Reading the Components of a Hybrid CDF
A cumulative distribution function can encode different kinds of probability in one formula. A jump represents probability concentrated at a single value, while a smooth increase can represent probability spread continuously over an interval. To identify the components correctly, read the CDF from left to right: compare its left and right values at candidate jump locations, then examine how it changes between those locations.
The key is to keep the two calculations distinct. At a point, the size of a jump gives the atom’s probability. On an interval where the CDF is absolutely continuous, its derivative gives the density (almost everywhere). Finally, add the atom probabilities and integrate the density to check that the total probability is one.
Right-continuity is part of the definition of a CDF, so \(F(x)\) gives the value after any jump at \(x\). The value just before that jump is \(F(x-)\). This distinction matters at boundaries as well as at interior points: if a CDF jumps at the left endpoint of a density interval, the jump contributes an atom, while the increase along the interval contributes continuous probability.
A Systematic Recovery Method
Suppose a CDF is specified by finitely many pieces. First list every point where the formula jumps, and compute \(F(x)-F(x-)\) at each one. Next, differentiate on each smooth piece that is absolutely continuous. The resulting density is zero outside the regions where the CDF increases continuously. Then verify that the atom sizes and integrated density account for one unit of probability.
At each candidate point \(x\), compare the CDF value \(F(x)\) with its left limit \(F(x-)\). A positive difference is an atom.
On each absolutely continuous interval, take \(F'(x)\) almost everywhere. Check that the derivative is nonnegative.
Add all jump sizes to the integrals of the density over its intervals. The sum must be one, and the CDF must be nondecreasing with limits zero and one.
Worked Example: Two Atoms and Two Density Regions
Consider the CDF
At \(x=0\), the left limit is \(F(0-)=0\), while \(F(0)=0.2\). Thus \(P(X=0)=0.2-0=0.2\). As \(x\) increases from 0 toward 1, the CDF rises from 0.2 toward \(0.2+0.3(1)=0.5\), so the density on \((0,1)\) is \(0.3\).
At \(x=1\), the left limit is \(0.2+0.3(1)=0.5\), and the CDF value is \(F(1)=0.7+0.15(0)=0.7\). The atom at 1 therefore has size \(0.7-0.5=0.2\). On \((1,3)\), the derivative is \(0.15\). The CDF reaches \(0.7+0.15(3-1)=1\) at 3, so it has no jump there.
The continuous probability is \(0.3(1-0)=0.3\) on \((0,1)\) and \(0.15(3-1)=0.3\) on \((1,3)\). Including the atoms gives \(0.2+0.2+0.3+0.3=1\), as required.
Why the Probability Check Works
The total-probability check is more than a convenient arithmetic test. Under the appropriate regularity conditions, it follows directly by adding the continuous increases and the jumps. Absolute continuity on the pieces ensures that each continuous increase equals the integral of the derivative there.
Proof. Nondecreasingness of a CDF implies that each jump \(F(x_i)-F(x_i-)\) is nonnegative. On each interval without a jump, absolute continuity and the fundamental theorem of calculus identify the integral of \(f\) with the increase in \(F\) across that interval. For finite endpoints \(a<b\) in such an interval, this gives \(\int_a^b f(x)\,dx=F(b)-F(a)\). Taking one-sided limits gives the corresponding identity up to a jump location; on the tails, the assumed improper-integral condition gives the identity using the limits 0 and 1.
Add these continuous increases over all the intervals and add the jumps at \(x_1,\ldots,x_k\). Each intermediate endpoint value cancels: the value after one interval is followed by the jump, if any, and then by the next interval’s increase. The sum therefore telescopes from \(F(-\infty)=0\) to \(F(\infty)=1\). The continuous contributions combine to \(\int_{-\infty}^{\infty}f(x)\,dx\), and each jump contributes \(F(x_i)-F(x_i-)\). This proves the identity. \(\square\)
With these assumptions, the recovered components describe the whole law: probability at each jump location is the jump size, and probability away from those atoms is obtained by integrating \(f\). Equivalently, for any Borel set \(A\),
The conditions matter. A CDF that is merely differentiable between jumps need not be absolutely continuous there; differentiability alone does not guarantee that integrating its derivative recovers the full increase. For the piecewise CDFs considered here, confirm that the stated pieces are absolutely continuous before interpreting their derivatives as the complete continuous component.
