Lévy Inversion for Interval Probabilities

Why Interval Inversion Needs Endpoint Care

A characteristic function contains information about an entire distribution, but extracting the probability of a particular interval requires care when the distribution has atoms. A formula that simply returned the probability of the open interval would miss any mass located exactly at its endpoints. Lévy inversion resolves this by giving each endpoint atom half weight. Once those contributions are stated precisely, the formula applies to both continuous and discrete laws.

Throughout, the characteristic function of a real-valued random variable $X$ is $\phi(t)=E[e^{itX}]$, for real $t$. The integral in the inversion formula is taken symmetrically over $[-T,T]$ and then $T$ tends to infinity. This symmetric limiting convention matters: the integral need not be absolutely integrable over the whole real line.

Theorem (Lévy inversion for intervals): Let $X$ be any real-valued random variable with characteristic function $\phi$, and let $a<b$ be finite real numbers. Then $$ P(a<X<b)+\frac{1}{2}P(X=a)+\frac{1}{2}P(X=b) = \lim_{T\to\infty}\frac{1}{2\pi}\int_{-T}^{T} \frac{e^{-ita}-e^{-itb}}{it}\,\phi(t)\,dt . $$ At $t=0$, the quotient is defined by continuity as $b-a$. The limit exists as a finite real number. No density or continuity assumption on the distribution of $X$ is required.

The half weights have a specific origin. The truncated integral acts like an approximation to the indicator of the interval: in the limit it is one strictly inside, zero strictly outside, and one half at either endpoint. The proof below establishes those values and justifies passing from the pointwise limit to an expectation.

Deriving the Formula

For $u\in\mathbb{R}$, define the sine integral $ \operatorname{Si}(u)=\int_0^u \sin(v)/v\,dv $, where the integrand at zero is defined by continuity as 1. For a fixed finite $T>0$, set $$ g_T(x)=\frac{1}{2\pi}\int_{-T}^{T} \frac{e^{it(x-a)}-e^{it(x-b)}}{it}\,dt. $$ The apparent singularity at $t=0$ is removable: the numerator is $it(b-a)+o(t)$, so the quotient tends to $b-a$. The integral is therefore an ordinary integral over a bounded interval.

For any real $u$, symmetry of the integration limits gives $$ \frac{1}{2\pi}\int_{-T}^{T}\frac{e^{itu}}{it}\,dt =\frac{1}{\pi}\int_0^T\frac{\sin(tu)}{t}\,dt =\frac{\operatorname{Si}(Tu)}{\pi}. $$ Indeed, writing $e^{itu}=\cos(tu)+i\sin(tu)$, the cosine term divided by $it$ is odd in $t$ and integrates to zero; the remaining term is $\sin(tu)/t$, which is even. Applying this identity to the two terms in $g_T$ yields $$ g_T(x)=\frac{\operatorname{Si}(T(x-a))-\operatorname{Si}(T(x-b))}{\pi}. $$

The Dirichlet integral gives $\lim_{u\to+\infty}\operatorname{Si}(u)=\pi/2$; since $\operatorname{Si}$ is odd, its limit at negative infinity is $-\pi/2$. Also, $\operatorname{Si}(0)=0$. Consequently, for each fixed $x$, the limit of $g_T(x)$ is 1 when $a<x<b$, 0 when $x<a$ or $x>b$, and $1/2$ when $x=a$ or $x=b$. Equivalently, $$ \lim_{T\to\infty} g_T(x) = \mathbf{1}_{(a,b)}(x)+\frac{1}{2}\mathbf{1}_{\{a\}}(x) +\frac{1}{2}\mathbf{1}_{\{b\}}(x). $$

To justify taking this limit inside an expectation, a bound independent of $T$ is needed. The function $\operatorname{Si}$ is bounded on the real line. For example, on $[0,1]$, its absolute value is at most 1. For $u\geq 1$, integration by parts gives $$ \int_1^u\frac{\sin v}{v}\,dv =-\frac{\cos u}{u}+\cos(1)-\int_1^u\frac{\cos v}{v^2}\,dv, $$ whose absolute value is at most $1+1+1=3$. Oddness supplies the corresponding bound for negative $u$. Thus $|g_T(x)|$ is bounded by a fixed finite constant for all $x$ and $T$.

