Sum Distributions by CDF Integration

From a Sum Event to a Region

For $S=X+Y$, the event $S\le t$ is the set of all pairs $(x,y)$ satisfying $x+y\le t$. In the plane, this is the half-plane below the line $y=t-x$. The probability of the sum being at most $t$ is therefore the probability mass of the joint distribution lying in that half-plane.

When $(X,Y)$ has a joint density, that probability is an area integral of the density over the part of the half-plane where the density can be nonzero. The essential task is not just drawing the line: the integration limits must also respect the support of the joint density. In particular, the support need not be the whole plane, a rectangle, or a product set.

Definition: Suppose $(X,Y)$ has joint density $f_{X,Y}(x,y)$, and write $S=X+Y$. The cumulative distribution function of $S$ is $$ F_S(t)=P(S\le t)=P(X+Y\le t) =\iint_{\{(x,y):\,x+y\le t\}} f_{X,Y}(x,y)\,dx\,dy. $$ The integral is over the entire indicated region; points outside the support contribute zero because the joint density is zero there.

This formula applies whether or not $X$ and $Y$ are independent. Independence is a special case in which the joint density factors as $f_{X,Y}(x,y)=f_X(x)f_Y(y)$ almost everywhere. Without independence, the joint density must be used as given.

The Region-Integration Formula

Theorem: If $(X,Y)$ has a joint density $f_{X,Y}$, then for every real $t$, $$ F_S(t)=\iint_{\{x+y\le t\}} f_{X,Y}(x,y)\,dx\,dy. $$ More generally, if $D$ is a measurable set outside which the density is zero almost everywhere, the integral may be restricted to $D$.

Proof. By the definition of a joint density, the probability that $(X,Y)$ belongs to any measurable set $A$ is the integral of $f_{X,Y}$ over $A$. Take $A=\{(x,y):x+y\le t\}$. Since $S=X+Y$, the event $\{S\le t\}$ is exactly the event $\{(X,Y)\in A\}$. Thus $$ P(S\le t)=P((X,Y)\in A)=\iint_A f_{X,Y}(x,y)\,dx\,dy. $$ If the density vanishes almost everywhere outside $D$, integrating over $A$ gives the same result as integrating over $A\cap D$. $\square$

To evaluate the double integral, it is often convenient to integrate in vertical slices: fix $x$, then allow $y$ to range over the support and below the line $y=t-x$. For a nonrectangular support, the available $y$-values can depend on $x$, so those limits must be taken from the support as well as from the event.

Vertical-slice formula: Let $D$ be a measurable support region for the joint density, and for each $x$ define its vertical section by $D_x=\{y:(x,y)\in D\}$. Then $$ F_S(t)=\int_{\mathbb R}\left(\int_{D_x\cap(-\infty,t-x]} f_{X,Y}(x,y)\,dy\right)dx. $$ An integral over an empty section is zero.

Justification. Write the region indicator as $\mathbf{1}_{\{x+y\le t\}}$, which is measurable because $(x,y)\mapsto x+y$ is continuous. The integrand $f_{X,Y}(x,y)\mathbf{1}_{\{x+y\le t\}}$ is nonnegative and measurable. Tonelli's theorem therefore permits integration first with respect to $y$, then with respect to $x$. At a fixed $x$, the event condition is $y\le t-x$; intersecting this with the support section $D_x$ gives exactly the inner integral above. $\square$

For example, if the density is supported on the rectangle $a\le x\le b,\ c\le y\le d$, the vertical-slice limits can be written as $$ F_S(t)=\int_a^{\min(b,t-c)} \left(\int_c^{\min(d,t-x)} f_{X,Y}(x,y)\,dy\right)dx, $$ when $t-c\ge a$; if $t-c<a$, the probability is zero. The outer upper limit ensures that $t-x\ge c$, so the inner interval is not empty. These limits still work if the density is not a product density.

Worked CDF Integrals

Worked Example: Two Independent Exponential Variables

Let $X$ and $Y$ be independent exponential random variables, each with rate $1$. Their joint density is $e^{-x-y}$ for $x\ge0,\ y\ge0$, and zero elsewhere. Find $F_S(t)$ for $t\ge0$, where $S=X+Y$.

The support requires $x\ge0$ and $y\ge0$, while the event requires $y\le t-x$. This is possible only for $0\le x\le t$. Thus $$ F_S(t)=\int_0^t\int_0^{t-x}e^{-x-y}\,dy\,dx. $$

Integrate first with respect to $y$: $$ \int_0^{t-x}e^{-x-y}\,dy =e^{-x}\left(1-e^{-(t-x)}\right) =e^{-x}-e^{-t}. $$ Therefore $$ F_S(t)=\int_0^t(e^{-x}-e^{-t})\,dx =1-e^{-t}-te^{-t} =1-(1+t)e^{-t}. $$ For $t<0$, the sum of two nonnegative variables cannot be negative, so $F_S(t)=0$. At $t=0$, the formula for $t\ge0$ also gives zero, as required.

