Why a Density Is Not a Point Probability

Density Height and Probability Are Different

A probability density can be positive at a value $x$ even though the probability that a random variable equals exactly $x$ is zero. This is not a contradiction: the density describes how probability is distributed across ranges of values, while a point is only one value and has no width. To find a probability, integrate the density over an interval.

The distinction matters whenever measurements are modeled continuously. A hypothetical claim duration might be recorded in days, a machine operating time in hours, or a financial loss in dollars. A density value has units inverse to the measurement units; a probability has no units and must lie between zero and one.

Definition: A random variable $X$ has probability density function $f_X$ if $f_X(x)\geq 0$, its integral over the real line is 1, and probabilities of intervals are given by integrating the density. In particular, for $a<b$, $$ P(a<X<b)=\int_a^b f_X(x)\,dx. $$ A density model is appropriate when the distribution is absolutely continuous. A density value $f_X(x)$ is not, in general, the probability $P(X=x)$.

A density may be specified by different formulas on different parts of its support, and it may be zero outside that support. Before using it, check that it is nonnegative and integrates to 1. Those are the requirements for a valid density; they do not make the density height itself a probability.

Why a Single Point Has Probability Zero

For a variable with a density, the probability assigned to an interval is the area under the density across that interval. A single point has zero width, so its integral is zero. More formally, for any real number $c$, the probability of the singleton set $\{c\}$ is the integral of the density over that set, and that integral is zero:

P(X=c)=\int_{\{c\}} f_X(x)\,dx=0.

The same result follows by integrating over intervals that shrink around $c$, when the density is bounded near $c$. If $f_X(x)\leq M$ near $c$, then for any sufficiently small $\varepsilon>0$,

0\leq P(c-\varepsilon<X<c+\varepsilon) =\int_{c-\varepsilon}^{c+\varepsilon} f_X(x)\,dx \leq 2M\varepsilon.

As $\varepsilon$ approaches zero, the upper bound approaches zero. This shrinking-interval argument is useful intuition, but the integral over the singleton gives the general result for a distribution with a density. For general integrable densities, integrals are understood in the Lebesgue sense; Riemann-sum approximations are valid under suitable regularity conditions, such as Riemann integrability, and should not be assumed for every density or every choice of sample points.

Theorem: If $X$ has a probability density, then $P(X=c)=0$ for every individual value $c$. Consequently, adding or removing an endpoint does not change an interval probability: $$ P(a<X<b)=P(a\leq X<b)=P(a<X\leq b)=P(a\leq X\leq b). $$

The statement depends on the model having a density. Some random variables can assign positive probability to particular points, in which case a density alone does not describe all their probability. Do not apply the zero-point-probability conclusion to a variable that has point masses.

Calculating Probabilities by Integration

To find the probability that $X$ falls between two values, integrate its density across that range. If the interval crosses a point where the formula for the density changes, split the integral at that point and add the pieces. The result is an area, not a density height.

Identify the interval and support.
Confirm which values of $X$ are possible and locate any changes in the density formula.

Set up the integral.
Integrate the appropriate density over the requested interval, splitting it wherever its formula changes.

Evaluate and check.
Add the integral pieces. The result must be between zero and one and must have no measurement units.

Worked Example: Operating Time Under a Uniform Density

A hypothetical sensor's operating time $X$, measured in hours, has density $f_X(x)=1/8$ for $0<x<8$, and zero otherwise. Find the probability that it operates for more than 2 hours but less than 5 hours, and find the probability that it operates for exactly 4 hours.

Method. Integrate the density over the requested interval. The density is constant on $(0,8)$, so the interval probability is its height multiplied by the interval width:

P(2<X<5)=\int_2^5 \frac18\,dx =\frac18(5-2)=\frac38=0.3750.

The interval has width $5-2=3$ hours, and the density is $1/8$ per hour. Their product is dimensionless, as a probability must be. For exactly 4 hours,

P(X=4)=\int_{\{4\}}\frac18\,dx=0.

Check. The probability $0.3750$ is between zero and one. The event $2<X<5$ is a subset of the full support $(0,8)$, and the point probability is zero. The common mistake to avoid is reporting $f_X(4)=1/8$ as the probability of operating for exactly 4 hours; $1/8$ is a density per hour, not a probability.

Density Values Carry Units

If $X$ is measured in hours, the density has units of inverse hours. If $X$ is measured in dollars, the density has units of inverse dollars. Integrating multiplies the density by a measurement interval, canceling the units and producing a probability. This explains why a density can exceed 1: its height is not a probability and need not be at most 1.

For a short interval around $x$, the probability is approximately the density near $x$ multiplied by the interval's width, when the density varies little over that interval. This is an approximation for a narrow interval, not a formula for the probability at a point. The exact probability for an interval is its integral.

