Uniform Density on a Planar Region
A bivariate distribution is called uniform on a region when its joint density is constant throughout that region and zero outside it. The geometry determines the constant: a larger region requires a smaller density so the total probability remains one. This idea gives a direct route from a support drawn in the plane to a valid joint model, its marginal densities, and probabilities of events described by subregions.
The models here are probability models, not financial or life-contingency calculations; no interest rate or mortality assumption is involved. Each example specifies the support and its constant density. Boundary lines can be included or excluded without changing probabilities, because a line has area zero.
The density is constant, but it is not usually equal to one. It must have the units of inverse area, and its integral over the support must equal one. To find a marginal density, integrate the joint density over the coordinate not retained. The limits of integration are the slice of the support at the fixed coordinate; they may vary with that coordinate.
Normalization and Marginal Densities
For a constant density \(c\) on \(R\), normalization requires \(\iint_R c\,dx\,dy=1\). Since \(c\) is constant, the integral is \(c|R|\), so \(c=1/|R|\). For example, a rectangle of width \(a\) and height \(b\) has area \(ab\), while a right triangle with perpendicular legs \(a\) and \(b\) has area \(ab/2\). The same normalization rule applies to both; only the support area changes.
Once the joint density is known, a marginal density is obtained from the relevant slices. If the vertical slice at \(x\) runs from \(y_{\min}(x)\) to \(y_{\max}(x)\), then \(f_X(x)=\int_{y_{\min}(x)}^{y_{\max}(x)}f_{X,Y}(x,y)\,dy\). The horizontal-slice formula for \(f_Y\) is analogous. Outside the coordinate range of the support, the corresponding marginal density is zero.
Worked Example: A Rectangle with a Slanted Event Boundary
Suppose \((X,Y)\) is uniform on \(R=\{(x,y):1\le x\le5,\ 0\le y\le2\}\). Find the joint density, both marginal densities, and \(P(X+Y\le4)\). The model is uniform with support parameters \(1\le x\le5\) and \(0\le y\le2\); no interest or mortality assumptions apply.
Normalize. The rectangle has width \(5-1=4\), height \(2-0=2\), and area \(4\cdot2=8\). Therefore its constant joint density is \(c=1/8\) on the rectangle and zero elsewhere.
Find the marginals. At each \(x\) from 1 to 5, the vertical slice has length 2. At each \(y\) from 0 to 2, the horizontal slice has length 4. Thus
Each marginal integrates to one: \((5-1)(1/4)=1\) and \((2-0)(1/2)=1\).
Find the event probability. The line \(y=4-x\) meets the rectangle's top edge \(y=2\) at \(x=2\), and it meets the bottom edge \(y=0\) at \(x=4\). The part of the rectangle below this line has vertical height \(4-x\) for \(1\le x\le2\), height 2 for \(2\le x\le4\), and height 0 for \(4\le x\le5\). Its area is
Dividing by the support area gives \(P(X+Y\le4)=(13/2)/8=13/16=0.8125\). The result is between zero and one, and the marginal checks confirm the normalization. The method is to find the area cut out by the event within the actual support; it avoids treating the slanted event boundary as if it covered the entire rectangle.
Triangular Supports and Changing Slice Lengths
On a triangle, the length of a slice generally changes as the fixed coordinate changes. Consequently, a uniform joint density on a triangle does not usually produce uniform marginal densities. The marginal density is proportional to the available slice length: slices near a wide part of the triangle contribute more density than slices near a narrow part.
Worked Example: The Triangle Below the Line \(x=y\)
Let \((X,Y)\) be uniform on \(R=\{(x,y):0\le x\le y\le2\}\). Find its density and marginal densities, then calculate \(P(Y-X\le1)\). The support parameters are \(0\le x\le y\le2\), and the distribution is uniform on this triangular region; no interest or mortality assumptions apply.
Normalize. The triangle has vertices \((0,0),(0,2),(2,2)\), so its area is \((2\cdot2)/2=2\). Hence \(f_{X,Y}(x,y)=1/2\) on \(R\), and zero outside.
Find \(f_X\). At fixed \(x\), the support allows \(y\) to range from \(x\) to 2. Therefore
Find \(f_Y\). At fixed \(y\), \(x\) ranges from 0 to \(y\), giving
Both marginals integrate to one: \(\int_0^2(2-x)/2\,dx=1\) and \(\int_0^2 y/2\,dy=1\).
Calculate the probability. Within the support, the event \(Y-X\le1\) is \(y\le x+1\). For \(0\le x\le1\), the vertical slice runs from \(y=x\) to \(y=x+1\), with length 1. For \(1\le x\le2\), it runs from \(y=x\) to \(y=2\), with length \(2-x\). The event area is
The support area is 2, so \(P(Y-X\le1)=(3/2)/2=3/4=0.7500\). The two slice ranges meet at \(x=1\), where the upper boundary switches from \(x+1\) to the triangle's boundary \(y=2\). Accounting for that switch is essential to the answer.