Worked Example: A Triangular Support

Suppose the joint density is $f_{X,Y}(x,y)=2$ on $0<x<y<1$, and zero elsewhere. First check that it is a density: the support has area $1/2$, so its total integral is $2(1/2)=1$. Find $P(X+Y\le 1/2)$.

For a fixed $x$, the support requires $y>x$, while the event requires $y\le 1/2-x$. These conditions can both hold only if $x<1/2-x$, or $0<x<1/4$. Hence the correct limits are $$ P(X+Y\le 1/2) =\int_0^{1/4}\int_x^{1/2-x}2\,dy\,dx. $$

The inner integral is $2(1/2-2x)=1-4x$. Thus $$ P(X+Y\le 1/2) =\int_0^{1/4}(1-4x)\,dx =\left[x-2x^2\right]_0^{1/4} =\frac14-\frac18 =\frac18. $$ The lower $y$-limit is $x$, not zero: the triangular support excludes points with $y\le x$.

Worked Example: A Dependent Pair on a Rectangle

Let the joint density be $f_{X,Y}(x,y)=x+y$ for $0<x<1,\ 0<y<1$, and zero elsewhere. It is normalized because $$ \int_0^1\int_0^1(x+y)\,dy\,dx =\int_0^1\left(x+\frac12\right)dx=1. $$ Find $P(X+Y\le1)$.

For each $x$ between 0 and 1, the event and support together give $0<y\le1-x$. Therefore $$ P(X+Y\le1)=\int_0^1\int_0^{1-x}(x+y)\,dy\,dx. $$

The inner integral is $$ \int_0^{1-x}(x+y)\,dy =x(1-x)+\frac{(1-x)^2}{2} =\frac{1-x^2}{2}. $$ Consequently, $$ P(X+Y\le1) =\int_0^1\frac{1-x^2}{2}\,dx =\frac12\left(1-\frac13\right) =\frac13. $$ The density does not factor into a function of $x$ times a function of $y$, so the calculation uses the joint density directly rather than replacing it with a product of marginal densities.

Choosing Limits Without Losing Support

A reliable setup begins with the support and the event as separate restrictions. For vertical slices, fix $x$, determine which $y$-values are allowed by the support, and then intersect those values with $y\le t-x$. Next identify which $x$-values give a nonempty intersection. If the resulting boundary changes across the range of $x$, split the integral at the point where it changes.

Write the support.
Record precisely where the joint density is nonzero, including any dependence of one coordinate's limits on the other.

Write the event boundary.
For a sum threshold, solve $x+y\le t$ for the inner variable, such as $y\le t-x$.

Intersect the restrictions.
At each fixed outer coordinate, keep only values allowed by both the support and the event; then determine the outer range where that intersection is nonempty.

Integrate and check.
Evaluate the integral, check that the result lies between zero and one, and verify endpoint behavior when it is a CDF calculation.

The support geometry is central: knowing only the marginal distributions is generally not enough to carry out this calculation, because the joint distribution determines how probability is arranged across pairs $(x,y)$. For independent variables, factorization simplifies the integrand, but it does not remove the need to set limits from both the event and the support.

Common Mistake: Integrating Over the Wrong Region

A tempting but incorrect setup is to integrate over all $x$ in the support and use $y$ from the support's lower boundary to $t-x$, without checking whether $t-x$ is above that lower boundary. This can create negative-length intervals or include points outside the support. Another error is to use only the condition $x+y\le t$ and forget restrictions such as $y>x$ in a triangular support.

The correction is to intersect the event region with the support before integrating. In the triangular example, $y$ had to satisfy both $y>x$ and $y\le1/2-x$; that intersection forced $x<1/4$. In the exponential example, nonnegative support forced $0\le x\le t$. Writing down both restrictions before calculating prevents these errors.

Key takeaway: The CDF of $X+Y$ is the joint probability under the line $x+y=t$. Correct integration limits describe the intersection of that half-plane with the actual joint support.

Check Your Understanding

For each question, identify the event region and pay particular attention to the support restrictions.

If a joint density is supported on $0<x<2,\ 0<y<3$, what vertical-slice limits describe $P(X+Y\le t)$ for $0<t<2$?
For support $0<x<y<1$, explain why the inner integral for a fixed $x$ cannot begin at $y=0$.
Does the region-integration formula require $X$ and $Y$ to be independent? State the assumption it does require.
For a density supported on $1<x<2,\ 1<y<2$, what is $P(X+Y\le t)$ when $t\le2$?
Describe one check that can reveal an incorrect CDF integration setup or calculation.