Recovering CDF Values from Interval Probabilities

From Probability Statements to CDF Values

The cumulative distribution function packages the probabilities of all events of the form \(X\le x\): by definition, \(F(x)=P(X\le x)\). Recovering a CDF value from a probability statement is therefore often direct. The care is needed when the statement uses a strict inequality, a right-hand interval, or an interval with endpoints included in a different way. Those choices determine whether the equation involves \(F(x)\) or its left limit.

For a real-valued random variable \(X\), the probability statements may be translated into equations and then solved. Sometimes the supplied statements determine a requested value uniquely; sometimes they determine only a difference between CDF values. The first step is to write each event with its exact endpoint conventions, rather than informally treating all nearby inequalities as equivalent.

Definition: Write \(F(x)=P(X\le x)\). The left limit of the CDF at \(x\) is \(F(x-)=\lim_{t\uparrow x}F(t)\); equivalently, \(F(x-)=P(X<x)\). The distinction between \(F(x-)\) and \(F(x)\) is the point probability \(P(X=x)=F(x)-F(x-)\).

The equivalence \(F(x-)=P(X<x)\) follows from continuity of probability for increasing events. For example, the events \(\{X\le x-1/n\}\) increase to \(\{X<x\}\), so their probabilities converge to \(P(X<x)\). This limit is exactly the left limit of the CDF.

Half-Lines Give Direct CDF Equations

The two left half-lines determine the CDF value and its left limit directly. The right half-lines give their complements. These identities are useful whenever a problem gives a tail probability rather than a probability written in CDF form.

Theorem: For every real \(x\), \(P(X\le x)=F(x)\), \(P(X<x)=F(x-)\), \(P(X>x)=1-F(x)\), and \(P(X\ge x)=1-F(x-)\).

The first two equations follow from the definitions of \(F(x)\) and \(F(x-)\). For the last two, the relevant events are complements: \(\{X>x\}\) is the complement of \(\{X\le x\}\), and \(\{X\ge x\}\) is the complement of \(\{X<x\}\). Thus, for example, \(P(X>x)=1-P(X\le x)=1-F(x)\). This also shows why replacing \(>\) by \(\ge\) in a tail statement can change the answer when \(P(X=x)>0\).

Worked Example: Recovering Values From Half-Lines

Suppose \(P(X\le 4)=0.72\) and \(P(X<4)=0.65\). Find \(F(4)\), \(F(4-)\), \(P(X=4)\), and \(P(X>4)\).

By definition, \(F(4)=P(X\le4)=0.72\), while \(F(4-)=P(X<4)=0.65\). Therefore the point probability is

$$ P(X=4)=F(4)-F(4-)=0.72-0.65=0.07. $$

The right-tail probability is the complement of \(X\le4\):

$$ P(X>4)=1-F(4)=1-0.72=0.28. $$

As a check, the events \(X<4\), \(X=4\), and \(X>4\) are disjoint and cover all outcomes. Their probabilities sum to \(0.65+0.07+0.28=1\).

Interval Probabilities and Endpoint Conventions

An interval probability is a difference between two cumulative probabilities. The exact difference depends on whether either endpoint is included. For real numbers \(a<b\), the four standard forms are:

Key identities: \(P(a<X\le b)=F(b)-F(a)\); \(P(a\le X\le b)=F(b)-F(a-)\); \(P(a<X<b)=F(b-)-F(a)\); and \(P(a\le X<b)=F(b-)-F(a-)\).

These identities can be derived by decomposing events, which makes the endpoint choices explicit. The event \(\{X\le b\}\) is the disjoint union of \(\{X\le a\}\) and \(\{a<X\le b\}\). Subtracting their probabilities gives \(P(a<X\le b)=F(b)-F(a)\). For an interval that includes \(a\), the part to remove is instead \(\{X<a\}\), whose probability is \(F(a-)\). At the right endpoint, including \(b\) uses \(F(b)\); excluding \(b\) uses \(F(b-)\). These decompositions prove all four identities.

The formulas can be read in reverse to recover a CDF value. For example, if \(F(a)\) is known and \(P(a<X\le b)=q\), then \(F(b)=F(a)+q\). But if that interval probability is the only information given, it determines \(F(b)-F(a)\), not generally either value separately.

Worked Example: An Interval With a Known Starting Value

Suppose \(F(-2)=0.18\) and \(P(-2<X\le1)=0.47\). Find \(F(1)\).

