Support Geometry for Sums, Products, and Ratios

Why Support Geometry Comes First

A density calculation for a sum, product, or ratio is an integral over a set of input values that can produce the requested output. Getting that set right is a geometric problem before it is an integration problem. If the input support has gaps or crosses zero, bounds copied from a familiar formula can include impossible values or omit possible ones.

The relevant object is usually the joint support of the inputs, not just their separate ranges. Dependence can restrict which pairs of values occur together. For example, knowing that both $X$ and $Y$ take values in $[0,1]$ does not imply that every pair in the square $[0,1]^2$ is possible.

Definition: The (topological) support $S$ of a random vector $(X,Y)$ is the set of points $(x,y)$ for which every open neighborhood has positive probability. The attainable set of a transformation $g$ is $g(S)$, with points excluded wherever $g$ is undefined. For a continuous transformation on its domain, the support of the transformed variable is the closure of the image of the input support on that domain.

The closure matters when the image is not closed. A ratio can approach arbitrarily large values as its denominator approaches zero without ever attaining an infinite value. For a ratio $X/Y$, the expression is defined as a random variable only if $P(Y=0)=0$, unless a separate convention is explicitly imposed on that event.

Support Images and Density Fibers

For a point $(x,y)$ in the joint support, the output values are $x+y$, $xy$, and $x/y$, respectively, when $y\ne 0$. Thus the support geometry is obtained by mapping the joint region through the relevant function. In density calculations, each output value corresponds to a fiber: the set of input points that map to it.

Theorem: Suppose $(X,Y)$ has joint density $f_{X,Y}$. Then, for almost every $z$, density versions for the sum $S=X+Y$, product $P=XY$, and ratio $R=X/Y$ are $$ f_S(z)=\int_{\mathbb R} f_{X,Y}(x,z-x)\,dx, $$ $$ f_P(z)=\int_{x\ne 0} \frac{1}{|x|}f_{X,Y}\left(x,\frac{z}{x}\right)\,dx, \qquad f_R(z)=\int_{y\ne 0}|y|f_{X,Y}(zy,y)\,dy. $$ In each integral, the joint density is zero outside its support, so only input points in that support contribute.

These formulas do not assume independence. If $X$ and $Y$ are independent, the joint density can additionally be written as $f_X(x)f_Y(y)$. The factors and integration regions follow from the inverse transformations, rather than from a rule that can be applied without checking the support.

Proof. For the sum, use coordinates $(x,z)=(x,x+y)$, whose inverse is $(x,y)=(x,z-x)$ and whose absolute Jacobian determinant is 1. Integrating the resulting joint density of $(X,S)$ over $x$ gives the sum formula.

For the product, on the region $x\ne0$, use $(x,z)=(x,xy)$. The inverse is $(x,y)=(x,z/x)$, and its absolute Jacobian determinant is $1/|x|$. The line $x=0$ has probability zero under a joint density. Integrating over $x$ gives the product formula for almost every $z$.

For the ratio, on $y\ne0$, use $(z,y)=(x/y,y)$. Its inverse is $(x,y)=(zy,y)$, with absolute Jacobian determinant $|y|$. Since a joint density gives probability zero to the line $y=0$, integrating over $y$ gives the ratio formula for almost every $z$. In all three cases, change of variables followed by integration of a nonnegative density justifies the marginalization. $\square$

The formulas describe the density calculation, but the support geometry tells where the integrands can be nonzero. For example, at a fixed product value $z$, the product fiber is $(x,z/x)$ for $x\ne0$; the integral must include only those $x$ for which that pair belongs to the joint support.

Worked Examples with Disconnected and Sign-Changing Supports

Worked Example: A Sum with a Gap in Its Support

Let $X$ have density $1/2$ on $[0,1]\cup[3,4]$, and let $Y$ be independent and uniform on $[0,1]$. Find the support of $S=X+Y$ and its density.

If $X\in[0,1]$, then $S\in[0,2]$; if $X\in[3,4]$, then $S\in[3,5]$. These intervals do not overlap, so the attainable support is $[0,2]\cup[3,5]$. The gap $(2,3)$ cannot contain a sum.

For the first component of $X$, the density integral is half the length of $[0,1]\cap[z-1,z]$, because $Y=z-X$ must be in $[0,1]$. That length increases from $z$ to 1 and then decreases from $2-z$ to 0. The second component has the same shape shifted by 3. Thus

f_S(z)= \begin{cases} z/2, & 0<z<1,\\ (2-z)/2, & 1\le z<2,\\ 0, & 2<z<3,\\ (z-3)/2, & 3<z<4,\\ (5-z)/2, & 4\le z<5,\\ 0, & \text{otherwise}. \end{cases}

The value at any single endpoint can be chosen without changing the density distribution. Each triangular component has area $1/2$: the two together integrate to 1, as required.

Worked Example: A Product with Both Signs and a Gap

Let $X$ be uniform on $[-2,-1]\cup[1,2]$, and let $Y$ be independent and uniform on $[1,2]$. Find the product support and density.

The density of $X$ is $1/2$ on its two intervals, and that of $Y$ is 1 on $[1,2]$. Multiplication by a positive value in $[1,2]$ maps the negative component of $X$ to $[-4,-1]$ and the positive component to $[1,4]$. Therefore $XY$ has support $[-4,-1]\cup[1,4]$, not the whole interval $[-4,4]$.

