Fourier Inversion for Densities

Why Fourier Inversion Matters

A characteristic function can encode a distribution in a compact formula, but recovering probabilities or a density from it requires an inversion argument. The key condition here is absolute integrability: when the characteristic function is integrable over the real line, its inverse Fourier transform is not merely a formal expression. It is a bounded continuous density for the distribution.

We use the convention $\varphi_X(t)=E[e^{itX}]$. With this sign in the characteristic function, the inverse formula has $e^{-itx}$ in its integrand. Keeping the two signs and the factor $1/(2\pi)$ consistent is essential.

Definition: The characteristic function of a real-valued random variable $X$ is $\varphi_X(t)=E[e^{itX}]$, for $t\in\mathbb{R}$. It is integrable if $\int_{\mathbb{R}}|\varphi_X(t)|\,dt<\infty$. The inverse Fourier transform used here is $g(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}\varphi_X(t)\,dt$.

The Fourier Inversion Theorem for Densities

Theorem (Fourier inversion for densities): Let $X$ be a real-valued random variable with characteristic function $\varphi_X$. If $\int_{\mathbb{R}}|\varphi_X(t)|\,dt<\infty$, then the distribution of $X$ has a bounded continuous density $g$, given at every $x\in\mathbb{R}$ by $$ g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-itx}\varphi_X(t)\,dt. $$ In particular, $g(x)\geq 0$ and $\int_{\mathbb{R}}g(x)\,dx=1$.

The absolute-integrability assumption makes the defining integral finite and also controls its dependence on $x$. It does not assume that a density already exists; the theorem establishes that the distribution is absolutely continuous.

Smooth the distribution.
Add an independent centered Gaussian with variance $\varepsilon>0$. The smoothed distribution has a density whose inverse-transform formula includes the factor $e^{-\varepsilon t^2/2}$.

Remove the smoothing in the inverse transform.
Integrability of $\varphi_X$ implies that the smoothed densities converge uniformly to the proposed inverse transform.

Identify the limiting distribution.
Against continuous functions of compact support, the smoothed distributions converge to the original distribution. This identifies the inverse transform as its density.

Proof by Gaussian Smoothing

For $\varepsilon>0$, let $Z_\varepsilon$ be independent of $X$ with distribution $N(0,\varepsilon)$. Its density is $q_\varepsilon(u)=(2\pi\varepsilon)^{-1/2}e^{-u^2/(2\varepsilon)}$, and its characteristic function is $e^{-\varepsilon t^2/2}$. The Gaussian Fourier formula gives $$ q_\varepsilon(u)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-itu}e^{-\varepsilon t^2/2}\,dt. $$

The random variable $X+Z_\varepsilon$ has density $p_\varepsilon(x)=E[q_\varepsilon(x-X)]$. Substituting the Gaussian Fourier formula and interchanging expectation and integration gives $$ p_\varepsilon(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}\varphi_X(t)e^{-\varepsilon t^2/2}\,dt. $$

This interchange is justified because the absolute value of the integrand before taking expectation is bounded by $e^{-\varepsilon t^2/2}$, which is integrable over $\mathbb{R}$. The expectation of $e^{itX}$ is $\varphi_X(t)$, giving the displayed expression.

Define $g(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}\varphi_X(t)\,dt$, which is absolutely convergent by assumption. Then $$ \sup_{x\in\mathbb{R}}|p_\varepsilon(x)-g(x)| \leq \frac{1}{2\pi}\int_{\mathbb{R}}|\varphi_X(t)|\,|e^{-\varepsilon t^2/2}-1|\,dt \longrightarrow 0. $$

The convergence follows from dominated convergence: the integrand tends pointwise to zero and is bounded by $2|\varphi_X(t)|$, an integrable function. Thus $p_\varepsilon$ converges uniformly to $g$. Also, $$ |g(x)|\leq \frac{1}{2\pi}\int_{\mathbb{R}}|\varphi_X(t)|\,dt, $$ so $g$ is bounded. If $x_m\to x$, then $e^{-itx_m}\to e^{-itx}$ for each $t$, and the integrands are bounded in absolute value by $|\varphi_X(t)|$. Dominated convergence therefore gives $g(x_m)\to g(x)$, proving continuity.

Each $p_\varepsilon$ is nonnegative. Uniform convergence implies $g(x)\geq 0$ at every $x$: a negative value at any point would contradict the nonnegativity of $p_\varepsilon$ there for sufficiently small $\varepsilon$.

It remains to identify the distribution. Let $h$ be any continuous function with compact support. Uniform convergence, together with the finite length of the support, implies $$ \int_{\mathbb{R}}h(x)p_\varepsilon(x)\,dx\longrightarrow \int_{\mathbb{R}}h(x)g(x)\,dx. $$

On the other hand, the left-hand side is $E[h(X+Z_\varepsilon)]$. For each fixed value $y$ of $X$, $E[h(y+Z_\varepsilon)]\to h(y)$: continuity gives convergence as $Z_\varepsilon$ tends to zero in probability, and boundedness of $h$ controls the expectations. Applying dominated convergence over the distribution of $X$ shows that $E[h(X+Z_\varepsilon)]\to E[h(X)]$. Consequently, $$ E[h(X)]=\int_{\mathbb{R}}h(x)g(x)\,dx \qquad\text{for every continuous compactly supported }h. $$

Continuous compactly supported functions determine finite Borel measures on $\mathbb{R}$. Since $g\geq 0$, the identity shows that the law of $X$ is the measure with density $g$; in particular, that density integrates to one. This proves the theorem. $\square$