The interval is open at \(-2\) and closed at \(1\), so its probability is \(F(1)-F(-2)\). Substitute the given values and solve:

0.47=F(1)-0.18 \qquad\Longrightarrow\qquad F(1)=0.47+0.18=0.65.

The result is within the required range \([0,1]\), and it satisfies \(F(1)\ge F(-2)\), as a CDF must. The endpoint convention matters: using \(F(-2-)\) here would correspond to including \(-2\), which this interval does not do.

Solving a Finite Collection of Statements

When several probability statements are given, assign a symbol to each unknown CDF value and translate every statement into an equation. Solve the resulting system, then check that the answers obey the basic CDF constraints: each value lies between zero and one, and CDF values cannot decrease as their arguments increase. Endpoint statements involving \(<\) may require left-limit values as additional unknowns.

Worked Example: Solving for Several CDF Values

Suppose the following probabilities are known: \(P(X\le0)=0.25\), \(P(0<X\le2)=0.40\), \(P(2<X\le5)=0.20\), and \(P(X>5)=0.15\). Find \(F(0)\), \(F(2)\), and \(F(5)\), and verify the probabilities are consistent.

The first statement gives \(F(0)=0.25\). The second interval has the form \((0,2]\), so \(P(0<X\le2)=F(2)-F(0)\). Hence

$$ F(2)=0.40+F(0)=0.40+0.25=0.65. $$

The third interval similarly gives \(P(2<X\le5)=F(5)-F(2)\), so

$$ F(5)=0.20+F(2)=0.20+0.65=0.85. $$

Finally, the complement identity gives \(P(X>5)=1-F(5)=1-0.85=0.15\), agreeing with the fourth statement. The recovered values satisfy \(0.25\le0.65\le0.85\), so they pass the monotonicity check as well as the probability check.

Takeaway: A finite set of probability statements becomes a finite set of equations after each event is translated with its correct endpoints. Solving the equations recovers requested values only when the information is sufficient to determine them; probability bounds and CDF monotonicity provide essential checks.

When the Information Is Not Enough

An interval probability by itself usually gives a difference, not an absolute CDF value. If \(P(a<X\le b)=q\), the equation is \(F(b)-F(a)=q\). Without another statement anchoring either endpoint, this one equation does not specify both unknown values. Do not silently assume \(F(a)=0\); that would require additional information, such as \(X>a\) with probability one.

Worked Example: Recognizing an Undetermined CDF Value

Suppose \(P(1<X\le3)=0.50\), and no other probability information is provided. Can \(F(3)\) be determined?

The interval identity gives

$$ F(3)-F(1)=0.50. $$

This specifies a difference, not either term alone. For example, the pairs \(F(1)=0.10,\ F(3)=0.60\) and \(F(1)=0.30,\ F(3)=0.80\) both satisfy the equation and the bounds \(0\le F(1)\le F(3)\le1\). They give different values of \(F(3)\). Thus the supplied statement does not determine \(F(3)\); another probability statement is needed.

The same issue arises with endpoint information. If \(P(X<1)=0.20\), \(P(1\le X\le3)=0.60\), and \(P(X>3)=0.20\), then \(F(1-)=0.20\), \(F(3)=0.80\), and \(F(3)-F(1-)=0.60\). These statements still do not determine \(F(1)\), because they do not specify how much probability is concentrated at \(1\).

Common Mistake: Dropping the Endpoint

A frequent error is to treat \(P(X>x)\) and \(P(X\ge x)\) as interchangeable, or to use \(F(x)\) whenever an endpoint \(x\) appears. That reasoning is only valid when the point probability \(P(X=x)\) is zero. In general,

P(X\ge x)=1-F(x-),\qquad P(X>x)=1-F(x),\qquad F(x)-F(x-)=P(X=x).

For example, if \(P(X<2)=0.30\) and \(P(X=2)=0.15\), then \(F(2)=0.45\), not \(0.30\). The correction is to preserve each inequality exactly, translate it to \(F(x)\) or \(F(x-)\), and only then perform the algebra.

Check Your Understanding

For each question, write the relevant probability equation before solving or deciding what can be known.

If \(P(X\le3)=0.62\), what are \(F(3)\) and \(P(X>3)\)?
If \(F(1)=0.40\) and \(P(1<X\le4)=0.35\), find \(F(4)\).
Express \(P(a\le X<b)\) using \(F\) and left limits.
If \(P(2<X\le5)=0.30\) is the only information supplied, can \(F(5)\) be determined? Explain.
If \(P(X<6)=0.72\) and \(P(X\le6)=0.80\), find \(P(X=6)\).