For $z>0$, the product formula receives contributions only from $1\le x\le2$, with $z/x\in[1,2]$. Equivalently, $z/2\le x\le z$. Including the support restriction on $x$, the integration bounds are $\max(1,z/2)$ and $\min(2,z)$. Hence, for $1<z<4$,

f_{XY}(z)=\frac12 \int_{\max(1,z/2)}^{\min(2,z)}\frac{dx}{x}.

For $1<z<2$, the bounds are 1 and $z$, giving $\tfrac12\log z$. For $2\le z<4$, they are $z/2$ and 2, giving $\tfrac12\log(4/z)$. The negative component gives the reflected density, so the complete density is

f_{XY}(z)= \begin{cases} \frac12\log |z|, & 1<|z|<2,\\ \frac12\log(4/|z|), & 2\le |z|<4,\\ 0, & \text{otherwise}. \end{cases}

For a check, the positive part has total probability $\int_1^2 \tfrac12\log z\,dz+\int_2^4\tfrac12\log(4/z)\,dz=1/2$; symmetry gives the other half.

Worked Example: A Ratio with an Unbounded, Disconnected Support

Let $X$ be uniform on $[1,2]$, and independently let $Y$ be uniform on $[-1,1]$. Find the support and density of $R=X/Y$. Since $P(Y=0)=0$, the ratio is defined almost surely.

For $Y>0$, the smallest possible ratio is $1/1=1$, and ratios grow without bound as $Y$ approaches zero. For $Y<0$, the ratios are at most $-1$ and extend without bound in the negative direction. Thus the support is $(-\infty,-1]\cup[1,\infty)$; values strictly between $-1$ and 1 are impossible.

The joint density is $1/2$ on $[1,2]\times[-1,1]$. For $z\ge1$, the ratio formula requires $zy\in[1,2]$, so $y$ ranges from $1/z$ to $\min(2/z,1)$. It follows that

f_R(z)=\frac12\int_{1/z}^{\min(2/z,1)}y\,dy = \begin{cases} \frac14(1-z^{-2}), & 1\le z<2,\\ \frac{3}{4z^2}, & z\ge2. \end{cases}

For $z\le-1$, symmetry of $Y$ gives the same formula with $|z|$ in place of $z$; for $|z|<1$, the density is zero. As a normalization check, the positive tail has mass $\int_1^2 (1-z^{-2})/4\,dz+\int_2^\infty 3/(4z^2)\,dz=1/8+3/8=1/2$. The negative tail has the same mass.

How to Read the Geometry

When the joint support is a product of intervals, familiar endpoint calculations can often find the image range. For a sum over a rectangle, the smallest and largest sums come from the lower and upper corners. For a product over a rectangle, extrema occur at corners because the function $xy$ is linear in either coordinate when the other is fixed. A ratio over a rectangle whose denominator interval stays strictly on one side of zero also has corner extrema: for fixed nonzero $y$, it is linear in $x$, and for fixed $x$, it is monotone in $y$.

These shortcuts do not replace mapping the actual support. For disconnected support, find the image of each component and then take their union. For dependent inputs, the joint support may be a curve or a restricted region rather than a rectangle. In that case, applying corner bounds to the marginal ranges can invent unattainable values.

For density integration, the same geometry is expressed through the fiber at a fixed output. The sum fiber is $(x,z-x)$, the product fiber is $(x,z/x)$, and the ratio fiber is $(zy,y)$. Restrict the parameter of each fiber to values for which the resulting point lies in the joint support. This procedure works whether the support is one interval, several pieces, or a sign-crossing region.

Common Pitfall: Assuming a Ratio Must Be Bounded

A tempting but incorrect argument is: “Both inputs are bounded, so their ratio is bounded.” Bounded numerators and denominators do not suffice. In the ratio example, $X$ stays between 1 and 2, but $Y$ can be arbitrarily close to zero, so $|X/Y|$ can be arbitrarily large.

The correction is to inspect the denominator values available in the joint support. If $|Y|$ is bounded below by a positive constant, then bounded $X$ implies bounded $X/Y$. If the support includes points with $y$ arbitrarily close to zero while $x$ remains separated from zero, the ratio image is unbounded. If both coordinates approach zero together, the answer depends on how the joint support approaches that point; marginal bounds alone do not settle it.

Key takeaway: First map the joint support to find attainable outputs. Then integrate along the corresponding fibers, retaining only input points in that support. For ratios, check the denominator-zero boundary separately: it determines whether the ratio is defined and whether its support can be unbounded.

Check Your Understanding

For each question, reason from the joint support before setting up a density integral.

If $X$ and $Y$ are independent uniform variables on $[0,1]$, what are the support and the sum-fiber values of $x$ for a fixed sum $z$?
If $X$ is supported on $[-3,-2]\cup[2,3]$ and $Y$ is supported on $[1,2]$, which intervals can contain $XY$?
Why can marginal supports $[0,1]$ for both inputs fail to imply that every point in $[0,1]^2$ is attainable?
What condition on the denominator support guarantees that a bounded numerator produces a bounded ratio?
For a ratio $X/Y=z$, which inverse-coordinate pair must be checked against the joint support in the density integral?