Left and Two-Sided Censoring Distributions

From Censoring Rules to Distributions

Censoring replaces an unobserved value beyond a reporting limit with the limit itself. A measurement below a lower limit \(a\), for example, may be recorded as \(a\). If both a lower and an upper limit are imposed, values are reported only within the interval \([a,b]\): anything below \(a\) is recorded at \(a\), and anything above \(b\) at \(b\). These rules do not discard the affected observations. They move their probability to the boundary values, creating atoms there.

The calculation must account for the full original law. In particular, the probability recorded at a boundary includes any probability the original variable already had exactly at that boundary. The formulas below apply whether the original distribution is continuous, discrete, or mixed.

Definition: Let \(X\) have cumulative distribution function \(F(x)=P(X\leq x)\), and write \(F(x-)=P(X<x)\). For a fixed threshold \(a\), the variable censored below at \(a\) is \(Y=\max(X,a)\). For fixed \(a<b\), the variable clipped to \([a,b]\) is \(Z=\min(\max(X,a),b)\). Thus values below \(a\) become \(a\), values above \(b\) become \(b\), and values strictly between the thresholds are unchanged.

Left Censoring at a Lower Threshold

For \(Y=\max(X,a)\), every value \(X\leq a\) is recorded as \(a\). Every value \(X>a\) is left as it is. The mass at \(a\) is therefore \(P(X\leq a)=F(a)\), including any original atom at \(a\). For \(y>a\), the event \(Y\leq y\) is exactly the event \(X\leq y\), so the original CDF is retained above the threshold.

Theorem: The CDF of \(Y=\max(X,a)\) is

F_Y(y)= \begin{cases} 0, & y<a,\\ F(y), & y\geq a. \end{cases}

In particular, \(P(Y=a)=F(a)\). For any Borel set \(A\subset(a,\infty)\), \(P(Y\in A)=P(X\in A)\).

To verify the CDF, if \(y<a\), then \(Y\geq a>y\), so \(P(Y\leq y)=0\). If \(y\geq a\), then

\{Y\leq y\}=\{\max(X,a)\leq y\}=\{X\leq y\},

and hence \(F_Y(y)=F(y)\). Its jump at \(a\) is \(F_Y(a)-F_Y(a-)=F(a)-0=F(a)\). For a Borel set \(A\subset(a,\infty)\), \(Y\in A\) can occur only when \(X>a\), where \(Y=X\); thus the events \(\{Y\in A\}\) and \(\{X\in A\}\) coincide. \(\square\)

Worked Example: A Normal Variable Censored Below Zero

Suppose \(X\sim N(0,1)\) and record \(Y=\max(X,0)\). Let \(\Phi\) and \(\phi\) denote the standard normal CDF and density. The censored variable has an atom at zero with probability

P(Y=0)=P(X\leq 0)=\Phi(0)=\frac12.

For \(y<0\), \(F_Y(y)=0\). For \(y\geq0\), \(F_Y(y)=\Phi(y)\). On the positive half-line the density is therefore \(\phi(y)\), unchanged from the original density. Its integral there is \(\int_0^\infty \phi(y)\,dy=1/2\); the remaining probability \(1/2\) is the atom at zero. The density on the positive half-line is not, by itself, a probability density for the entire censored variable.

Clipping to a Finite Interval

For \(Z=\min(\max(X,a),b)\), the preimage of the lower endpoint is all values \(X\leq a\), while the preimage of the upper endpoint is all values \(X\geq b\). In between, the transformation leaves values unchanged. Since \(P(X\geq b)=1-F(b-)\), the left limit—not necessarily \(F(b)\)—gives the upper endpoint mass. This distinction matters when \(X\) has an atom at \(b\).

Theorem: For \(a<b\), the CDF of \(Z=\min(\max(X,a),b)\) is

F_Z(y)= \begin{cases} 0, & y<a,\\ F(y), & a\leq y<b,\\ 1, & y\geq b. \end{cases}

The boundary masses are \(P(Z=a)=F(a)\) and \(P(Z=b)=1-F(b-)\). For every Borel set \(A\subset(a,b)\), \(P(Z\in A)=P(X\in A)\).

For \(y<a\), \(Z\geq a\), so the CDF is zero. If \(a\leq y<b\), then \(Z\leq y\) holds exactly when \(X\leq y\): values at or below \(a\) are sent to \(a\leq y\), values in \((a,y]\) are unchanged, and values above \(y\) cannot be sent below \(y\). Thus \(F_Z(y)=F(y)\). For \(y\geq b\), \(Z\leq b\leq y\) always, so the CDF is one. The jump at \(a\) is \(F(a)\). The jump at \(b\) is

F_Z(b)-F_Z(b-)=1-F(b-)=P(X\geq b).

Finally, if \(A\subset(a,b)\), the transformation can produce a value in \(A\) only from an \(X\)-value in \(A\), and then \(Z=X\). This proves the probability-preservation statement for interior Borel sets. \(\square\)

These statements describe the entire law, not just its CDF values. If \(\mu\) is the distribution of \(X\), then for any Borel set \(B\),

P(Z\in B) = F(a)\mathbf{1}_{\{a\in B\}} +\mu\bigl(B\cap(a,b)\bigr) +\bigl[1-F(b-)\bigr]\mathbf{1}_{\{b\in B\}}.

Here \(\mathbf{1}_{\{a\in B\}}\) equals one when \(a\in B\) and zero otherwise, and likewise for \(b\). This measure formula shows that any atoms or other probability structure strictly inside the interval are preserved, while all probability from outside or at the endpoints is collected into the corresponding boundary masses.

