Choosing Parameters in a Candidate CDF

From Boundary Checks to Parameter Constraints

A piecewise formula can look like a CDF while failing at one boundary: a segment might decrease, the function might jump downward, or its values might exceed 1. When parameters appear in the formula, checking one convenient parameter value is not enough. The task is to find all parameter values for which the whole function has the defining properties of a CDF.

The earlier material on checking whether a function can be a CDF established the basic requirements: a CDF is nondecreasing and right-continuous, with limits 0 at negative infinity and 1 at positive infinity. For piecewise functions, the parameter constraints come from checking monotonicity inside every piece and comparing the left-hand limit at each breakpoint with the value assigned there. A jump upward is allowed; a jump downward is not.

Definition: A function \(F:\mathbb{R}\to[0,1]\) is a CDF if it is nondecreasing, right-continuous, and satisfies \(\lim_{x\to-\infty}F(x)=0\) and \(\lim_{x\to\infty}F(x)=1\). To choose parameters in a candidate CDF, each of these requirements must hold for every \(x\), not just at the displayed breakpoints.

A Boundary Test for Piecewise Candidates

Suppose a candidate is 0 to the left of a first breakpoint \(t_0\), equals a function \(g_i(x)\) on each interval \([t_i,t_{i+1})\), and is 1 from a last breakpoint \(t_k\) onward. Assume each \(g_i\) is continuous on its interval, including its left endpoint, and the formula assigns the value at each breakpoint using the piece to its right. Right-continuity at the breakpoints is then built into the definition. The remaining checks can be stated as inequalities.

Theorem: Under these assumptions, the candidate is a CDF exactly when:

\(g_i\) is nondecreasing on each interval \([t_i,t_{i+1})\);
\(g_0(t_0)\geq 0\);
at every internal breakpoint \(t_{i+1}\), \(g_{i+1}(t_{i+1})\geq \lim_{x\uparrow t_{i+1}}g_i(x)\); and
\(\lim_{x\uparrow t_k}g_{k-1}(x)\leq 1\).

The breakpoint inequality compares the value at the breakpoint with the limit from the left. It does not require equality: a strict inequality represents an upward jump. If the pieces are linear or otherwise have simple endpoint limits, these checks reduce directly to inequalities in the parameters.

Check each piece.
Impose the parameter conditions that make the formula nondecreasing on its own interval.

Check the boundaries.
At each breakpoint, require the value assigned there to be at least the limit from the left. At the first and last boundaries, compare with 0 and 1.

Combine the constraints.
The valid parameter set is the intersection of all the inequalities, including parameter restrictions needed for the within-piece checks.

Proof of the theorem. Necessity follows from the definition of a CDF. A nondecreasing function must be nondecreasing on each piece. Its value at \(t_0\) cannot be less than the constant value 0 to the left. At an internal breakpoint, monotonicity requires the assigned value to be at least every value immediately to its left, and hence at least the left-hand limit. The same reasoning at \(t_k\) requires the constant value 1 to be at least the limit from the preceding piece.

For sufficiency, the within-piece conditions ensure that the function does not decrease inside any interval. The boundary inequalities ensure that moving from one interval to the next cannot produce a decrease. Together with the constant tails, the function is nondecreasing on all of \(\mathbb{R}\). The piecewise definition using \([t_i,t_{i+1})\) assigns each internal breakpoint the value from the piece to its right; continuity of that piece at its left endpoint gives right-continuity there. The constant tails give right-continuity elsewhere. The left tail is 0 and the right tail is 1, so the required limits hold. Thus the candidate is a CDF. \(\square\)

Worked Parameter Choices

Worked Example: Three Constant Levels

For a real parameter \(\theta\), consider

F_\theta(x)= \begin{cases} 0, & x<0,\\ \theta, & 0\leq x<1,\\ 2\theta, & 1\leq x<2,\\ 1, & x\geq 2. \end{cases}

Find exactly which values of \(\theta\) make this a CDF, then calculate its jumps when \(\theta=0.3\).

Each piece is constant, so there is no within-piece decrease. At 0, the value \(\theta\) must be at least the left-tail value 0. At 1, the value \(2\theta\) must be at least the left-hand limit \(\theta\). At 2, the value 1 must be at least the left-hand limit \(2\theta\). These conditions are

\theta\geq 0,\qquad 2\theta\geq\theta,\qquad 1\geq 2\theta.

Together they give \(0\leq\theta\leq \tfrac12\). For \(\theta=0.3\), the jumps at 0, 1, and 2 are respectively

F(0)-F(0-)=0.3-0=0.3,\qquad F(1)-F(1-)=0.6-0.3=0.3,\qquad F(2)-F(2-)=1-0.6=0.4.

The jumps are nonnegative and sum to 1. For example, \(P(0\leq X<2)=F(2-)-F(0-)=0.6-0=0.6\). The interval includes the jumps at 0 and 1, but excludes the jump at 2.

