Moments Under Continuous Truncation

Why Truncation Changes Moments

When a continuous random variable is observed only if it falls in a specified region, the retained observations no longer follow the original distribution. Their density is the original density restricted to that region and then renormalized to integrate to one. Consequently, their mean and variance must be calculated from that conditional density; the original moments generally do not apply.

Let \(X\) have density \(f\), and let \(A\) be a measurable retained region with positive probability. For example, \(A\) might be an interval or the part of the support below a cutoff. The key calculation is to integrate \(x^r f(x)\) over \(A\) and divide by the probability of landing in \(A\). This tutorial develops that formula and applies it to several continuous distributions.

Definition: Let \(X\) have density \(f\), and let \(A\) be a measurable subset of its support with \(p_A=P(X\in A)>0\). The density conditional on retention in \(A\) is \(f_A(x)=f(x)\mathbf{1}_A(x)/p_A\), where \(\mathbf{1}_A(x)\) is 1 on \(A\) and 0 elsewhere. For an integer \(r\geq 0\), the \(r\)th raw moment under truncation is \(m_r(A)=E[X^r\mid X\in A]\), provided \(\int_A |x|^r f(x)\,dx<\infty\).

The Normalized Raw-Moment Formula

The normalizing probability is \(p_A=\int_A f(x)\,dx\). Thus, substituting the conditional density into the definition of expectation gives \(m_r(A)=\int_A x^r f(x)\,dx/p_A\). The denominator is essential: it converts the mass of the retained part of the original distribution into a probability distribution with total mass one.

Theorem: Suppose \(X\) has density \(f\), \(A\) is measurable, \(p_A=P(X\in A)>0\), and \(\int_A |X|^2 f(x)\,dx<\infty\). Then the conditional mean and variance given retention in \(A\) are \(E[X\mid X\in A]=m_1(A)\) and \(\operatorname{Var}(X\mid X\in A)=m_2(A)-m_1(A)^2\), where \(m_r(A)=\int_A x^r f(x)\,dx/p_A\) for \(r=1,2\).

Proof. For every Borel set \(B\), the definition of conditional probability gives \(P(X\in B\mid X\in A)=P(X\in B\cap A)/p_A=\int_{B\cap A}f(x)\,dx/p_A\). Therefore the conditional distribution has density \(f_A(x)=f(x)\mathbf{1}_A(x)/p_A\). Integrating \(x^r\) against that density gives \(E[X^r\mid X\in A]=\int_A x^r f(x)\,dx/p_A\) for \(r=1,2\). These integrals are finite under the stated second-moment condition; in particular, the first absolute moment is finite on \(A\) by the Cauchy–Schwarz inequality. By the definition of variance, \(\operatorname{Var}(X\mid X\in A)=E[X^2\mid X\in A]-(E[X\mid X\in A])^2=m_2(A)-m_1(A)^2\). \(\square\)

The case \(r=0\) gives \(m_0(A)=1\), since its numerator is exactly \(p_A\). This is a useful normalization check. If a proposed conditional density or moment calculation does not yield total probability one, the retained probability has likely been omitted or miscalculated.

Worked Examples

Worked Example: Right-Truncated Exponential Distribution

Let \(X\sim\operatorname{Exp}(\lambda)\), with density \(f(x)=\lambda e^{-\lambda x}\) for \(x>0\), where \(\lambda>0\). Retain observations with \(X<c\), for a fixed \(c>0\). This is right truncation: values above the upper cutoff are not retained. Find the conditional mean and variance.

The probability of retention is

p_A=\int_0^c \lambda e^{-\lambda x}\,dx =1-e^{-\lambda c}.

For the first raw moment, integration by parts gives

\begin{aligned} \int_0^c x\lambda e^{-\lambda x}\,dx &=\left[-xe^{-\lambda x}\right]_0^c+\int_0^c e^{-\lambda x}\,dx\\ &=-ce^{-\lambda c}+\frac{1-e^{-\lambda c}}{\lambda}. \end{aligned}

Dividing by \(p_A\) yields

m_1(A)=\frac{1}{\lambda}-\frac{ce^{-\lambda c}}{1-e^{-\lambda c}} =\frac{1}{\lambda}-\frac{c}{e^{\lambda c}-1}.

For the second raw moment, integrate by parts once and then evaluate the remaining first-moment integral:

\begin{aligned} \int_0^c x^2\lambda e^{-\lambda x}\,dx &=\left[-x^2e^{-\lambda x}\right]_0^c +2\int_0^c xe^{-\lambda x}\,dx\\ &=-c^2e^{-\lambda c} +\frac{2}{\lambda^2}\left(1-e^{-\lambda c}(1+\lambda c)\right). \end{aligned}

After division by \(1-e^{-\lambda c}\), this becomes \(m_2(A)=2/\lambda^2-(c^2+2c/\lambda)/(e^{\lambda c}-1)\). Subtracting the square of the mean gives

\begin{aligned} \operatorname{Var}(X\mid X<c) &=m_2(A)-m_1(A)^2\\ &=\frac{1}{\lambda^2} -\frac{c^2e^{\lambda c}}{(e^{\lambda c}-1)^2}. \end{aligned}

For the simplification, put \(q=e^{-\lambda c}\) and \(p=1-q\). The terms involving \(c^2\) combine as \(-c^2q/p-c^2q^2/p^2=-c^2q(p+q)/p^2=-c^2q/p^2\), which is the displayed correction term.