Worked Example: A Mixed Law Clipped to Its Thresholds

Let \(X\) have probability \(0.2\) at \(0\), probability \(0.1\) at \(2\), and probability \(0.7\) from a \(N(1,1)\) component. Clip \(X\) to \([0,2]\), producing \(Z=\min(\max(X,0),2)\). The original atoms at the thresholds must be included in the corresponding clipped atoms.

At zero, the point mass already there combines with the normal component's probability below zero:

P(Z=0)=P(X\leq0)=0.2+0.7\Phi(-1).

At two, the original atom at two combines with the normal component's probability above two. The normal component has no atom at two, so

P(Z=2)=P(X\geq2)=0.1+0.7\bigl[1-\Phi(1)\bigr].

For \(0<z<2\), only the normal component contributes a density, and it is unchanged:

f_Z(z)=0.7\phi(z-1),\qquad 0<z<2.

As a check, the total probability is

\begin{aligned} P(Z=0)+P(0<Z<2)+P(Z=2) &=0.2+0.7\Phi(-1) +0.7\bigl[\Phi(1)-\Phi(-1)\bigr]\\ &\quad+0.1+0.7\bigl[1-\Phi(1)\bigr]\\ &=0.3+0.7=1. \end{aligned}

The normal component's interior probability is \(0.7[\Phi(1)-\Phi(-1)]\); its remaining probability is transferred to the two endpoints. The original point masses at zero and two remain at those locations and are included in the endpoint totals.

Density Formula When the Original Law Is Continuous

When \(X\) has a density \(f\), its CDF is continuous, so \(F(b-)=F(b)\). The clipped law then has a particularly direct decomposition: two endpoint atoms plus the original density restricted to the open interval. The restricted density is not renormalized.

Theorem: If \(X\) has density \(f\) on the real line, then the distribution of \(Z=\min(\max(X,a),b)\) has measure

P(Z\in B) = F(a)\mathbf{1}_{\{a\in B\}} +\int_{B\cap(a,b)} f(x)\,dx +\bigl[1-F(b)\bigr]\mathbf{1}_{\{b\in B\}}

for every Borel set \(B\). Thus \(Z\) has density \(f(x)\) on \((a,b)\), and atoms \(F(a)\) at \(a\) and \(1-F(b)\) at \(b\).

Because \(X\) has a density, \(P(X=a)=P(X=b)=0\). The events \(X\leq a\), \(a<X<b\), and \(X\geq b\) partition the sample space, up to no omitted cases. On the first event, \(Z=a\); on the middle event, \(Z=X\); on the last event, \(Z=b\). For any Borel set \(B\), splitting \(P(Z\in B)\) across these three events gives

\begin{aligned} P(Z\in B) &=P(X\leq a)\mathbf{1}_{\{a\in B\}} +P(X\in B\cap(a,b)) +P(X\geq b)\mathbf{1}_{\{b\in B\}}\\ &=F(a)\mathbf{1}_{\{a\in B\}} +\int_{B\cap(a,b)}f(x)\,dx +[1-F(b)]\mathbf{1}_{\{b\in B\}}. \end{aligned}

This proves the asserted measure formula. In particular, the interior density integrates to \(P(a<X<b)\), which need not equal one; the two remaining probabilities are the endpoint atoms. \(\square\)

Worked Example: Clipping a Discrete Distribution

Suppose \(X\) takes the values \(-2,0,1,2,4\) with probabilities \(0.15,0.10,0.25,0.20,0.30\), respectively. Clip to \([0,2]\). The transformation sends \(-2\) and \(0\) to \(0\), leaves \(1\) and \(2\) unchanged, and sends \(4\) to \(2\). Therefore,

P(Z=0)=0.15+0.10=0.25,\qquad P(Z=1)=0.25,\qquad P(Z=2)=0.20+0.30=0.50.

The masses sum to \(0.25+0.25+0.50=1\). In particular, the atom originally at the upper threshold \(2\) is part of the clipped mass at \(2\), just as the original atom at \(0\) is part of the clipped mass at \(0\). This example also shows why “interior unchanged” does not mean that the whole distribution has a density: the interior value \(1\) remains an atom.

Common Mistakes and Why the Distinction Matters

Counting only values strictly beyond a threshold. The lower atom is \(P(X\leq a)\), not just \(P(X<a)\); the upper atom is \(P(X\geq b)\), not just \(P(X>b)\). Existing atoms at the thresholds are included.
Renormalizing the interior density. Censoring transfers probability to endpoints; it does not condition on \(a<X<b\). The interior density remains the original density there, with integral generally less than one.
Assuming the interior must have a density. The interior law is the original law restricted to that interval. If \(X\) has interior atoms, they remain atoms after clipping.
Using \(F(b)\) for the upper mass in a general law. The correct expression is \(1-F(b-)\), since it includes any atom at \(b\). It simplifies to \(1-F(b)\) when \(X\) has a density.

Keeping the transformed variable separate from the original one is useful whenever recorded measurements are limited by detection or reporting thresholds. The boundary masses describe how often a recorded value equals a limit; the interior probabilities describe the part of the original distribution that remains directly visible. Neither component should be mistaken for the conditional distribution obtained by retaining only observations inside an interval.

Check Your Understanding

Use the event definitions and boundary-mass formulas to answer these questions.

If \(Y=\max(X,a)\), express \(P(Y=a)\) in terms of \(F\). Does this formula include an atom of \(X\) at \(a\)?
For clipping to \([a,b]\), why is the upper endpoint mass \(1-F(b-)\) rather than \(1-F(b)\) in the general case?
If \(X\) has density \(f\), what is the density of the clipped variable on \((a,b)\), and why does it generally not integrate to one?
A distribution has an atom inside \((a,b)\). What happens to that atom under clipping to \([a,b]\)?
Explain why censoring at thresholds is not the same operation as conditioning on \(a<X<b\).