Recovering a Density from a Differentiable CDF

Why Recover a Density from a CDF?

A CDF records accumulated probability: $F_X(x)=P(X\leq x)$. When the distribution is continuous in the appropriate sense, the rate at which this accumulation changes gives the density. The qualification matters. Differentiating a CDF is not automatically a way to find every part of a distribution: a jump represents point probability, while a singular continuous component can escape the derivative almost everywhere.

The key question is therefore not merely whether $F_X$ has a derivative at some points. It is whether the distribution is absolutely continuous on the region being studied. If it is, the derivative recovers the density almost everywhere on that region.

Definition: A random variable $X$ has a probability density function $f_X$ if probabilities over intervals are obtained by integration: for all $a<b$, $$ P(a<X\leq b)=\int_a^b f_X(x)\,dx, $$ with $f_X(x)\geq 0$ and total integral $1$. Its CDF is absolutely continuous when it can be written as the integral of an integrable density. In that case, $F_X'(x)=f_X(x)$ for almost every $x$, and $F_X(b)-F_X(a)=\int_a^b f_X(x)\,dx$.

“Almost every” means that the equality may fail on a set of $x$-values whose length is zero. Changing a density at isolated points, or on any set of length zero, does not change the probabilities its integrals give. Thus a density is determined by the CDF derivative only up to such changes.

When Differentiation Works

If a CDF is continuously differentiable on an interval, the fundamental theorem of calculus gives the density there by differentiation. More generally, absolute continuity on the interval is sufficient, even if the derivative is not continuous. In either case, the derivative must be nonnegative almost everywhere because a CDF is nondecreasing.

Here is the justification. Suppose $F_X$ is absolutely continuous on $[a,b]$. The fundamental theorem for absolutely continuous functions gives

F_X(b)-F_X(a)=\int_a^b F_X'(x)\,dx.

The left side is $P(a<X\leq b)$. Therefore the interval probability is the integral of $F_X'$, so $F_X'$ serves as a density on that interval. This reasoning depends on absolute continuity. Differentiability alone does not guarantee the integral identity.

If the whole distribution is absolutely continuous, the same result applies across its support, and

F_X(x)=\int_{-\infty}^{x} f_X(t)\,dt, \qquad f_X(x)=F_X'(x)\ \text{almost everywhere}.

A useful check is that the density integrates to the probability represented by the CDF increase. For a fully continuous distribution, that total probability is $1$. If the derivative integrates to less than $1$, or the CDF has jumps, do not treat the derivative as a complete description of the distribution.

Worked Example: Differentiating a Piecewise CDF

A hypothetical device has a random calibration error $X$, measured in units. Its CDF is

F_X(x)= \begin{cases} 0, & x<0,\\ x^2/8, & 0\leq x<2,\\ 1-(4-x)^2/8, & 2\leq x<4,\\ 1, & x\geq 4. \end{cases}

Find the density on each interval, check whether this CDF describes an absolutely continuous distribution, and calculate $P(1<X\leq 3)$.

Method and interval derivatives. Each formula is continuously differentiable inside its interval. For $0<x<2$, $$ F_X'(x)=\frac{2x}{8}=\frac{x}{4}. $$ For $2<x<4$, $$ F_X'(x)=-\frac{2(4-x)(-1)}{8}=\frac{4-x}{4}. $$ Outside $[0,4]$, the CDF is constant, so its derivative is $0$.

Check the joins. At $x=0$, the CDF starts at $0$. At $x=2$, the left-hand value is $2^2/8=1/2$, and the right-hand value is $1-(4-2)^2/8=1-4/8=1/2$. At $x=4$, the left-hand value is $1$, matching the constant value to the right. There are no jumps. Each piece is smooth and the pieces join continuously, so this CDF is absolutely continuous.