For a numerical check, take \(\lambda=1\) and \(c=\ln 2\). The retention probability is \(1-e^{-\ln 2}=1/2\). The mean is \(1-\ln 2\approx0.3069\), and the variance is \(1-2(\ln 2)^2\approx0.0391\). Both are positive, and the conditional mean is below the cutoff, as expected.

Worked Example: Interval-Truncated Uniform Distribution

Let \(X\sim\operatorname{Uniform}(0,10)\), and retain observations in \(A=[2,6]\). The original density is \(1/10\) on \((0,10)\), so the probability of retention is \(p_A=\int_2^6 (1/10)\,dx=4/10=2/5\). The conditional density on \([2,6]\) is therefore \((1/10)/(2/5)=1/4\).

The first raw moment is

m_1(A)=\int_2^6 x\left(\frac14\right)\,dx =\frac14\left[\frac{x^2}{2}\right]_2^6 =\frac{36-4}{8}=4.

The second raw moment is

m_2(A)=\int_2^6 x^2\left(\frac14\right)\,dx =\frac14\left[\frac{x^3}{3}\right]_2^6 =\frac{216-8}{12}=\frac{52}{3}.

Hence \(\operatorname{Var}(X\mid X\in[2,6])=52/3-4^2=4/3\). The conditional mean is the interval midpoint, and the variance is the variance of a uniform distribution over an interval of width 4.

Worked Example: Interval Truncation of a Triangular Density

Let \(X\) have density \(f(x)=2x\) for \(0<x<1\), and retain observations in \(A=[1/2,1]\). This example shows why the midpoint of the retained interval need not be the conditional mean when the original density is not constant.

First calculate the retention probability:

p_A=\int_{1/2}^1 2x\,dx =\left[x^2\right]_{1/2}^1 =1-\frac14=\frac34.

The first raw moment is

m_1(A)=\frac{1}{3/4}\int_{1/2}^1 x(2x)\,dx =\frac43\left[\frac{2x^3}{3}\right]_{1/2}^1 =\frac43\cdot\frac{7}{12} =\frac79.

The second raw moment is

m_2(A)=\frac{1}{3/4}\int_{1/2}^1 x^2(2x)\,dx =\frac43\left[\frac{x^4}{2}\right]_{1/2}^1 =\frac43\cdot\frac{15}{32} =\frac58.

Therefore the conditional variance is \(\frac58-(\frac79)^2=\frac{405-392}{648}=\frac{13}{648}\). The mean \(7/9\) exceeds the interval midpoint \(3/4\), because the original density assigns more weight to values nearer 1.

Using the Formulas Carefully

Truncation changes the relative weighting of values: observations outside \(A\) contribute nothing, and the retained density is rescaled by \(1/p_A\). This can move the mean and change the variance even when the retained region is simple. In the uniform example, the conditional mean is the midpoint because the density is constant across the retained interval. In the triangular example, the density increases with \(x\), so the conditional mean is pulled toward the upper endpoint.

A common mistake is to integrate \(x^r f(x)\) over the retained region and call the result a conditional moment without dividing by \(P(X\in A)\). That integral is an unnormalized contribution from the original distribution, not an expectation under the conditional law. In the uniform example, the unnormalized first-moment integral is \(\int_2^6 x(1/10)\,dx=8/5\), not the conditional mean 4. Dividing by the retention probability \(2/5\) gives the correct result.

Another error is to keep the original mean or variance simply because the same variable \(X\) is being studied. Those quantities describe the full distribution, including values that are not retained. The conditional raw moments must be computed over the retained region and normalized by its probability. Before calculating a variance, it is also worth checking that the retained probability is positive and that the second-moment integral is finite.

Key takeaway: For continuous truncation to a region \(A\), each raw moment is the integral of \(x^r f(x)\) over \(A\), divided by \(P(X\in A)\). The conditional variance is the second raw moment minus the square of the first.

Check Your Understanding

Use the normalized integral formula to answer these questions.

For a density \(f\) and retained region \(A\), why must the probability \(P(X\in A)\) be positive before defining the conditional density?
For right truncation of an exponential distribution at \(c\), what is the retention probability?
In the uniform example, what conditional mean would result if the retention probability were omitted, and why would it not be a conditional expectation?
Why is the mean under interval truncation of the triangular density not the midpoint of the retained interval?
State the condition on the second-moment integral that ensures the truncated variance is finite.

Moments Under Continuous Truncation

What You'll Learn