The Distribution Created by Right Censoring

What an Upper Censoring Limit Does

Suppose a measurement can exceed a reporting limit $c$, but the recorded value is never allowed to exceed $c$. A value below the limit is reported as measured; a value at or above the limit is recorded as $c$. The resulting random variable is not generally continuous: all of the original upper-tail probability is concentrated at the single reported value $c$.

This tutorial studies the distribution of that recorded value alone. In other settings, a censoring record may also include an indicator stating whether the original value exceeded the limit. That additional information is not part of the random variable defined here.

Definition: Let $X$ be a real-valued random variable with an absolutely continuous distribution, CDF $F$, and density $f$. For a fixed finite upper limit $c$, define the upper-censored variable $$ Z=\min(X,c). $$ Thus $Z=X$ when $X<c$, and $Z=c$ when $X\ge c$. Absolute continuity implies $P(X=c)=0$, so $P(X\ge c)=1-F(c)$.

The transformation $x\mapsto\min(x,c)$ leaves values below $c$ unchanged and maps every value at or above $c$ to $c$. It is this many-to-one mapping that creates a point mass, even though the original variable has no point masses. The distribution of $Z$ has a continuous part below $c$ and, when $P(X\ge c)>0$, an atom at $c$.

The CDF and the Point Mass

For any real $z<c$, the event that the recorded value is at most $z$ is exactly the event that the original value is at most $z$. Once $z$ reaches $c$, every possible recorded value is at most $z$. These two facts give the CDF.

Theorem: If $X$ is absolutely continuous with CDF $F$ and $Z=\min(X,c)$, then $$ F_Z(z)=P(Z\le z)= \begin{cases} F(z), & z<c,\\ 1, & z\ge c. \end{cases} $$ The jump at $c$ has size $$ P(Z=c)=F_Z(c)-\lim_{z\uparrow c}F_Z(z)=1-F(c). $$ For $z<c$, the continuous part of the law has density $f(z)$; it has no continuous part above $c$.

Proof. If $z<c$, then $Z\le z$ holds if and only if $X\le z$: values $X\ge c$ are recorded as $c$, which is larger than $z$. Therefore $F_Z(z)=F(z)$. If $z\ge c$, then $Z\le c\le z$ always, so $F_Z(z)=1$. Since $F$ is continuous, the left limit at $c$ is $F(c)$, while $F_Z(c)=1$. The jump is consequently $1-F(c)$. Below $c$, $F_Z$ agrees with $F$, so its derivative is $f$ almost everywhere there. Above $c$, $F_Z$ is constant. $\square$

The word density needs care here. The function $f(z)$ below $c$ describes the continuous part of the censored law; it does not describe the entire law by itself. The entire law also includes the probability $1-F(c)$ at exactly $c$. In particular, an ordinary density with respect to length alone cannot represent this distribution whenever that probability is positive.

Distribution as a measure: For every Borel set $A\subseteq\mathbb{R}$, $$ P(Z\in A)=\int_{A\cap(-\infty,c)} f(x)\,dx +\mathbf{1}_{\{c\in A\}}\{1-F(c)\}. $$ The first term is the continuous part below the limit; the second is the point mass at the limit.

This measure formula follows by splitting the original sample space into $X<c$ and $X\ge c$. On the first event, $Z=X$, so the contribution to $P(Z\in A)$ is $P(X\in A,\ X<c)=\int_{A\cap(-\infty,c)}f(x)\,dx$. On the second event, $Z=c$, so it contributes $P(X\ge c)=1-F(c)$ exactly when $c\in A$, and contributes zero otherwise. Adding the two disjoint contributions gives the formula.

As a useful consequence, for any bounded measurable function $g$, $$ E[g(Z)]=\int_{-\infty}^{c}g(x)f(x)\,dx+g(c)\{1-F(c)\}. $$ The endpoint $c$ may be included in the integral without changing its value, since an integral over a single point is zero. This identity makes explicit that the tail of $X$ is represented by the single value $g(c)$, rather than by its original range of values.

Worked Distributions

Worked Example: Exponential Variable Capped at 3

Let $X\sim\operatorname{Exp}(0.5)$, with rate $0.5$, and set $Z=\min(X,3)$. For $x\ge0$, $f(x)=0.5e^{-0.5x}$ and $F(x)=1-e^{-0.5x}$; for $x<0$, $F(x)=0$.

Below the cap, the recorded value equals the original value. Hence the continuous density is $0.5e^{-0.5z}$ for $0\le z<3$, and zero for $z<0$. At the cap, the probability is the original upper-tail probability: $$ P(Z=3)=P(X\ge3)=1-F(3)=e^{-1.5}\approx0.2231. $$

The CDF, including its boundary behavior, is $$ F_Z(z)= \begin{cases} 0, & z<0,\\ 1-e^{-0.5z}, & 0\le z<3,\\ 1, & z\ge3. \end{cases} $$ The continuous component has total probability $\int_0^3 0.5e^{-0.5z}\,dz=1-e^{-1.5}\approx0.7769$. Adding the atom gives $0.7769+0.2231=1$, as required.