Worked Example: A Boundary Atom and a Continuous Endpoint
Let
At the left boundary \(x=-2\), \(F(-2-)=0\) and \(F(-2)=0.1\), giving an atom of size \(0.1\). On \((-2,0)\), the density is \(0.2\); its integrated probability is \(0.2(0-(-2))=0.4\). The left limit at zero is \(0.1+0.2(0+2)=0.5\), whereas \(F(0)=0.75\), so there is an atom of size \(0.75-0.5=0.25\).
On \((0,2)\), the density is \(0.125\), and its integral is \(0.125(2-0)=0.25\). At \(x=2\), the left limit is \(0.75+0.125(2)=1\), equal to \(F(2)=1\), so there is no atom at 2. The total is \(0.1+0.4+0.25+0.25=1\). This also shows why a boundary value must be compared with the left limit rather than judged from the formula’s endpoint appearance.
Worked Example: A Curved Piece and a Final Accounting Check
Consider
At zero the jump is \(0.1-0=0.1\). At one, the left limit is \(0.1+0.2+0.1=0.4\), while \(F(1)=0.5\), giving a jump of \(0.1\). At two, the left limit is \(0.5+0.2+0.1=0.8\), while \(F(2)=0.85\), giving a jump of \(0.05\). At three the left limit is \(0.85+0.15(1)=1\), so there is no jump.
Differentiating the three continuous pieces gives
with values at the finitely many endpoints irrelevant to the density. The first integral is \(\int_0^1(0.2+0.2x)\,dx=0.2+0.1=0.3\). The second is \(\int_1^2[0.2+0.2(x-1)]\,dx=0.2+0.1=0.3\), and the third is \(\int_2^3 0.15\,dx=0.15\). The atom total is \(0.1+0.1+0.05=0.25\). Thus the full probability is \(0.25+0.3+0.3+0.15=1\). Each derivative is nonnegative on its interval, and each piece joins the next by either an upward jump or the appropriate endpoint limit, so the formula is consistent with a nondecreasing CDF.
Common Mistakes and Practical Checks
A frequent error is to treat the CDF value at a point as the atom size. It is the difference between the value after the jump and the left limit that gives the atom. For example, in the first worked example \(F(1)=0.7\), but the atom at 1 is \(0.7-0.5=0.2\), not \(0.7\). Another error is to call the derivative at a point the probability of that point. The derivative describes continuous probability per unit length; its integral over an interval gives probability there. A point’s atom instead comes from a jump.
When checking a proposed decomposition, verify both the local and global conditions. Locally, jump sizes and densities must be nonnegative, and the CDF must not decrease between pieces. Globally, the CDF must start at limit zero, end at limit one, and the sum of all atom sizes and density integrals must equal one. A mismatch signals a missed jump, an incorrect endpoint limit, a derivative or integration error, or an invalid CDF.
Check Your Understanding
Use the jump-and-derivative method, and show the probability accounting in each case.
- If \(F(a-)=0.35\) and \(F(a)=0.52\), what is \(P(X=a)\)?
- A CDF rises linearly from 0.2 at \(x=1\) to 0.8 at \(x=4\), with no jump at either endpoint. What is its density on \((1,4)\), and how much probability lies in that interval?
- A piecewise CDF has atom sizes 0.15 and 0.10, and its density integrates to 0.70. What does the total tell you to check?
- Why is differentiability between jumps alone not enough to guarantee that the atoms and derivative account for all probability?
- If \(F(b-)=1\) and \(F(b)=1\), is there an atom at \(b\)? Explain using the jump formula.