Worked Inverse Transforms

Worked Example: Recovering a Gaussian Density

Suppose $X\sim N(\mu,\sigma^2)$, where $\sigma>0$. Its characteristic function is $\varphi_X(t)=e^{i\mu t}e^{-\sigma^2t^2/2}$, which is integrable. Apply inversion: $$ g(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}e^{i\mu t}e^{-\sigma^2t^2/2}\,dt =\frac{1}{2\pi}\int_{\mathbb{R}}e^{-it(x-\mu)}e^{-\sigma^2t^2/2}\,dt. $$

The Gaussian Fourier integral is $\int_{\mathbb{R}}e^{-itu}e^{-\sigma^2t^2/2}\,dt =\frac{\sqrt{2\pi}}{\sigma}e^{-u^2/(2\sigma^2)}$. Taking $u=x-\mu$ gives $$ g(x)=\frac{1}{2\pi}\frac{\sqrt{2\pi}}{\sigma}e^{-(x-\mu)^2/(2\sigma^2)} =\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x-\mu)^2/(2\sigma^2)}. $$

This is the $N(\mu,\sigma^2)$ density. The calculation checks both the inverse-transform sign and the normalization constant.

Worked Example: Recovering a Cauchy Density

Let $\varphi_X(t)=e^{-a|t|}$, with $a>0$. This function is integrable, since $\int_{\mathbb{R}}e^{-a|t|}\,dt=2/a$. Its inverse transform is real and even, so $$ g(x)=\frac{1}{\pi}\int_0^\infty e^{-at}\cos(tx)\,dt. $$

For $a>0$, integrating $e^{(-a+ix)t}$ from zero to infinity gives $1/(a-ix)$. Taking real parts yields $$ \int_0^\infty e^{-at}\cos(tx)\,dt =\operatorname{Re}\left(\frac{1}{a-ix}\right) =\operatorname{Re}\left(\frac{a+ix}{a^2+x^2}\right) =\frac{a}{a^2+x^2}. $$

Therefore $$ g(x)=\frac{a}{\pi(a^2+x^2)}. $$

This is a Cauchy density with scale $a$. The inversion theorem guarantees that this nonnegative function integrates to one, since it is the density of the probability distribution having the stated characteristic function.

Worked Example: Recovering a Laplace Density

Consider $\varphi_X(t)=1/(1+t^2)$. It is integrable. Inversion reduces to the integral $$ g(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{-itx}}{1+t^2}\,dt. $$

For $x>0$, integrate $e^{-izx}/(1+z^2)$ around a semicircle in the lower half-plane. The exponential has absolute value at most one there, and the semicircular contribution tends to zero because the rational factor is of order $R^{-2}$ on a semicircle of radius $R$. The contour is clockwise and encloses the pole $z=-i$, whose residue is $$ \operatorname{Res}_{z=-i}\frac{e^{-izx}}{1+z^2} =\frac{e^{-x}}{-2i}. $$

The residue theorem gives the real-line integral as $-2\pi i$ times this residue, or $\pi e^{-x}$. For $x<0$, close the contour in the upper half-plane, counterclockwise. The pole at $z=i$ has residue $e^x/(2i)$, so the integral is $\pi e^x$. At $x=0$, direct integration gives $\int_{\mathbb{R}}(1+t^2)^{-1}\,dt=\pi$. Thus, for every $x$, $$ g(x)=\frac{1}{2}e^{-|x|}. $$

This is the Laplace density with location zero and scale one. The separate calculation at $x=0$ also confirms that the two expressions from the half-plane contours meet continuously.

Interpretation and a Common Pitfall

The inversion theorem is useful when a characteristic function is easier to obtain or manipulate than a density. Once its absolute integrability is checked, the formula recovers a continuous density at every point, not just almost everywhere. It also supplies the bound $$ \sup_{x\in\mathbb{R}}g(x)\leq \frac{1}{2\pi}\int_{\mathbb{R}}|\varphi_X(t)|\,dt, $$ which can be useful even when the inverse integral has no simple closed form.

A common mistake is to treat integrability of the characteristic function as a necessary condition for a density. It is sufficient, not necessary. For example, a uniform random variable on $[-1,1]$ has a density, but its characteristic function is $\sin(t)/t$ (with value one at zero), which is not absolutely integrable: the absolute oscillations have a divergent integral. The theorem does not apply in that case, but the density still exists.

Another frequent error is to reverse the sign in the exponential without changing the characteristic-function convention. Under $\varphi_X(t)=E[e^{itX}]$, the inversion factor is $e^{-itx}$, as in every calculation above. A reversed sign can produce an incorrect reflected expression when the distribution is not symmetric.

Key takeaway: If $\varphi_X\in L^1(\mathbb{R})$, then its inverse Fourier transform, with factor $1/(2\pi)$ and exponential $e^{-itx}$, is the bounded continuous density of $X$. The condition guarantees inversion; it is not required for every distribution that has a density.

Check Your Understanding

Use the inversion convention and the proof strategy from this tutorial to answer the following.

State the integrability condition and write the Fourier inversion formula for the convention $\varphi_X(t)=E[e^{itX}]$.
Why does integrability of $\varphi_X$ imply that the inverse-transform function is bounded?
In the Gaussian-smoothing proof, what role does the factor $e^{-\varepsilon t^2/2}$ play?
What density results from inverting $\varphi_X(t)=e^{-a|t|}$ for $a>0$?
Why does failure of $\int_{\mathbb{R}}|\varphi_X(t)|\,dt<\infty$ not prove that a random variable lacks a density?