Check total probability. $$ \int_0^2 \frac{x}{4}\,dx =\left[\frac{x^2}{8}\right]_0^2 =\frac{4}{8}=\frac12, \qquad \int_2^4 \frac{4-x}{4}\,dx =\left[-\frac{(4-x)^2}{8}\right]_2^4 =\frac12. $$ The total integral is $1/2+1/2=1$, as required.

Calculate the interval probability. Since the distribution is continuous at $1$ and $3$, $$ P(1<X\leq 3)=F_X(3)-F_X(1). $$ The values are $F_X(3)=1-(4-3)^2/8=7/8$ and $F_X(1)=1/8$, so the probability is $7/8-1/8=6/8=3/4$. Integrating the density gives the same result: $$ \int_1^2 \frac{x}{4}\,dx+\int_2^3 \frac{4-x}{4}\,dx =\frac{3}{8}+\frac{3}{8}=\frac34. $$

Check: the density integrates to $1$, and its integral over $(1,3]$ agrees with the CDF difference. A common mistake is to differentiate only one piece and use that formula over the entire support. The derivative changes at $2$, so each interval must be handled using its own CDF formula.

Jumps and Missing Point Probability

A CDF jump at $c$ has size $F_X(c)-F_X(c-)=P(X=c)$, where $F_X(c-)$ is the left limit. A derivative on either side of $c$ does not record this point probability. The distribution can therefore have a density for its continuous part and still fail to have a density for its entire probability.

To check a CDF with jumps, compute the derivative on its smooth intervals, then separately identify each jump. The integrals of the derivatives account for the absolutely continuous probability; the jump sizes account for the atoms. These components must not be confused or counted twice.

Worked Example: A Density Plus an Atom

A hypothetical reimbursement amount $X$, in hundreds of dollars, has CDF

F_X(x)= \begin{cases} 0, & x<0,\\ 0.35x, & 0\leq x<1,\\ 0.35x+0.30, & 1\leq x<2,\\ 1, & x\geq 2. \end{cases}

Identify the density away from the jump, find the point mass, and verify that all probability is accounted for.

Method: differentiate between jumps and calculate jump sizes separately. On both $(0,1)$ and $(1,2)$, the derivative is $0.35$. It is $0$ outside $[0,2]$. Thus the absolutely continuous part has density $f_X(x)=0.35$ on $(0,2)$, except that this formula alone does not describe the entire distribution.

At $x=1$, the left limit is $$ F_X(1-)=0.35(1)=0.35, $$ while $F_X(1)=0.35(1)+0.30=0.65$. The jump is $0.65-0.35=0.30$, so $P(X=1)=0.30$.

The continuous probability is $$ \int_0^2 0.35\,dx=0.35(2)=0.70. $$ Adding the atom gives $0.70+0.30=1.00$. The total probability is accounted for, but the full distribution has no density alone because it places positive probability at exactly $1$.

For example, the continuous contribution to $P(0<X\leq 1)$ is $0.35(1)=0.35$, and the atom at $1$ adds $0.30$, giving $P(0<X\leq 1)=0.65$. This agrees with $F_X(1)-F_X(0)=0.65-0=0.65$. A common mistake is to report the derivative $0.35$ as a complete density and overlook the jump; integrating that derivative gives only $0.70$, not $1$.

Why a Derivative Almost Everywhere May Still Be Incomplete

Not every CDF is a mixture of smooth pieces and jumps. A distribution can also be singular continuous: its CDF is continuous, so it has no jumps, but its probability is concentrated on a set of length zero. In that case the CDF derivative can be zero almost everywhere even though the distribution has total probability $1$.

The standard example is the Cantor distribution. Its CDF is continuous and nondecreasing, and its derivative is $0$ almost everywhere. If that derivative were a density for the whole distribution, its integral would be $0$, implying total probability $0$. But a CDF increases from $0$ to $1$, so the distribution has total probability $1$. The contradiction shows that the CDF is not absolutely continuous and its derivative does not recover its probability.

