Left, Right, and Interval Truncation Formulas

From a Truncation Rule to a Conditional Distribution

Truncation changes which values are observed: a value is retained only if it satisfies a specified rule. The distribution of retained values is therefore the original distribution conditioned on satisfying that rule. The key calculation is to divide probabilities on the retained region by the probability of entering it. The region’s endpoints and its normalizing probability determine the resulting CDF and density.

Here, left truncation means retaining values above a lower threshold, and right truncation means retaining values below an upper threshold. Some applications use different naming conventions, so the retained event will always be stated explicitly. For an absolutely continuous random variable, including or excluding any single finite endpoint makes no difference to these conditional distributions.

Definition: A random variable $X$ is absolutely continuous if its distribution has a density $f$ with respect to Lebesgue measure; equivalently, its CDF satisfies $F(x)=\int_{-\infty}^{x} f(t)\,dt$ for every $x$, for some nonnegative measurable $f$ integrating to one. In particular, absolute continuity implies $P(X=x)=0$ for every $x$. The converse is false: having no point masses alone does not guarantee a density.

Let $F$ and $f$ denote the CDF and density of an absolutely continuous $X$. Each truncation formula below requires that the retained event have strictly positive probability. Otherwise the conditional distribution is not defined by conditioning on that event.

The Three Truncation Formulas

The conditioning event determines the denominator. Once that probability is known, the conditional CDF is the original probability accumulated inside the retained region, divided by the denominator.

Theorem: Let $X$ have an absolutely continuous distribution with CDF $F$ and density $f$.

Retain above $c$: If $P(X>c)=1-F(c)>0$, then $F_{X\mid X>c}(x)= \begin{cases} 0, & x\le c,\\[2pt] \dfrac{F(x)-F(c)}{1-F(c)}, & x>c, \end{cases} \qquad f_{X\mid X>c}(x)=\frac{f(x)}{1-F(c)}\,\mathbf{1}_{\{x>c\}} \quad\text{a.e.}$
Retain below $c$: If $P(X<c)=F(c)>0$, then $F_{X\mid X<c}(x)= \begin{cases} \dfrac{F(x)}{F(c)}, & x<c,\\[2pt] 1, & x\ge c, \end{cases} \qquad f_{X\mid X<c}(x)=\frac{f(x)}{F(c)}\,\mathbf{1}_{\{x<c\}} \quad\text{a.e.}$
Retain within $(a,b)$: If $a<b$ and $P(a<X<b)=F(b)-F(a)>0$, then $F_{X\mid a<X<b}(x)= \begin{cases} 0, & x\le a,\\[2pt] \dfrac{F(x)-F(a)}{F(b)-F(a)}, & a<x<b,\\[6pt] 1, & x\ge b, \end{cases} \qquad f_{X\mid a<X<b}(x)= \frac{f(x)}{F(b)-F(a)}\,\mathbf{1}_{\{a<x<b\}} \quad\text{a.e.}$

The formulas are derived directly from conditional probabilities. For example, when $x>c$, the event $\{X\le x\}$ intersects $\{X>c\}$ in $\{c<X\le x\}$; its probability is $F(x)-F(c)$. When $x\le c$, that intersection has probability zero. The other cases follow by intersecting $\{X\le x\}$ with their respective retained events.

Worked Example: Exponential Values Retained Above a Threshold

Suppose $X\sim \operatorname{Exponential}(2)$, with rate $2$, so $F(x)=1-e^{-2x}$ and $f(x)=2e^{-2x}$ for $x\ge0$, and both are zero as appropriate below zero. Retain observations with $X>1.5$. First calculate the normalizing probability:

P(X>1.5)=1-F(1.5)=e^{-2(1.5)}=e^{-3}>0.

For $x\le1.5$, the conditional CDF is zero. For $x>1.5$, substitute into the left-truncation formula:

\begin{aligned} F_{X\mid X>1.5}(x) &=\frac{F(x)-F(1.5)}{1-F(1.5)}\\ &=\frac{(1-e^{-2x})-(1-e^{-3})}{e^{-3}}\\ &=1-e^{-2(x-1.5)}. \end{aligned}

The density is $2e^{-2x}/e^{-3}=2e^{-2(x-1.5)}$ for $x>1.5$, and zero for $x<1.5$. It integrates to one:

\int_{1.5}^{\infty}2e^{-2(x-1.5)}\,dx =\left[-e^{-2(x-1.5)}\right]_{1.5}^{\infty}=1.

The retained values have support $[1.5,\infty)$, in the sense that the conditional distribution assigns positive probability to every neighborhood of a point in that set. The density’s value at the single endpoint can be chosen without changing the distribution.

Worked Example: Uniform Values Retained Below a Threshold

Let $X\sim\operatorname{Uniform}(0,10)$, so $F(x)=x/10$ on $0\le x\le10$. Retain values satisfying $X<6$. The normalizing probability is $F(6)=6/10=0.6$, which is positive. For $0\le x<6$,

F_{X\mid X<6}(x)=\frac{F(x)}{F(6)} =\frac{x/10}{6/10}=\frac{x}{6}.

The full conditional CDF is zero for $x<0$, $x/6$ for $0\le x<6$, and one for $x\ge6$. Its density is

f_{X\mid X<6}(x)=\frac{1/10}{0.6}\,\mathbf{1}_{\{0<x<6\}} =\frac{1}{6}\,\mathbf{1}_{\{0<x<6\}}.

