Why This Concept Exists
This node addresses the single most common source of lost marks in PTS2: double integration over non-rectangular support regions. Examiners consistently test this because it separates students who understand integration from those who merely recognise formulas. The PTS2 papers from 2013–2025 overwhelmingly feature non-rectangular supports — triangular regions, regions bounded by parabolas, absolute value boundaries, and implicit region descriptions.
The distinction between treating a support as rectangular (constant limits) versus non-rectangular (variable limits) is the difference between gaining 15 marks and 0 marks on a single question. It is the most high-stakes technical skill in the course.
High-stakes: Treating a non-rectangular support as rectangular is the single most costly error in PTS2. Students who use constant limits when the support is triangular can be confident, elegant, and wrong throughout an entire question.
Prerequisites
- N1 & N2 complete: You must be comfortable with joint PDFs, marginals, conditional distributions, and covariance.
- Double integration: Comfort with iterated integrals, understanding that \(\displaystyle\int_a^b \int_c^d f(x,y)\,dy\,dx\) means "for each fixed x from a to b, integrate y from c to d."
- Sketching regions in \(\mathbb{R}^2\): Ability to sketch lines (\(y = x\)), curves (\(y = x^2\), \(y = \sqrt{x}\)), and identify the region bounded by them.
- Changing integration order: Understanding that \(\int_0^1\int_0^{x}f(x,y)\,dy\,dx \neq \int_0^1\int_0^{1}f(x,y)\,dy\,dx\), and being able to reverse limits: \(\int_0^1\int_0^{x} \cdots dy\,dx = \int_0^1\int_{y}^{1} \cdots dx\,dy\).
- Standard integrals: Polynomials, exponentials, and simple substitutions. No advanced techniques required, but accuracy is essential.
Core Exposition
3.1 The Support: Why It Matters
The support of a joint PDF is the set of all points \((x,y)\) where the density is positive. For PTS2, supports fall into three categories:
Type 1 — Rectangular: \(a \leq x \leq b,\; c \leq y \leq d\). Integration limits are constants. This is the only case where independence is possible.
Type 2 — Triangular (vertex-based): e.g., \(0 \leq x \leq y \leq 1\), or \(0 \leq y \leq x \leq 1\). One variable is bounded by a fraction involving the other. Integration limits are linear functions.
Type 3 — Curved: e.g., \(0 \leq y \leq \sqrt{x} \leq 1\), or \(0 \leq x \leq 1,\; x^2 \leq y \leq 1\). Boundaries involve curves. Integration limits are non-linear functions of the outer variable.
Type 2 — Triangular (vertex-based): e.g., \(0 \leq x \leq y \leq 1\), or \(0 \leq y \leq x \leq 1\). One variable is bounded by a fraction involving the other. Integration limits are linear functions.
Type 3 — Curved: e.g., \(0 \leq y \leq \sqrt{x} \leq 1\), or \(0 \leq x \leq 1,\; x^2 \leq y \leq 1\). Boundaries involve curves. Integration limits are non-linear functions of the outer variable.
Universal rule: For a non-rectangular support, the inner integral's limits are always functions of the outer variable. The outer integral's limits are always constants. This is a fundamental property of iterated integrals over regions in \(\mathbb{R}^2\).
3.2 Sketching the Support
Before computing anything, always sketch the support. The procedure:
- Identify all boundary curves from the support description.
- Draw each boundary on a common set of axes.
- Identify the enclosed region (often a triangle or the area under/between curves).
- For each variable, determine: "as a function of the other, where does it range?"
3.3 Computing Probabilities: Setting Up the Integral
To compute \(P((X,Y) \in A)\) where A is some sub-region of the support:
\(P((X,Y) \in A) = \displaystyle\iint_{A \cap \text{support}} f_{X,Y}(x,y)\,dy\,dx\)
The region of integration is the intersection of A with the support. You must carefully determine the bounds of this intersection. This is where the most marks are lost.