Worked Example: Testing Whether a Derivative Is a Complete Density

A candidate examines the Cantor distribution and proposes $f_X(x)=F_X'(x)$ as its density. The CDF is continuous, increases from $0$ to $1$, and has derivative $0$ almost everywhere. Determine whether the proposal is valid.

Method: compare the derivative integral with the total CDF increase. Since $F_X'(x)=0$ almost everywhere, the integral of that derivative over the real line is $0$. If it were a density, the total probability would therefore be $$ \int_{-\infty}^{\infty} f_X(x)\,dx =\int_{-\infty}^{\infty}0\,dx=0. $$

However, the CDF increases from $0$ to $1$, so the distribution's total probability is $1$, not $0$. The proposed derivative cannot be a density for the whole distribution. The missing probability is singular continuous probability: there are no jumps to identify as atoms, but the CDF is not an integral of its derivative.

Check: continuity alone does not ensure absolute continuity. The common mistake is to infer that a continuous CDF with derivative zero almost everywhere must describe a distribution with no probability. The CDF's total increase rules that out.

A Reliable Procedure and Common Traps

Check the CDF's structure.
Locate intervals with differentiable formulas, possible jumps, and any stated or suspected singular component.

Differentiate on the smooth intervals.
Use the formula for the relevant interval, and check that the derivative is nonnegative almost everywhere.

Check absolute continuity.
If the CDF is absolutely continuous on the region, its derivative gives the density almost everywhere there. Differentiability at individual points is not enough.

Account for jumps separately.
At each jump $c$, calculate $F_X(c)-F_X(c-)$. This is point probability, not density.

Verify probability.
Integrate the density and add any atom probabilities. For a complete distribution, these components must sum to $1$.

The most important trap is treating “has a derivative” as equivalent to “has a density.” The valid implication is narrower: for an absolutely continuous CDF, its derivative is a density almost everywhere. A jump needs a separate point-mass calculation, and a singular continuous component requires more than differentiating the CDF. This distinction is aligned with the SOA VEE Mathematical Statistics syllabus.

Key takeaway: Differentiate a CDF to recover a density only when absolute continuity justifies the integral relationship. Check the derivative on each interval, check for jumps, and verify that all probability—not just the derivative's integral—has been accounted for.

Check Your Understanding

For each problem, distinguish the derivative on smooth intervals from any probability not represented by that derivative.

A CDF is $0$ for $x<0$, $x^3/27$ for $0\leq x<3$, and $1$ for $x\geq 3$. Find the density on $(0,3)$ and $P(1<X\leq 2)$. (Answer: $f_X(x)=x^2/9$ on $(0,3)$; $P(1<X\leq 2)=7/27$.)
A CDF is the mixture of weight $0.8$ on a uniform distribution over $[0,4]$ and weight $0.2$ at $x=2$. Find its density away from the atom, its jump size, and the continuous probability. (Answer: density $0.2$ on $(0,4)$; jump $0.2$ at $2$; continuous probability $0.8$.)
A CDF is $0$ for $x<0$, $x/3$ for $0\leq x<1$, $1/3+(x-1)/3$ for $1\leq x<3$, and $1$ for $x\geq 3$. Find its density and check its total integral. (Answer: $f_X(x)=1/3$ on $(0,3)$; total integral $1$.)
A continuous CDF increases from $0$ to $1$, but its derivative is $0$ almost everywhere. Can the derivative be a density for the entire distribution? Explain using the total integral condition. (Answer: No; the derivative integrates to $0$, while a density for the entire distribution must integrate to $1$.)
A CDF is $0$ for $x<0$, $x/4$ for $0\leq x<2$, $0.75+(x-2)/4$ for $2\leq x<3$, and $1$ for $x\geq 3$. Find its density on the smooth intervals and any point mass. (Answer: density $1/4$ on $(0,2)$ and $(2,3)$; point mass $0.25$ at $2$.)

What You'll Learn