An Exponential Example With Truncation and Censoring

Entry and Censoring in One Model

Suppose a study records lifetimes only for units that survive long enough to enter, and follows each retained unit only until a fixed study limit. These are two different operations: entry truncation selects which units appear, while right censoring limits how much of a selected unit's lifetime is observed. Combining them changes the observed distribution in a specific way.

Let $X\sim\operatorname{Exp}(\lambda)$, where $\lambda>0$ is the rate, so $X$ has density $\lambda e^{-\lambda x}$ for $x\geq 0$. Let $a$ be the entry threshold and $b$ the later censoring time, with $0\leq a<b$. A unit is included only if $X>a$. For an included unit, the recorded value is $W=\min(X,b)$. All distributions and expectations below are conditional on inclusion, $X>a$.

Definition: The conditionally observed lifetime is $W=\min(X,b)$ given $X>a$. The entry threshold $a$ determines which lifetimes are retained; the censoring time $b$ determines the largest recorded value. Since $b>a$, $W$ takes values in $(a,b]$, with an ordinary continuous component below $b$ and a possible point mass at $b$.

The exponential density gives $P(X>a)=e^{-\lambda a}>0$, so conditioning on entry is well-defined. Among retained units, lifetimes just above $a$ remain exact observations, whereas every lifetime at least $b$ is recorded as $b$. The resulting law is therefore not purely continuous.

Deriving the Conditional Observed Distribution

First condition on entry, before applying the censoring rule. For $x>a$, the conditional density of the retained lifetime is

f_{X\mid X>a}(x) =\frac{\lambda e^{-\lambda x}}{P(X>a)} =\frac{\lambda e^{-\lambda x}}{e^{-\lambda a}} =\lambda e^{-\lambda(x-a)}.

This conditional density integrates to one on $(a,\infty)$. Censoring then maps every value $X\geq b$ to the single recorded value $W=b$, while values $a<X<b$ remain unchanged.

Theorem: For $0\leq a<b$ and $\lambda>0$, the conditional distribution function of $W=\min(X,b)$ given $X>a$ is $$ F_W(w)= \begin{cases} 0, & w\leq a,\\ 1-e^{-\lambda(w-a)}, & a<w<b,\\ 1, & w\geq b. \end{cases} $$ Its continuous density on $(a,b)$ is $f_W(w)=\lambda e^{-\lambda(w-a)}$. There is an atom at the censoring time with probability $$ P(W=b\mid X>a)=e^{-\lambda(b-a)}. $$ There is no atom at $a$.

To prove the distribution formula, take $a<w<b$. On the event $X>a$, the inequality $W\leq w$ is equivalent to $X\leq w$: if $X\geq b$, then $W=b>w$, and if $X<b$, then $W=X$. Therefore

P(W\leq w\mid X>a) =\frac{P(a<X\leq w)}{P(X>a)} =\frac{e^{-\lambda a}-e^{-\lambda w}}{e^{-\lambda a}} =1-e^{-\lambda(w-a)}.

For $w\leq a$, $W\leq w$ is impossible for an included unit. For $w\geq b$, it is certain. The jump in the distribution function at $b$ is its value at $b$ minus its left limit:

P(W=b\mid X>a) =F_W(b)-\lim_{w\uparrow b}F_W(w) =1-\left(1-e^{-\lambda(b-a)}\right) =e^{-\lambda(b-a)}.

Equivalently, $W=b$ exactly when $X\geq b$, apart from the probability-zero event $X=b$. Differentiating the distribution function on $(a,b)$ gives the stated density. There is no jump at $a$, since the distribution function approaches zero there and $X$ is continuous. Finally, the continuous part has total probability $1-e^{-\lambda(b-a)}$, so it and the atom at $b$ sum to one. $\square$

Worked Example: Separating the Continuous Probability from the Censoring Atom

Let $X\sim\operatorname{Exp}(2)$, retain only lifetimes above $a=1$, and censor at $b=1.4$. Find the probability that the recorded value lies strictly between $1.1$ and $1.4$, and compare it with the probability that it lies in $(1.1,1.4]$.