For each finite $T$, Fubini's theorem applies to the defining integral for $g_T(X)$: the integrand, with its removable value at zero, is bounded on $[-T,T]$, and the probability measure of $X$ is finite. Hence $$ E[g_T(X)] =\frac{1}{2\pi}\int_{-T}^{T} \frac{e^{-ita}-e^{-itb}}{it}\,E[e^{itX}]\,dt =\frac{1}{2\pi}\int_{-T}^{T} \frac{e^{-ita}-e^{-itb}}{it}\,\phi(t)\,dt. $$ The uniform bound on $g_T$ and the fact that $P$ is a probability measure allow the dominated convergence theorem to be applied as $T\to\infty$. Taking expectations of the pointwise limit gives exactly the weighted interval probability in the theorem. This also proves that the displayed symmetric limit exists. $\square$

Reading the Endpoint Weights

Write $C(a,b)$ for the limit of the symmetric integral in Lévy inversion. The theorem says $C(a,b)=P(a<X<b)+\tfrac12P(X=a)+\tfrac12P(X=b)$. It does not, by itself, give the probability of every choice of endpoint inclusion. If the endpoint masses are known, the desired interval probability follows by adding or subtracting the appropriate half weights.

Interval conversions: For $a<b$, $$ \begin{aligned} P(a<X<b)&=C(a,b)-\frac{P(X=a)+P(X=b)}{2},\\ P(a\leq X\leq b)&=C(a,b)+\frac{P(X=a)+P(X=b)}{2},\\ P(a<X\leq b)&=C(a,b)+\frac{P(X=b)-P(X=a)}{2},\\ P(a\leq X<b)&=C(a,b)+\frac{P(X=a)-P(X=b)}{2}. \end{aligned} $$ If both endpoints have probability zero, all four interval conventions give the same probability, equal to $C(a,b)$.

These conversions are simple consequences of separating the interior event from the two endpoint events. They are useful because many familiar formulas are stated for an open interval, while applications may ask for a closed or half-open one. The characteristic-function integral is the same; only the endpoint corrections change.

Worked Applications

Worked Example: A Uniform Distribution with No Endpoint Atoms

Let $X\sim\operatorname{Uniform}[-1,1]$. Its characteristic function is $\phi(t)=\sin(t)/t$, with $\phi(0)=1$. Find the probability of $(-1,1)$ by Lévy inversion.

Use $a=-1$ and $b=1$. Then $e^{-ita}-e^{-itb}=e^{it}-e^{-it}=2i\sin t$. The integrand in the formula is therefore $2\sin(t)\phi(t)/t=2\sin^2(t)/t^2$, including its continuous limiting value at zero. Thus $$ C(-1,1) =\lim_{T\to\infty}\frac{1}{\pi}\int_{-T}^{T}\frac{\sin^2(t)}{t^2}\,dt =\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin^2(t)}{t^2}\,dt =1. $$ For the last equality, the standard Dirichlet integral $\int_0^\infty \sin^2(t)/t^2\,dt=\pi/2$ gives an integral of $\pi$ over the full real line. This integral is absolutely convergent, so the symmetric limit equals the ordinary improper integral. A uniform distribution has no atoms at $-1$ or $1$, and the answer is $P(-1<X<1)=1$.

Worked Example: A Two-Point Distribution with Endpoint Atoms

Suppose $P(X=0)=p$ and $P(X=2)=1-p$, where $0\leq p\leq1$. Its characteristic function is $\phi(t)=p+(1-p)e^{2it}$. Find the Lévy inversion value for the interval with endpoints 0 and 2, and compare it with the open-interval probability.

