Probability Densities

A probability density is not a probability at a point. It is a function that must be integrated over a region to give a probability.

For a real-valued random variable $X$, a density $f_X(x)$ satisfies

$$\mathbb P(X\in\Delta) = \int_\Delta f_X(x)\,dx$$

for allowed sets of values $\Delta$. The density tells how probability is distributed with respect to the reference measure $dx$.

This page explains the mathematical rules for densities. The physical Born-rule page for continuous observables is Born Rule for Continuous Spectra, and the wave-mechanics page is Wavefunctions and Probability Density.

Density with Respect to a Measure

Let $X$ be a real-valued random variable. A probability density for $X$ with respect to $dx$ is a nonnegative function $f_X$ such that

$$\mathbb P_X(\Delta) = \int_\Delta f_X(x)\,dx.$$

Here $\mathbb P_X$ is the distribution, or law, of $X$:

$$\mathbb P_X(\Delta) = \mathbb P(X\in\Delta).$$

The phrase “with respect to $dx$” matters. A density is always density relative to a chosen measure. In elementary real-variable examples the reference measure is usually length $dx$. In three-dimensional position space it is volume $d^3r$. In spherical coordinates the same volume measure becomes

$$d^3r = r^2\sin\theta\,dr\,d\theta\,d\phi.$$

Changing the reference measure changes the function that deserves to be called the density.

Normalization

A density must satisfy

$$f_X(x)\ge0$$

and

$$\int_{-\infty}^{\infty} f_X(x)\,dx=1$$

for a real-valued variable on the full line. More generally, over a domain $D$,

$$\int_D f(x)\,dx=1.$$

For a three-dimensional density $\rho(\mathbf r)$,

$$\int_{\mathbb R^3} \rho(\mathbf r)\,d^3r = 1.$$

Normalization is a probability statement, not an arbitrary scale convention. In quantum mechanics, normalizing a wavefunction ensures that the total probability of finding the particle somewhere in the configuration space is one.

Density Values Are Not Probabilities

For an ordinary continuous density,

$$\mathbb P(X=x_0)=0.$$

Instead, probabilities come from intervals:

$$\mathbb P(a\le X\le b) = \int_a^b f_X(x)\,dx.$$

For a small interval centered at $x_0$,

$$\mathbb P \left( x_0-\frac{\Delta x}{2} \le X\le x_0+\frac{\Delta x}{2} \right) \approx f_X(x_0)\Delta x$$

when $f_X$ varies slowly over the interval.

This is the operational meaning of density: it gives the leading probability per unit width for sufficiently small bins.

Units

A density carries inverse units of its variable. If $X$ has units of length, then $f_X$ has units of inverse length. This makes

$$f_X(x)\,dx$$

dimensionless.

For a one-dimensional wavefunction,

$$\rho(x) = \lvert\psi(x)\rvert^2$$

has units of inverse length, so $\psi(x)$ has units of length$^{-1/2}$. In three dimensions, $\lvert\psi(\mathbf r)\rvert^2$ has units of inverse volume, and $\psi(\mathbf r)$ has units of length$^{-3/2}$.

Units are a useful guardrail when changing variables. If $p=\hbar k$, then a momentum density and a wave-number density cannot be the same numerical function without a Jacobian.

Cumulative Distribution Function

Every real-valued random variable has a cumulative distribution function

$$F_X(x) = \mathbb P(X\le x).$$

If $X$ has a density $f_X$, then

$$F_X(x) = \int_{-\infty}^{x} f_X(t)\,dt.$$

At points where $F_X$ is differentiable,

$$f_X(x) = \frac{dF_X}{dx}.$$

This relation is often the easiest way to derive a density from a known distribution function. It also shows why a density can be changed at isolated points without changing any probabilities: intervals see integrals, not point values.

Change of Variables

Suppose $Y=g(X)$ and $g$ is one-to-one and differentiable. Conservation of probability gives

$$f_Y(y)\,dy = f_X(x)\,dx.$$

Since $x=g^{-1}(y)$,

$$f_Y(y) = f_X(g^{-1}(y)) \left\lvert \frac{d}{dy}g^{-1}(y) \right\rvert.$$

Equivalently,

$$f_Y(y) = \frac{f_X(x)}{\lvert g'(x)\rvert} \qquad \text{with } y=g(x).$$

The absolute value is essential. Densities must stay nonnegative even when a coordinate map reverses orientation.