A Second Triangular Geometry
A common triangular support is bounded by the coordinate axes and a descending line. Its slices shrink to zero at the line's intercepts. This gives a useful contrast with the preceding triangle, where one marginal density increases and the other decreases.
Worked Example: A Simplex and a Corner-Region Probability
Suppose \((X,Y)\) is uniform on \(R=\{(x,y):x\ge0,\ y\ge0,\ x+y\le3\}\). Find the joint density and both marginals, then evaluate \(P(X>1,\ Y>1)\). The support is the stated right triangle and the distribution is uniform there; there are no interest or mortality assumptions.
Normalize. The triangle has intercepts 3 on both axes, so its area is \(3\cdot3/2=9/2\). Thus \(c=1/(9/2)=2/9\), and the density is \(2/9\) inside the triangle and zero outside.
Find \(f_X\). For a fixed \(x\) between 0 and 3, \(y\) ranges from 0 to \(3-x\). Hence
By symmetry of the support and density, or by integrating over \(x\) from 0 to \(3-y\), the other marginal is
For either marginal, the integral is \(\int_0^3 2(3-x)/9\,dx=(2/9)(9-9/2)=1\).
Calculate the corner probability. The event \(X>1,\ Y>1\) is the part of the triangle above and to the right of \((1,1)\). Translate coordinates by setting \(u=x-1\) and \(v=y-1\). Then \(u\ge0,\ v\ge0\), and the support condition becomes \(u+v\le1\). This is a right triangle of area \(1\cdot1/2=1/2\). Therefore
The translation preserves area, and the event triangle lies entirely within the original support. This area-ratio method avoids integrating over a region whose boundary is easier to identify after translation.
Checks, Interpretation, and an Exam Trap
For any proposed constant-density model, check that the density is zero outside the stated support, that its integral over the support is one, and that each marginal integrates to one. For an event probability, verify that the event region is intersected with the support before finding its area or setting integral limits. The bounding rectangle of a triangle is not itself the support.
On a rectangle with constant density, the vertical slice length is constant, so the \(X\) marginal is uniform across the rectangle's horizontal range; the same reasoning applies to \(Y\). A rectangular support and constant density also factor into the product of the two marginal densities. On a triangular support, slice lengths generally vary, and the joint density's support does not extend over every combination of the marginal ranges. In particular, multiplying marginal probabilities is not a general way to calculate event probabilities for a triangular model.
Write the inequalities that define the region and identify its geometric shape and area.
Set the constant equal to the reciprocal of the support area.
For a marginal or event probability, identify where each slice starts and ends, splitting the integral if a boundary changes.
Confirm that marginal densities integrate to one and probabilities are between zero and one.
The main practical benefit of the planar-uniform model is that many probability calculations become geometric. But geometry must be applied to the right region: the support sets which points are possible, while the event further restricts which of those points are counted.
Check Your Understanding
For each model, use the stated support, derive the requested quantity, and check that the event region stays within the support.
- \((X,Y)\) is uniform on \(0\le X\le2,\ 0\le Y\le5\). Find the joint density, \(f_X(x)\), and \(P(X\le1,\ Y\ge2)\). (Answer: \(f_{X,Y}=1/10\) on the rectangle; \(f_X(x)=1/2\) for \(0\le x\le2\); probability \(3/10\).)
- \((X,Y)\) is uniform on \(x\ge0,\ y\ge0,\ x+2y\le4\). Find the normalizing constant, the marginal density \(f_Y(y)\), and \(P(Y\le1)\). (Answer: \(c=1/4\); \(f_Y(y)=1-y/2\) for \(0\le y\le2\), zero otherwise; probability \(3/4\).)
- \((X,Y)\) is uniform on \(0\le x\le y\le3\). Find \(f_Y(y)\) and \(P(X<1)\). (Answer: \(f_Y(y)=2y/9\) for \(0\le y\le3\), zero otherwise; probability \(5/9\).)
- \((X,Y)\) is uniform on \(0\le X\le3,\ 0\le Y\le2\). Find the two marginal densities and \(P(X+Y\le2)\). (Answer: \(f_X(x)=1/3\) for \(0\le x\le3\), \(f_Y(y)=1/2\) for \(0\le y\le2\); probability \(1/3\).)
- \((X,Y)\) is uniform on \(x\ge0,\ y\ge0,\ x+y\le4\). Find \(f_X(x)\) and \(P(X>1,\ Y>1)\). (Answer: \(f_X(x)=(4-x)/8\) for \(0\le x\le4\), zero otherwise; probability \(1/4\).)