3.4 Computing Marginals on Non-Rectangular Supports
The marginal of X is always:
\(f_X(x) = \displaystyle\int_{y_{\min}(x)}^{y_{\max}(x)} f_{X,Y}(x,y)\,dy\)
where \(y_{\min}(x)\) and \(y_{\max}(x)\) come from the support. For example:
If the support is \(0 \leq x \leq y \leq 1\):
For fixed x: y ranges from \(x\) to \(1\). So \(f_X(x) = \displaystyle\int_x^1 f_{X,Y}(x,y)\,dy\).
For fixed y: x ranges from \(0\) to \(y\). So \(f_Y(y) = \displaystyle\int_0^y f_{X,Y}(x,y)\,dx\).
For fixed x: y ranges from \(x\) to \(1\). So \(f_X(x) = \displaystyle\int_x^1 f_{X,Y}(x,y)\,dy\).
For fixed y: x ranges from \(0\) to \(y\). So \(f_Y(y) = \displaystyle\int_0^y f_{X,Y}(x,y)\,dx\).
3.5 Changing the Order of Integration
Sometimes you need to switch the order of integration. The procedure:
- Step 1: Sketch the region.
- Step 2: Read off the limits in the old order (e.g., "dx dy" means inner is x, outer is y).
- Step 3: For the new order, fix the new outer variable first and find its range (constants). Then find the range of the new inner variable as a function of the outer variable.
[INTERACTIVE: Support region sketcher — will be added later]
Worked Examples
Example 1: Triangular Support
Let \(f_{X,Y}(x,y) = c\) on the triangle \(0 \leq x \leq y \leq 1\) (uniform on a triangle). Find c, the marginals, and \(P(X + Y \geq 1)\).
Area of the triangle
The triangle has vertices at (0,0), (0,1), and (1,1). Area = 1/2.
For a uniform distribution: \(c \times \text{area} = 1 \implies c = 2\).
Verify: \(\displaystyle\int_0^1\int_x^1 2\,dy\,dx = \int_0^1 2(1-x)\,dx = 2\left[x - \frac{x^2}{2}\right]_0^1 = 2(1/2) = 1\) \(\checkmark\)
For a uniform distribution: \(c \times \text{area} = 1 \implies c = 2\).
Verify: \(\displaystyle\int_0^1\int_x^1 2\,dy\,dx = \int_0^1 2(1-x)\,dx = 2\left[x - \frac{x^2}{2}\right]_0^1 = 2(1/2) = 1\) \(\checkmark\)
Marginal of X
For fixed \(x\), \(y\) ranges from \(x\) to \(1\):
\(f_X(x) = \displaystyle\int_x^1 2\,dy = 2(1-x)\), for \(0 \leq x \leq 1\).
Check: \(\displaystyle\int_0^1 2(1-x)\,dx = 2 - 1 = 1\) \(\checkmark\)
\(f_X(x) = \displaystyle\int_x^1 2\,dy = 2(1-x)\), for \(0 \leq x \leq 1\).
Check: \(\displaystyle\int_0^1 2(1-x)\,dx = 2 - 1 = 1\) \(\checkmark\)
Marginal of Y
For fixed \(y\), \(x\) ranges from \(0\) to \(y\):
\(f_Y(y) = \displaystyle\int_0^y 2\,dx = 2y\), for \(0 \leq y \leq 1\).
Check: \(\displaystyle\int_0^1 2y\,dy = 1\) \(\checkmark\)
\(f_Y(y) = \displaystyle\int_0^y 2\,dx = 2y\), for \(0 \leq y \leq 1\).
Check: \(\displaystyle\int_0^1 2y\,dy = 1\) \(\checkmark\)
P(X + Y ≥ 1)
The event \(x + y \geq 1\) intersects the support. The support is \(x \leq y,\; 0 \leq x \leq 1,\; 0 \leq y \leq 1\).
The intersection with \(x + y \geq 1\): for \(y\) from \(1/2\) to \(1\), \(x\) ranges from \(\max(0, 1-y)\) to \(y\).