Worked Example: Uniform Variable Capped at 6

Let $X\sim\operatorname{Uniform}(0,10)$ and record $Z=\min(X,6)$. The original density is $0.1$ on $(0,10)$, and its CDF on $[0,10]$ is $F(x)=x/10$.

For $0\le z<6$, the censored density is still $0.1$, because those values are not changed by the cap. The mass at the cap is $$ P(Z=6)=P(X\ge6)=1-F(6)=1-0.6=0.4. $$ The CDF is zero for $z<0$, equals $z/10$ for $0\le z<6$, and equals $1$ for $z\ge6$. Its left limit at 6 is $0.6$, so the jump there is $1-0.6=0.4$.

For example, $P(2<Z\le6)=F_Z(6)-F_Z(2)=1-0.2=0.8$. This consists of the continuous probability $P(2<X<6)=0.4$ and the cap probability $P(X\ge6)=0.4$. The two contributions add to $0.8$.

Worked Example: Normal Variable Capped at 12

Let $X\sim N(10,4)$, so its standard deviation is $2$, and set $Z=\min(X,12)$. Write $\Phi$ for the standard normal CDF. The original density is $$ f(x)=\frac{1}{2}\phi\!\left(\frac{x-10}{2}\right), $$ where $\phi$ is the standard normal density.

For every $z<12$, the continuous part of the censored distribution has density $f(z)$. At the cap, $$ P(Z=12)=P(X\ge12) =1-\Phi\!\left(\frac{12-10}{2}\right) =1-\Phi(1)\approx0.1587. $$ The CDF is $\Phi((z-10)/2)$ for $z<12$, and $1$ for $z\ge12$. Its left limit at 12 is $\Phi(1)\approx0.8413$, confirming that its jump is $1-0.8413\approx0.1587$.

A further check is $P(Z\le11)=P(X\le11)=\Phi(0.5)\approx0.6915$, since 11 is below the cap. By contrast, $P(Z\le12)=1$: the event includes both original values below the cap and all original values above it, which are recorded at 12.

Reading the Distribution Correctly

The CDF is continuous at every point below $c$ under the stated absolute-continuity assumption, then jumps at $c$ if the original upper tail is nonempty. It remains equal to one thereafter. In probability calculations, the jump is part of the law: for $a<c$, $$ P(a<Z\le c)=1-F(a). $$ This includes both the original values in $(a,c)$ and all original values at or above $c$. In contrast, if $a<b<c$, then $P(a<Z\le b)=F(b)-F(a)$, just as for the uncensored variable.

The formula for the atom also handles limits outside the effective support. If $c$ lies above all possible values of $X$, then $F(c)=1$ and the atom has size zero: censoring changes nothing. If $c$ is below part of the support, the upper-tail probability is pooled at $c$, as in the examples.

Common Mistakes

Using $f(c)$ as the mass at the cap. A density value is probability per unit of measurement, not a point probability. The atom is $P(X\ge c)=1-F(c)$, obtained by integrating the original density over the upper tail.
Renormalizing the density below $c$. The density $f(x)$ for $x<c$ is not divided by $F(c)$. Such division would describe a conditional distribution restricted to values below $c$, not the censored variable, which retains the upper-tail observations as records equal to $c$.
Giving the CDF value $F(c)$ at the cap. $F(c)$ is the left limit of the censored CDF, not its value at $c$. Since $Z\le c$ always, $F_Z(c)=1$.
Calling the whole law continuous because $X$ is continuous. The transformation maps a whole interval of possible original values, $X\ge c$, to one recorded value. That creates an atom even when $X$ has none.

Key takeaway: For $Z=\min(X,c)$, the observed law keeps the original density below $c$ and places the entire original upper-tail probability $1-F(c)$ at $c$. Its CDF is $F(z)$ below the cap and $1$ at and above it.

Check Your Understanding

Use the definition $Z=\min(X,c)$ and distinguish the continuous part from the mass at the cap.

If $F(c)=0.72$, what is $P(Z=c)$, and what is the left limit of $F_Z$ at $c$?
For $z<c$, why does $F_Z(z)=F(z)$? State the event equivalence that gives the result.
If $X\sim\operatorname{Uniform}(0,8)$ and $c=5$, find the mass at the cap and the continuous density below it.
Explain why $f(c)$ is not, in general, the probability that the recorded value equals $c$.
What is $F_Z(c)$, and why can it differ from $F(c)$?