If the map is many-to-one, sum over all roots $x_i$ satisfying $g(x_i)=y$:

$$f_Y(y) = \sum_{x_i:g(x_i)=y} \frac{f_X(x_i)}{\lvert g'(x_i)\rvert}.$$

This formula is the source of many square-root singularities in energy, radial, and density-of-states calculations.

Multivariable Densities

For a vector random variable $\mathbf X=(X_1,\ldots,X_n)$, a joint density $f_{\mathbf X}(\mathbf x)$ satisfies

$$\mathbb P(\mathbf X\in R) = \int_R f_{\mathbf X}(\mathbf x)\,d^n x.$$

If $\mathbf y=g(\mathbf x)$ is an invertible differentiable coordinate change, then

$$f_{\mathbf Y}(\mathbf y) = f_{\mathbf X}(g^{-1}(\mathbf y)) \left\lvert \det \frac{\partial \mathbf x}{\partial \mathbf y} \right\rvert.$$

The determinant is the multivariable Jacobian. Omitting it changes probabilities.

For example, in three-dimensional position space,

$$d^3r = r^2\sin\theta\,dr\,d\theta\,d\phi.$$

If $\rho(\mathbf r)$ is the density per unit volume, then the probability in a region is

$$\int \rho(r,\theta,\phi) r^2\sin\theta\,dr\,d\theta\,d\phi,$$

not just the integral of $\rho$ over $dr\,d\theta\,d\phi$.

Marginal Densities

For two continuous random variables with joint density $f_{X,Y}(x,y)$, the marginal density of $X$ is obtained by integrating out $Y$:

$$f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dy.$$

Similarly,

$$f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y)\,dx.$$

The joint density contains correlation information that the marginals do not. Two different joint densities can have the same $f_X$ and $f_Y$ but different covariance; see Variance and Covariance.

Conditional densities are developed in Conditional Probability, but the basic formula, when $f_Y(y)>0$, is

$$f_{X\mid Y}(x\mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}.$$

This formula belongs to classical probability. Quantum measurement update has analogies to conditioning, but it is not just this formula applied to pre-existing values of all observables.

Wavefunction Probability Density

In one-dimensional wave mechanics, a normalized position-space wavefunction has

$$\int_{-\infty}^{\infty} \lvert\psi(x)\rvert^2\,dx = 1.$$

The position density is

$$\rho(x) = \lvert\psi(x)\rvert^2.$$

Thus

$$\mathbb P(X\in[a,b]) = \int_a^b \lvert\psi(x)\rvert^2\,dx.$$

The wavefunction $\psi$ is a probability amplitude. The density is its squared magnitude. The phase of $\psi$ is invisible in $\rho(x)$ at one instant, but it still matters for interference, current, and momentum-space structure.

In momentum representation, a normalized momentum-space wavefunction $\phi(p)$ gives

$$\mathbb P(P\in[p_1,p_2]) = \int_{p_1}^{p_2} \lvert\phi(p)\rvert^2\,dp.$$

The same state can have different density functions in different representations because the measured variable and reference measure have changed.

Radial Densities

A common source of mistakes is the difference between a density per unit volume and a density per unit radius.

For a three-dimensional wavefunction $\psi(\mathbf r)$,

$$\mathbb P(\mathbf r\in R) = \int_R \lvert\psi(\mathbf r)\rvert^2\,d^3r.$$

If the state is spherically symmetric, then the radial probability density $p_r(r)$ is defined so that

$$\mathbb P(r\in[a,b]) = \int_a^b p_r(r)\,dr.$$

The angular integration gives

$$p_r(r) = 4\pi r^2\lvert\psi(r)\rvert^2$$

for a spherically symmetric wavefunction. The extra $r^2$ is not optional; it is the volume element.

For angular-momentum eigenstates written as $R_{n\ell}(r)Y_{\ell m}(\theta,\phi)$ with normalized spherical harmonics, the radial density is

$$p_r(r) = r^2\lvert R_{n\ell}(r)\rvert^2.$$

The convention has changed because $R_{n\ell}$ is separated from a normalized angular factor.