Since \(1-y \geq 0\) when \(y \leq 1\), and \(1-y \leq y\) when \(y \geq 1/2\):
\(P = \displaystyle\int_{1/2}^1 \int_{1-y}^y 2\,dx\,dy = \int_{1/2}^1 2[y - (1-y)]\,dy = \int_{1/2}^1 2(2y-1)\,dy\)
\(= \left[2y^2 - 2y\right]_{1/2}^1 = (2-2) - (1/2 - 1) = 0 - (-1/2) = \boxed{\frac{1}{2}}\)
The intersection with \(x + y \geq 1\): for \(y\) from \(1/2\) to \(1\), \(x\) ranges from \(\max(0, 1-y)\) to \(y\).
Since \(1-y \geq 0\) when \(y \leq 1\), and \(1-y \leq y\) when \(y \geq 1/2\):
\(P = \displaystyle\int_{1/2}^1 \int_{1-y}^y 2\,dx\,dy = \int_{1/2}^1 2[y - (1-y)]\,dy = \int_{1/2}^1 2(2y-1)\,dy\)
\(= \left[2y^2 - 2y\right]_{1/2}^1 = (2-2) - (1/2 - 1) = 0 - (-1/2) = \boxed{\frac{1}{2}}\)
Example 2: Curved Support
Let \(f_{X,Y}(x,y) = c \cdot xy\) for \(0 \leq x \leq 1\) and \(0 \leq y \leq \sqrt{x}\). Find c and \(P(Y \leq X)\).
Find c
Support: for each x from 0 to 1, y ranges from 0 to \(\sqrt{x}\).
\[\int_0^1\int_0^{\sqrt{x}} c\,xy\,dy\,dx = c\int_0^1 x\left[\frac{y^2}{2}\right]_0^{\sqrt{x}}dx = c\int_0^1 x\cdot\frac{x}{2}\,dx = \frac{c}{2}\int_0^1 x^2\,dx = \frac{c}{6}\] Set = 1: \(c = 6\).
\[\int_0^1\int_0^{\sqrt{x}} c\,xy\,dy\,dx = c\int_0^1 x\left[\frac{y^2}{2}\right]_0^{\sqrt{x}}dx = c\int_0^1 x\cdot\frac{x}{2}\,dx = \frac{c}{2}\int_0^1 x^2\,dx = \frac{c}{6}\] Set = 1: \(c = 6\).
P(Y ≤ X)
The event \(y \leq x\) within the support \(0 \leq y \leq \sqrt{x},\; 0 \leq x \leq 1\).
When \(x \in [0,1]\): \(\sqrt{x} \geq x\) (since \(x \geq x^2\) for \(x \in [0,1]\)).
So the condition \(y \leq x\) further restricts: \(0 \leq y \leq \min(x, \sqrt{x}) = x\) for \(x \in [0,1]\).
\[P(Y \leq X) = \int_0^1\int_0^x 6xy\,dy\,dx = \int_0^1 6x\left[\frac{y^2}{2}\right]_0^x dx = \int_0^1 6x\cdot\frac{x^2}{2}\,dx = 3\int_0^1 x^3\,dx = \frac{3}{4}\]
When \(x \in [0,1]\): \(\sqrt{x} \geq x\) (since \(x \geq x^2\) for \(x \in [0,1]\)).
So the condition \(y \leq x\) further restricts: \(0 \leq y \leq \min(x, \sqrt{x}) = x\) for \(x \in [0,1]\).
\[P(Y \leq X) = \int_0^1\int_0^x 6xy\,dy\,dx = \int_0^1 6x\left[\frac{y^2}{2}\right]_0^x dx = \int_0^1 6x\cdot\frac{x^2}{2}\,dx = 3\int_0^1 x^3\,dx = \frac{3}{4}\]
Example 3: Changing Integration Order
Evaluate: \(\displaystyle\int_0^1\int_0^x f_{X,Y}(x,y)\,dy\,dx\) where the support is \(0 \leq y \leq x \leq 1\).
The region is the lower triangle. In the reverse order (dx dy):
For fixed \(y\), \(x\) ranges from \(y\) to \(1\). And \(y\) ranges from \(0\) to \(1\).
\(\displaystyle\int_0^1\int_y^1 f_{X,Y}(x,y)\,dx\,dy\).