For $a=0,b=2$, the finite-$T$ integral equals $E[g_T(X)]$, where the kernel derived in the proof is $g_T(x)=[\operatorname{Si}(Tx)-\operatorname{Si}(T(x-2))]/\pi$. At $x=0$, this is $\operatorname{Si}(2T)/\pi$, which tends to $1/2$. At $x=2$, it is also $\operatorname{Si}(2T)/\pi$, and also tends to $1/2$. Therefore $$ C(0,2)=p\left(\frac12\right)+(1-p)\left(\frac12\right)=\frac12. $$ There is no probability strictly between 0 and 2, so $P(0<X<2)=0$. Lévy inversion still gives $1/2$, because each endpoint atom contributes half its mass: $\tfrac12p+\tfrac12(1-p)=\tfrac12$. The open-interval conversion subtracts those two half weights and correctly returns zero.

Worked Example: A Mixed Law and a Half-Open Interval

Let $X$ equal 0 with probability $0.4$; with the remaining probability $0.6$, let $X$ have the uniform distribution on $[0,2]$. Find $P(0<X\leq1)$ using interval inversion and endpoint corrections.

The characteristic function of the uniform component is $\int_0^2 e^{itx}\,dx/2=(e^{2it}-1)/(2it)$, with value 1 at $t=0$. Thus the mixture has $$ \phi(t)=0.4+0.6\frac{e^{2it}-1}{2it}, $$ again interpreting the quotient by continuity at zero. For the interval $(0,1)$, the continuous component contributes probability $1/2$ before mixing, and the atom at 0 contributes no interior probability. Hence $P(0<X<1)=0.6(1/2)=0.3$.

There is an atom of mass $0.4$ at the left endpoint, and no atom at 1. The inversion value is therefore $C(0,1)=0.3+\tfrac12(0.4)+\tfrac12(0)=0.5$. For the requested half-open interval, the conversion gives $$ P(0<X\leq1) =C(0,1)+\frac{P(X=1)-P(X=0)}{2} =0.5+\frac{0-0.4}{2}=0.3. $$ This agrees with the direct calculation: the part of the uniform component in $(0,1]$ has probability $1/2$, so its mixture contribution is $0.6(1/2)=0.3$; the atom at zero is excluded.

Common Mistake and Practical Use

A frequent error is to equate the Lévy inversion integral directly with $P(a<X<b)$, even when there may be atoms at $a$ or $b$. For example, in the two-point distribution above, the open-interval probability is zero but the inversion value is $1/2$. The correction is not to change the characteristic function or the integration limits: it is to account for the half mass at each endpoint. If endpoint masses are zero, the distinction disappears.

The formula is useful when the characteristic function is known explicitly but a distribution function or density is inconvenient to calculate. It also makes clear why inversion can recover interval probabilities without requiring a density: the proof works directly with the probability law, including point masses and mixtures. In applications, first identify the interval convention, then apply inversion, and finally make the endpoint correction required by that convention.

Key takeaway: Symmetric Fourier truncation recovers the probability in the interior of an interval and assigns half weight to each endpoint atom. To obtain a particular open, closed, or half-open interval probability, adjust that weighted value by the endpoint masses.

Check Your Understanding

For each question, pay attention to whether the endpoints can carry probability.

State the Lévy inversion formula for $a<b$, including its symmetric limiting convention.
If $P(X=a)=0.2$, $P(X=b)=0$, and $P(a<X<b)=0.5$, what is the Lévy inversion value $C(a,b)$?
Why can the dominated convergence theorem be used in the proof even when $X$ has atoms?
Suppose the inversion value is $0.7$, with endpoint masses $P(X=a)=0.2$ and $P(X=b)=0.1$. Find $P(a<X\leq b)$.
When do the probabilities of the open, closed, and half-open versions of an interval all agree?