Densities and Delta Notation

Some distributions are not represented by ordinary functions. A point mass at $x=a$ can be written formally as

$$\delta(x-a)$$

inside integrals, meaning

$$\int_{-\infty}^{\infty} \varphi(x)\delta(x-a)\,dx = \varphi(a).$$

This is useful notation, especially in quantum mechanics, but the delta function is not an ordinary probability density function. The distribution-theory treatment is Delta Function and Distributions.

Mixed distributions can have both a discrete part and a continuous-density part. For example, a detector model may include an atom at “no click” plus a continuous density over measured positions. In such cases, a single ordinary density with respect to $dx$ is not the whole probability law.

Common Mistakes

• Treating $f(x_0)$ as $\mathbb P(X=x_0)$.
• Forgetting the Jacobian when changing variables.
• Omitting volume elements such as $r^2\sin\theta$ in spherical coordinates.
• Comparing densities in different variables as if they had the same units.
• Assuming every distribution has an ordinary density.
• Confusing $\lvert\psi\rvert^2$ with the wavefunction itself.
• Inferring the full quantum state from a single position probability density.

Cross-Links

• Probability Spaces, Light Version
• Random Variables
• Expectation Values
• Variance and Covariance
• Conditional Probability
• Gaussian Distributions
• Entropy
• Born Rule for Continuous Spectra
• Wavefunctions and Probability Density
• Momentum Representation
• Delta Function
• Distributions

References

• P. Billingsley, Probability and Measure, 3rd ed., Wiley, 1995.
• R. Durrett, Probability: Theory and Examples, 5th ed., Cambridge University Press, 2019.
• W. Feller, An Introduction to Probability Theory and Its Applications, Volume II, 2nd ed., Wiley, 1971.
• G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Wiley, 1999.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Let $f(x)=Ce^{-ax}$ for $x\geq0$ and $f(x)=0$ for $x\lt0$, with $a\gt0$. Find $C$.

Solution

Normalize:

$$1 = \int_0^\infty Ce^{-ax}\,dx = \frac{C}{a}.$$

Thus

$$C=a.$$

2. If $X$ has density $f_X(x)$ and $Y=2X$, find $f_Y(y)$.

Solution

Here $x=y/2$, so

$$f_Y(y) = f_X(y/2) \left\lvert \frac{d}{dy}\frac{y}{2} \right\rvert = \frac12 f_X(y/2).$$

The factor $1/2$ keeps the total probability normalized.

3. Let $X$ be uniformly distributed on $[-1,1]$, and let $Y=X^2$. Find the density of $Y$.

Solution

The density of $X$ is $f_X(x)=1/2$ on $[-1,1]$. For $0\lt y\lt1$, the roots of $x^2=y$ are $x=\sqrt y$ and $x=-\sqrt y$. Since $g'(x)=2x$,

$$f_Y(y) = \frac{1/2}{2\sqrt y} + \frac{1/2}{2\sqrt y} = \frac{1}{2\sqrt y}.$$

The density is supported on $0\le y\le1$. The singularity at $y=0$ is integrable.

4. A three-dimensional spherically symmetric wavefunction is normalized as $\psi(r)=Ae^{-r/a}$ with $a>0$. Find $A$ for a real positive normalization.

Solution

Use the volume element:

$$1 = 4\pi A^2 \int_0^\infty r^2e^{-2r/a}\,dr.$$

The integral is

$$\int_0^\infty r^2e^{-\lambda r}\,dr = \frac{2}{\lambda^3}, \qquad \lambda=\frac{2}{a}.$$

Thus

$$\int_0^\infty r^2e^{-2r/a}\,dr = \frac{a^3}{4}.$$

So

$$1 = \pi A^2a^3, \qquad A = \frac{1}{\sqrt{\pi a^3}}.$$

5. Explain why changing the value of a continuous density at one isolated point does not change any interval probabilities.

Solution

Interval probabilities are integrals of the density. Changing a function at one point changes its integral by zero with respect to ordinary length measure. Therefore the probability of every interval remains the same.