For \(f_{X,Y}(x,y) = 2\) on this region:
Old order: \(\displaystyle\int_0^1\int_0^x 2\,dy\,dx = \int_0^1 2x\,dx = 1\). \(\checkmark\)
Reversed: \(\displaystyle\int_0^1\int_y^1 2\,dx\,dy = \int_0^1 2(1-y)\,dy = 1\). Same result. \(\checkmark\)
Pattern Recognition & Examiner Traps
Trap 1: Using constant limits for non-rectangular supports
This is the #1 error on PTS2 papers. When the support is \(0 \leq x \leq y \leq 1\), integrating y from 0 to 1 (instead of x to 1) gives a completely wrong marginal.
WRONG\(f_X(x) = \displaystyle\int_0^1 8xy\,dy = 4x\) — using [0,1] constants.
RIGHT\(f_X(x) = \displaystyle\int_x^1 8xy\,dy = 4x(1-x^2)\) — limits must reflect the support.
Trap 2: Wrong intersection region for probability queries
When asked for \(P(g(X,Y) \in B)\), students sometimes integrate over the entire support instead of the intersection of the support with the event region.
Trap 3: Not identifying the correct range of the outer variable
For a marginal, the outer variable's range comes from the projection of the support onto that axis. Students sometimes use an outer limit that is too wide or too narrow.
Examiner patterns to recognise:
- "Sketch the region of positive density" — do this before ANY calculation. First marks are for the sketch.
- "Find P(X + Y ≤ 1)" — requires careful identification of the intersection region with the support.
- "By changing the order of integration..." — signals you need to reverse the dy dx / dx dy order.
- "for 0 ≤ x ≤ y ≤ 1" — triangular support. Inner limits are functions of outer variable.
Connections
Where N3 sits in the PTS2 architecture:
- ← From N1 & N2: The marginal and conditional distribution machinery from N1/N2 is applied to non-rectangular supports here.
- → To N4 (Transformations): When you transform variables using the Jacobian method, the new support is almost always non-rectangular. N3 skills are essential.
- → To N5 (Order Statistics): The joint distribution of order statistics has a triangular support \(x_{(1)} \leq x_{(2)} \leq \cdots \leq x_{(n)}\).
- → To Transformation proofs: Many transformation results require integrating over non-standard supports to normalise the resulting density.
Summary Table
| Support Type | Example | Inner Limits | Outer Limits | Can X⊥Y? |
|---|---|---|---|---|
| Rectangular | \(0\leq x\leq 1,\;0\leq y\leq 1\) | Constants | Constants | Possible |
| Triangular (lower) | \(0\leq x\leq y\leq 1\) | Linear functions | Constants | Never |
| Triangular (upper) | \(0\leq y\leq x\leq 1\) | Linear functions | Constants | Never |
| Curved (lower) | \(0\leq y\leq \sqrt{x}\leq 1\) | Curved functions | Constants | Never |
| Curved (upper) | \(x^2\leq y\leq 1,\;-1\leq x\leq 1\) | Curved functions | Constants | Never |
| Implicit | \(x+y\leq 1,\;x\geq 0,\;y\geq 0\) | Depends on form | Constants | Never |
Self-Assessment
Checklist — Can you do all of these?
- Sketch the support region given an algebraic description (triangle, parabola, implicit).
- Set up the correct iterated integral for a marginal on a non-rectangular support.
- Compute \(P(g(X,Y) \in B)\) by identifying the correct intersection region.
- Change the order of integration for a given iterated integral.
- Normalise a PDF on a triangular or curved support.
- Compute \(E[g(X,Y)]\) over a non-rectangular support.
- Verify that your marginal integrates to 1.
Practice problems to attempt independently
- If \(f_{X,Y}(x,y) = c\) on \(0 \leq x \leq 2,\; 0 \leq y \leq x^2\), find c and the marginals. [Area = \(\int_0^2 x^2 dx = 8/3\), so \(c = 3/8\).]
- If \(f_{X,Y}(x,y) = 6(1-y)\) on \(0 \leq x \leq y \leq 1\), find \(P(X + Y \leq 1)\). [Sketch first, then integrate.]
- For \(f_{X,Y}(x,y) = 4xy\) on \(0 \leq x \leq 1,\; 0 \leq y \leq \sqrt{x}\), find \(f_X(x)\). [Answer: \(2x^2\).]