As a normalization check, $\int_0^6(1/6)\,dx=1$. The effective support is $[0,6]$; the formula does not assign any density above 6, and the CDF reaches one at the truncation boundary.

Worked Example: Standard Normal Values Retained Inside an Interval

Let $X\sim N(0,1)$, with density $\phi(x)=(2\pi)^{-1/2}e^{-x^2/2}$ and CDF $\Phi$. Retain values satisfying $-1<X<1$. Symmetry gives the normalizing probability

P(-1<X<1)=\Phi(1)-\Phi(-1)=2\Phi(1)-1\approx 0.682689.

The conditional CDF is zero for $x\le-1$, equals $[\Phi(x)-\Phi(-1)]/[2\Phi(1)-1]$ for $-1<x<1$, and equals one for $x\ge1$. The conditional density on the retained interval is

f_{X\mid -1<X<1}(x)= \frac{(2\pi)^{-1/2}e^{-x^2/2}}{2\Phi(1)-1} \,\mathbf{1}_{\{-1<x<1\}}.

The denominator is not $\Phi(1)$: it must account for both excluded tails, so it is $\Phi(1)-\Phi(-1)$. Integrating the displayed density over $(-1,1)$ gives $[\Phi(1)-\Phi(-1)]/[2\Phi(1)-1]=1$, as required.

Why the Density Formula Is Correct

The CDF expressions identify the conditional laws. The corresponding density formulas can also be verified independently by integrating them over any interval. This makes explicit both their normalization and the role of the support restriction.

Result: For any of the three retained events $A$ above, with $P(A)>0$, the conditional density is $f(x)\mathbf{1}_{A}(x)/P(A)$, where $\mathbf{1}_{A}(x)$ indicates whether the value $x$ lies in the retained region.

To prove the result, let $u<v$. By the definition of conditional probability and the density of $X$,

P(u<X\le v\mid A) =\frac{P(\{u<X\le v\}\cap A)}{P(A)} =\frac{1}{P(A)}\int_u^v f(x)\mathbf{1}_{A}(x)\,dx.

Thus the function $f(x)\mathbf{1}_{A}(x)/P(A)$ gives the conditional probabilities of intervals. It is nonnegative and integrates over the real line to

\int_{-\infty}^{\infty}\frac{f(x)\mathbf{1}_{A}(x)}{P(A)}\,dx =\frac{P(A)}{P(A)}=1.

It is therefore a density for the conditional distribution. In particular, it vanishes outside the retained region, as well as wherever the original density vanishes, up to sets of Lebesgue measure zero. $\square$

Support, Endpoints, and Checks

Truncation restricts the possible values; it does not generally fill gaps in the original support. More precisely, the conditional distribution has positive probability only in the intersection of the original support with the retained region. Its support is the closure of the points in that region that have positive conditional probability in every neighborhood. This accounts for boundary points even when the retention rule uses strict inequalities.

For an absolutely continuous variable, $P(X=c)=0$ at any fixed $c$. Consequently $X>c$ and $X\ge c$ produce the same conditional distribution whenever their normalizing probability is positive; likewise for $X<c$ versus $X\le c$, and for open versus closed interval endpoints. This boundary equivalence depends on absolute continuity. The fact that a distribution has no point masses, by itself, is not enough to justify the density formulas: a singular continuous distribution can have no point masses and still have no density with respect to Lebesgue measure.

A useful check is to verify both the denominator and the total conditional mass. Above a threshold, the denominator is the upper-tail probability $1-F(c)$; below it, the denominator is $F(c)$; on an interval, it is the probability between the endpoints $F(b)-F(a)$. Dividing by any other probability changes the conditional law.

Common Mistakes

Using the wrong tail probability. Retaining values above $c$ requires division by $1-F(c)$, not $F(c)$. The denominator is the probability of the event actually retained.
Forgetting the piecewise CDF. The conditional CDF is zero below the retained region and one above it. The ratio formula applies only where the event $\{X\le x\}$ intersects that region partially.
Claiming that no point masses implies a density. Absolute continuity implies no point masses, but the converse fails. A density-based truncation formula requires absolute continuity, not merely $P(X=x)=0$ for each $x$.
Ignoring the original support. A truncation boundary restricts the support but cannot create density where the original $f$ was zero.

Key takeaway: State the retained event, calculate its positive probability, divide the probability accumulated within the retained region by that normalizer, and restrict the density to that region. The CDF’s boundary values and the density’s support then follow directly.

Check Your Understanding

For each question, identify the conditioning event and check the normalizing probability before writing a conditional CDF or density.

If $F(c)=0.35$, what is the normalizing probability when retaining values above $c$, and what is the conditional CDF for $x>c$?
For a continuous distribution with $P(X<d)=0.8$, what is the conditional density after retaining values below $d$?
Write the normalizing probability and the middle piece of the conditional CDF when values are retained in $(a,b)$.
Why does the interval $(a,b)$ have the same conditional distribution as $[a,b]$ for an absolutely continuous $X$, when the conditioning probability is positive?
Does $P(X=x)=0$ for every $x$ guarantee that a density-based truncation formula applies? Explain.