Worked Example: A Triangular Density

A hypothetical claim-review time $X$, in days, has density $f_X(x)=x/8$ for $0<x<4$, and zero otherwise. Find the probability that the review takes between 1 and 3 days, and the probability that it takes exactly 2 days.

Check the model. The density is nonnegative on its support. Its total integral is

\int_0^4 \frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_0^4 =\frac{16}{16}-0=1.

Thus it is a valid density. Integrate from 1 to 3 for the interval probability:

P(1<X<3)=\int_1^3 \frac{x}{8}\,dx =\left[\frac{x^2}{16}\right]_1^3 =\frac{9}{16}-\frac{1}{16} =\frac{8}{16} =\frac12=0.5000.

At the point $x=2$, the density height is $f_X(2)=2/8=1/4$ per day. But the point probability is

P(X=2)=\int_{\{2\}}f_X(x)\,dx=0.

Check. The interval probability is $0.5000$, a valid probability, while $1/4$ per day is a density with units. The common mistake is to say $P(X=2)=1/4$ because the density at 2 equals $1/4$. The zero-width point has probability zero; a nonzero interval around 2 is needed to obtain a positive probability.

Intervals That Cross a Change in the Density

A piecewise density requires attention to the interval's location. If the requested interval crosses the boundary between two formulas, calculate the area on each side using its own formula. The value of the density at the boundary itself does not change the integral or create a point probability.

Worked Example: Integrating Across a Density Boundary

A hypothetical device lifetime $T$, measured in hours, has density $f_T(t)=t/6$ for $0<t<2$, $f_T(t)=2/3$ for $2\leq t<3$, and zero otherwise. Find $P(1<T<2.5)$ and $P(T=2)$.

Check the model. The total integral over the support is

\int_0^2 \frac{t}{6}\,dt+\int_2^3 \frac23\,dt =\left[\frac{t^2}{12}\right]_0^2+\frac23(3-2) =\frac{4}{12}+\frac23 =\frac13+\frac23=1.

The requested interval crosses the boundary at 2, so split it there:

\begin{aligned} P(1<T<2.5) &=\int_1^2 \frac{t}{6}\,dt+\int_2^{2.5}\frac23\,dt\\ &=\left[\frac{t^2}{12}\right]_1^2+\frac23(2.5-2)\\ &=\frac{4-1}{12}+\frac23(0.5)\\ &=\frac14+\frac13=\frac7{12}=0.5833. \end{aligned}

For the boundary point itself, the variable has a density, so

$$ P(T=2)=0. $$

Check. The total density integrates to 1, and the interval probability is $7/12$, between zero and one. The common mistake is to use the second density formula across the entire interval from 1 to 2.5. That would ignore the actual formula $t/6$ on $(1,2)$; splitting at 2 gives the correct result.

Common Mistakes and Exam Checks

Using a density height as a point probability. A value such as $f_X(c)=0.4$ is not $P(X=c)$. For a variable with a density, $P(X=c)=0$.
Forgetting the interval width. For a constant density $k$ across an interval of width $w$, the probability is $kw$, not $k$.
Using one formula across a piecewise boundary. Split the integral wherever the density changes.
Ignoring units. A density has inverse measurement units; an integrated probability does not.
Assuming every continuous-valued model has zero point probabilities. Check that the model is absolutely continuous with a density. A distribution with point masses needs separate treatment.

Key takeaway: A density measures probability per unit of measurement. Find probabilities by integrating over intervals; for a random variable with a density, every individual point has probability zero, regardless of the density height there.

Check Your Understanding

For each model, verify that it is a valid density where requested, then distinguish density heights from probabilities.

A hypothetical inspection time $X$, in hours, has density $1/10$ for $0<x<10$, and zero otherwise. Find $P(3<X<7)$ and $P(X=4)$. (Answer: $0.4000$ and $0$, respectively.)
A hypothetical processing delay $Y$ has density $y/18$ for $0<y<6$, and zero otherwise. Find $P(2<Y<4)$ and $P(Y=3)$. (Answer: $1/3=0.3333$ and $0$, respectively.)
A hypothetical repair time $Z$, in days, has density $1/4$ for $0<z<2$, $1/8$ for $2\leq z<6$, and zero otherwise. Verify the total integral and find $P(1<Z<4)$. (Answer: total integral $1$; probability $0.5000$.)
A hypothetical review duration $W$ has density $2w/25$ for $0<w<5$, and zero otherwise. Find $P(1<W<3)$, and state whether $f_W(2)$ is the probability of exactly 2 units. (Answer: $0.3200$; no, $f_W(2)=0.1600$ per unit while $P(W=2)=0$.)
A hypothetical equipment lifetime $R$, in years, has density $1/6$ for $0<r<2$, $1/3$ for $2\leq r<4$, and zero otherwise. Verify the density and find $P(0.5<R<3)$ and $P(R=2)$. (Answer: total integral $1$; probability $7/12=0.5833$; point probability $0$.)

What You'll Learn