Worked Example: Linear Pieces With Three Boundary Checks

Consider

F(x)= \begin{cases} 0, & x<0,\\ \alpha+\beta x, & 0\leq x<1,\\ \gamma+\delta(x-1), & 1\leq x<2,\\ 1, & x\geq 2. \end{cases}

Find the parameter constraints. Then use \((\alpha,\beta,\gamma,\delta)=(0.1,0.2,0.5,0.3)\) to find the jumps and \(P(0<X\leq 1)\).

The first line segment is nondecreasing exactly when \(\beta\geq0\); the second is nondecreasing exactly when \(\delta\geq0\). At 0, \(\alpha\geq0\). At 1, the value from the second piece must be at least the limit from the first, so \(\gamma\geq\alpha+\beta\). At 2, the terminal value 1 must be at least the limit from the second piece, so \(\gamma+\delta\leq1\). Thus the full conditions are

\alpha\geq0,\qquad \beta\geq0,\qquad \gamma\geq\alpha+\beta,\qquad \delta\geq0,\qquad \gamma+\delta\leq1.

For the specified values, the two within-piece slopes are positive, \(0.5\geq0.1+0.2\), and \(0.5+0.3=0.8\leq1\); the constraints hold. The left-hand limits and assigned values at the breakpoints give jumps

\begin{aligned} F(0)-F(0-) &= 0.1-0=0.1,\\ F(1)-F(1-) &= 0.5-(0.1+0.2)=0.2,\\ F(2)-F(2-) &= 1-\bigl(0.5+0.3(2-1)\bigr)=0.2. \end{aligned}

For the requested interval, \(F(1)=0.5\) and \(F(0)=0.1\), so

P(0<X\leq1)=F(1)-F(0)=0.5-0.1=0.4.

The endpoint convention matters: the event excludes a possible point mass at 0 and includes any point mass at 1.

Worked Example: A Quadratic Segment

For a real parameter \(p\), let

F_p(x)= \begin{cases} 0, & x<-1,\\ p(x+1)^2, & -1\leq x<0,\\ p+(1-p)x, & 0\leq x<1,\\ 1, & x\geq1. \end{cases}

Determine the allowed values of \(p\). For \(p=0.4\), calculate \(P(-0.5<X\leq0.5)\).

On \([-1,0)\), the derivative is \(2p(x+1)\). Since \(x+1\geq0\) there, this piece is nondecreasing when \(p\geq0\). Its limit at 0 is \(p\), which matches the value assigned by the next piece at 0. The second piece has slope \(1-p\), so it is nondecreasing when \(p\leq1\). Its limit at 1 is \(p+(1-p)=1\), matching the final piece. Therefore both pieces and all boundaries satisfy the CDF conditions exactly when \(0\leq p\leq1\).

For \(p=0.4\),

F(-0.5)=0.4(0.5)^2=0.1,\qquad F(0.5)=0.4+0.6(0.5)=0.7.

The interval \((-0.5,0.5]\) has probability \(F(0.5)-F(-0.5)\), giving \(0.7-0.1=0.6\).

Jumps and Interval Probabilities After the Check

Once the parameter constraints are satisfied, the candidate can be used as a CDF. At any breakpoint \(c\), its jump is \(F(c)-F(c-)\), where \(F(c-)=\lim_{x\uparrow c}F(x)\). The earlier material on CDF jumps and point probabilities identifies this jump with \(P(X=c)\). For \(a<b\), the interval probability is \(P(a<X\leq b)=F(b)-F(a)\); changing whether an endpoint is included changes which endpoint jumps contribute.

These calculations should come after the validity check. A difference between two displayed values may look like a probability, but it has no such interpretation if the candidate decreases somewhere or violates a boundary condition.

Common Mistake: Forcing Continuity at Every Breakpoint

A frequent error is to set the value at each breakpoint equal to the left-hand limit, as if every valid CDF had to be continuous. This incorrectly rules out valid upward jumps. For example, in the three-level example, \(\theta=0.3\) gives a valid CDF with a jump of 0.3 at \(x=1\). The correct condition is that the value at the breakpoint be at least the left-hand limit. Equality means no jump there; a strict inequality means a positive jump. The opposite error—checking only the values assigned at the breakpoints—can miss a decreasing segment between them, so the within-piece check is essential too.

Key takeaway: For a parameterized piecewise CDF, check monotonicity inside every piece, compare each boundary value with the left-hand limit, and verify the two tail limits. Only after the resulting parameter inequalities hold should jumps and interval probabilities be interpreted as probabilities.

Check Your Understanding

Use the CDF conditions and endpoint conventions to answer each question.

For \(F(x)=0\) when \(x<0\), \(F(x)=a\) when \(0\leq x<1\), and \(F(x)=1\) when \(x\geq1\), what values of \(a\) make \(F\) a CDF?
In the linear-piece example, which inequality prevents a downward jump at \(x=1\)?
For a valid CDF with \(F(c-)=0.42\) and \(F(c)=0.57\), what is \(P(X=c)\)?
Explain why a positive jump at a breakpoint does not violate monotonicity.
If \(F(a)=0.2\) and \(F(b)=0.75\) for \(a<b\), what is \(P(a<X\leq b)\)?