- Reverse the order: \(\int_0^2\int_{x^2}^{4} f(x,y)\,dy\,dx = \int_0^4\int_0^{\sqrt{y}}f(x,y)\,dx\,dy\).
HLQ: Exam-Style Question with Worked Solution
14 MARKS
FINAL 2024 Q1(a-d)
DISTINCTION
The joint PDF of \(X\) and \(Y\) is given by:
\[f_{X,Y}(x,y) = k(x^2 + y) \quad \text{for } 0 \leq y \leq x \leq 1\]
(a) Find the value of \(k\). (3 marks)
(b) Find the marginal PDFs of \(X\) and \(Y\). (4 marks)
(c) Compute \(P\!\left(X \geq \frac{1}{2}\right)\). (2 marks)
(d) Compute the conditional expectation \(E[Y|X = x]\). (3 marks)
(e) Are X and Y independent? Justify your answer. (2 marks)
Part (a): Finding k
The support is \(0 \leq y \leq x \leq 1\): for each \(x\) from 0 to 1, \(y\) ranges from 0 to \(x\).
\[\int_0^1\int_0^x k(x^2 + y)\,dy\,dx = k\int_0^1\left[x^2 y + \frac{y^2}{2}\right]_0^x\,dx\] \[= k\int_0^1\left(x^3 + \frac{x^2}{2}\right)dx = k\left[\frac{x^4}{4} + \frac{x^3}{6}\right]_0^1 = k\left(\frac{1}{4} + \frac{1}{6}\right) = k\cdot\frac{5}{12}\] Set equal to 1: \(k = \dfrac{12}{5}\).
\[\int_0^1\int_0^x k(x^2 + y)\,dy\,dx = k\int_0^1\left[x^2 y + \frac{y^2}{2}\right]_0^x\,dx\] \[= k\int_0^1\left(x^3 + \frac{x^2}{2}\right)dx = k\left[\frac{x^4}{4} + \frac{x^3}{6}\right]_0^1 = k\left(\frac{1}{4} + \frac{1}{6}\right) = k\cdot\frac{5}{12}\] Set equal to 1: \(k = \dfrac{12}{5}\).
Part (b): Marginals
Marginal of X: For fixed x, y ranges from 0 to x.
\[f_X(x) = \int_0^x \frac{12}{5}(x^2+y)\,dy = \frac{12}{5}\left[x^2y + \frac{y^2}{2}\right]_0^x = \frac{12}{5}\left(x^3 + \frac{x^2}{2}\right) = \frac{12x^3}{5} + \frac{6x^2}{5}\] for \(0 \leq x \leq 1\).
Marginal of Y: For fixed y, x ranges from y to 1 (since \(x \geq y\) and \(x \leq 1\)).
\[f_Y(y) = \int_y^1 \frac{12}{5}(x^2+y)\,dx = \frac{12}{5}\left[\frac{x^3}{3} + yx\right]_y^1 = \frac{12}{5}\left(\frac{1}{3} + y - \frac{y^3}{3} - y^2\right)\] for \(0 \leq y \leq 1\).
\[f_X(x) = \int_0^x \frac{12}{5}(x^2+y)\,dy = \frac{12}{5}\left[x^2y + \frac{y^2}{2}\right]_0^x = \frac{12}{5}\left(x^3 + \frac{x^2}{2}\right) = \frac{12x^3}{5} + \frac{6x^2}{5}\] for \(0 \leq x \leq 1\).
Marginal of Y: For fixed y, x ranges from y to 1 (since \(x \geq y\) and \(x \leq 1\)).
\[f_Y(y) = \int_y^1 \frac{12}{5}(x^2+y)\,dx = \frac{12}{5}\left[\frac{x^3}{3} + yx\right]_y^1 = \frac{12}{5}\left(\frac{1}{3} + y - \frac{y^3}{3} - y^2\right)\] for \(0 \leq y \leq 1\).
