and $\int_1^2 2(y-1)\,dy=1$. At $z=2.5$, $X$ requires $0\leq x\leq1$. The condition $1\leq2.5-x\leq2$ gives $0.5\leq x\leq1.5$. The intersection is therefore $0.5\leq x\leq1$, not the whole interval from either restriction alone.

Substitute $y=2.5-x$ into the product: $$ f_X(x)f_Y(2.5-x) =2x\cdot2(2.5-x-1) =4x(1.5-x)=6x-4x^2. $$ Then $$ f_Z(2.5)=\int_{0.5}^{1}(6x-4x^2)\,dx =\left[3x^2-\frac43x^3\right]_{0.5}^{1}. $$ At $x=1$, the antiderivative is $3-4/3=5/3$. At $x=0.5$, it is $3/4-1/6=7/12$. Therefore $$ f_Z(2.5)=\frac53-\frac7{12} =\frac{20}{12}-\frac7{12} =\frac{13}{12}\approx1.0833. $$

A density value can exceed $1$; it is the area over an interval, not the height at one point, that represents probability. The result is consistent with the support $[1,3]$, and the computed interval has positive length.

Boundary Changes and Reliable Checks

The maximum and minimum in the bounds are not merely notation: they specify which support restriction is active. A useful procedure is to write both candidate lower limits and both candidate upper limits before choosing an interval for a particular $z$. For a full density, solve where the competing expressions are equal. These are the points at which the formula may change.

For example, the lower bound $\max(a,z-d)$ switches from $a$ to $z-d$ when $z=a+d$. The upper bound $\min(b,z-c)$ switches from $z-c$ to $b$ when $z=b+c$. The switches may occur in either order, depending on the support widths. Do not assume the density always has a triangular shape or a constant middle section; those shapes depend on the supports and the densities.

Before accepting a result, check that it is zero outside the sum support $[a+c,b+d]$, that each nonzero integral uses a lower limit below its upper limit, and that the density integrates to $1$ when a full density has been found. At a boundary, adjacent formulas should ordinarily agree for these examples; a discrepancy is a strong signal to recheck the active limits or the integration.

A Common Exam Trap

For $X\sim U[0,4]$, $Y\sim U[2,5]$, and $z=5$, an incorrect calculation might use $2\leq x\leq4$ simply because the support of $Y$ begins at $2$ and the support of $X$ ends at $4$. That interval has length $2$, leading to the wrong value $(1/12)(2)=1/6$. The condition on $Y$ applies to $z-x$, not to $x$: $2\leq5-x\leq5$ gives $0\leq x\leq3$. Intersecting with $0\leq x\leq4$ gives $0\leq x\leq3$, so the correct density is $(1/12)(3)=1/4$.

Key takeaway: For a fixed sum $z$, keep only values of $x$ that satisfy both $x\in\operatorname{support}(X)$ and $z-x\in\operatorname{support}(Y)$. The lower limit is the maximum of the lower bounds, the upper limit is the minimum of the upper bounds, and an empty intersection contributes zero density.

Check Your Understanding

For each problem, determine the requested interval or density by intersecting the two support restrictions.

Independent $X\sim U[0,3]$ and $Y\sim U[1,2]$. At $z=2.5$, find the integration interval and $f_{X+Y}(2.5)$. (Answer: $0.5\leq x\leq1.5$; $f_{X+Y}(2.5)=1/3$.)
Independent $X\sim U[-1,2]$ and $Y\sim U[0,4]$. At $z=1$, find the integration interval and density. (Answer: $-1\leq x\leq1$; $f_{X+Y}(1)=1/6$.)
Independent $X$ with density $f_X(x)=2x$ on $[0,1]$, and $Y\sim U[0,2]$. Find the bounds and $f_{X+Y}(1.5)$. (Answer: $0\leq x\leq1$; $f_{X+Y}(1.5)=0.5000$.)
Independent $X\sim U[0,2]$ and $Y\sim U[0,1]$. Give the full density of $X+Y$, including its support. (Answer: $f_Z(z)=z/2$ for $0\leq z\leq1$, $f_Z(z)=1/2$ for $1\leq z\leq2$, $f_Z(z)=(3-z)/2$ for $2\leq z\leq3$, and $0$ otherwise.)
Independent $X$ with density $f_X(x)=x/2$ on $[0,2]$, and $Y\sim U[1,3]$. At $z=3.5$, find the integration interval and density. (Answer: $0.5\leq x\leq2$; $f_{X+Y}(3.5)=15/32=0.4688$.)

What You'll Learn