Part (c): P(X ≥ 1/2)
\(P(X \geq 1/2) = \displaystyle\int_{1/2}^1 f_X(x)\,dx = \int_{1/2}^1 \frac{12}{5}\left(x^3 + \frac{x^2}{2}\right)dx\)
\(= \frac{12}{5}\left[\frac{x^4}{4} + \frac{x^3}{6}\right]_{1/2}^1 = \frac{12}{5}\left[\left(\frac{1}{4}+\frac{1}{6}\right) - \left(\frac{1}{64}+\frac{1}{48}\right)\right]\)
\(= \frac{12}{5}\left[\frac{5}{12} - \frac{7}{192}\right] = \frac{12}{5}\cdot\frac{80-7}{192} = \frac{12\cdot 73}{5\cdot 192} = \frac{73}{80}\)
\(= \frac{12}{5}\left[\frac{x^4}{4} + \frac{x^3}{6}\right]_{1/2}^1 = \frac{12}{5}\left[\left(\frac{1}{4}+\frac{1}{6}\right) - \left(\frac{1}{64}+\frac{1}{48}\right)\right]\)
\(= \frac{12}{5}\left[\frac{5}{12} - \frac{7}{192}\right] = \frac{12}{5}\cdot\frac{80-7}{192} = \frac{12\cdot 73}{5\cdot 192} = \frac{73}{80}\)
Part (d): E[Y | X = x]
Conditional PDF: \(f_{Y|X}(y|x) = \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{\frac{12}{5}(x^2+y)}{\frac{12}{5}x^3 + \frac{6}{5}x^2} = \dfrac{x^2+y}{x^3 + x^2/2} = \dfrac{2(x^2+y)}{2x^3 + x^2}\)
for \(0 \leq y \leq x\).
\(E[Y|X=x] = \displaystyle\int_0^x y\cdot\frac{x^2+y}{x^3 + x^2/2}\,dy = \frac{1}{x^3 + x^2/2}\int_0^x (x^2y + y^2)\,dy\)
\(= \frac{1}{x^3 + x^2/2}\left[\frac{x^2\cdot x^2}{2} + \frac{x^3}{3}\right] = \frac{1}{x^3 + x^2/2}\left[\frac{x^4}{2} + \frac{x^3}{3}\right]\)
\(= \frac{x^3(\frac{x}{2} + \frac{1}{3})}{x^2(x + \frac{1}{2})} = \frac{x(\frac{x}{2} + \frac{1}{3})}{x + \frac{1}{2}} = \frac{x(3x+2)}{6x+3} = \frac{x(3x+2)}{3(2x+1)}\)
for \(0 \leq y \leq x\).
\(E[Y|X=x] = \displaystyle\int_0^x y\cdot\frac{x^2+y}{x^3 + x^2/2}\,dy = \frac{1}{x^3 + x^2/2}\int_0^x (x^2y + y^2)\,dy\)
\(= \frac{1}{x^3 + x^2/2}\left[\frac{x^2\cdot x^2}{2} + \frac{x^3}{3}\right] = \frac{1}{x^3 + x^2/2}\left[\frac{x^4}{2} + \frac{x^3}{3}\right]\)
\(= \frac{x^3(\frac{x}{2} + \frac{1}{3})}{x^2(x + \frac{1}{2})} = \frac{x(\frac{x}{2} + \frac{1}{3})}{x + \frac{1}{2}} = \frac{x(3x+2)}{6x+3} = \frac{x(3x+2)}{3(2x+1)}\)
Part (e): Independence
The support \(0 \leq y \leq x \leq 1\) is not a Cartesian product (it's a triangle).
Therefore X and Y are NOT independent.
(Additionally, \(f_X(x) \cdot f_Y(y) \neq f_{X,Y}(x,y)\) can be verified by substitution.)
Therefore X and Y are NOT independent.
(Additionally, \(f_X(x) \cdot f_Y(y) \neq f_{X,Y}(x,y)\) can be verified by substitution.)
Summary: (a) \(k = 12/5\). (b) \(f_X(x) = \frac{12x^3}{5} + \frac{6x^2}{5}\), \(f_Y(y) = \frac{12}{5}(\frac{1+y-y^2-y^3/3}{1})\)... see above. (c) \(73/80\). (d) \(E[Y|X=x] = \frac{x(3x+2)}{3(2x+1)}\). (e) Not independent (triangular support).