For the strict interval, a recorded value $W$ satisfies $1.1<W<1.4$ exactly when $1.1<X<1.4$. Conditional on $X>1$, this probability is

\begin{aligned} P(1.1<W<1.4\mid X>1) &=\frac{P(1.1<X<1.4)}{P(X>1)}\\ &=e^{-2(1.1-1)}-e^{-2(1.4-1)}\\ &=e^{-0.2}-e^{-0.8}\\ &\approx 0.818731-0.449329\\ &\approx 0.369402. \end{aligned}

The boundary atom at $1.4$ is

P(W=1.4\mid X>1)=e^{-2(1.4-1)}=e^{-0.8}\approx 0.449329.

Including that atom gives

P(1.1<W\leq1.4\mid X>1) =0.369402+0.449329 \approx 0.818731.

Thus the strict interval probability is the continuous contribution alone; the interval with a closed upper endpoint also includes all observations censored at $1.4$.

The Mean of the Observed Lifetime

Censoring caps the recorded values at $b$, so the mean must include both the continuous observations below $b$ and the units recorded at $b$. Write $c=b-a>0$ for the distance between entry and censoring. The density and atom give the expectation directly:

E[W\mid X>a] =\int_a^b w\lambda e^{-\lambda(w-a)}\,dw +b e^{-\lambda(b-a)}.

Theorem: Under the conditions above, $$ E[W\mid X>a] =a+\frac{1-e^{-\lambda(b-a)}}{\lambda}. $$

For a complete derivation, set $y=w-a$ and $c=b-a$. The integral and atom become

\begin{aligned} E[W\mid X>a] &=\int_0^c (a+y)\lambda e^{-\lambda y}\,dy +(a+c)e^{-\lambda c}\\ &=a\left(1-e^{-\lambda c}\right) +\int_0^c y\lambda e^{-\lambda y}\,dy +(a+c)e^{-\lambda c}. \end{aligned}

Integration by parts, taking $u=y$ and $dv=\lambda e^{-\lambda y}\,dy$, yields

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

Substitute this expression in the expectation. The terms in $a$ combine to $a$, and the terms in $c e^{-\lambda c}$ cancel:

\begin{aligned} E[W\mid X>a] &=a(1-e^{-\lambda c}) -c e^{-\lambda c} +\frac{1-e^{-\lambda c}}{\lambda} +(a+c)e^{-\lambda c}\\ &=a+\frac{1-e^{-\lambda c}}{\lambda}\\ &=a+\frac{1-e^{-\lambda(b-a)}}{\lambda}. \end{aligned}

This proves the mean formula. It also shows why the censoring atom cannot be omitted: the cancellation involving $c e^{-\lambda c}$ depends on including its contribution to the expectation. $\square$

Worked Example: Calculating the Mean with a Later Censoring Time

Suppose $X\sim\operatorname{Exp}(0.5)$, units are retained when $X>2$, and follow-up ends at $b=5$. Here $c=5-2=3$, so the probability of being censored is

P(W=5\mid X>2)=e^{-0.5(3)}=e^{-1.5}\approx0.223130.

The continuous portion therefore has probability $1-0.223130=0.776870$. The conditional mean is

\begin{aligned} E[W\mid X>2] &=2+\frac{1-e^{-0.5(3)}}{0.5}\\ &=2+\frac{1-0.223130}{0.5}\\ &=2+\frac{0.776870}{0.5}\\ &\approx 2+1.553740\\ &\approx 3.553740. \end{aligned}

The result lies between the entry threshold $2$ and the censoring limit $5$, as it must. It is below $5$ because a positive fraction of retained lifetimes are observed before the censoring time.

Reading the Mixed Law in Practice

The conditional observed distribution is often easiest to use by separating exact observations from censored observations. For $a<w<b$, probabilities over intervals are calculated from the continuous density, or by differences of the distribution function. Any event that includes $b$ must additionally account for the point mass there. For example, $P(W\leq b\mid X>a)=1$, while $P(W<b\mid X>a)=1-e^{-\lambda(b-a)}$.

Worked Example: Evaluating the CDF, Density, and Censoring Probability

Let $X\sim\operatorname{Exp}(1.5)$, with entry threshold $a=0.5$ and censoring time $b=2$. The distance to censoring is $b-a=1.5$. The censoring probability is

P(W=2\mid X>0.5) =e^{-1.5(1.5)} =e^{-2.25} \approx0.105399.

At $w=1.25$, which is inside the continuous part of the distribution, the cumulative probability is

\begin{aligned} P(W\leq1.25\mid X>0.5) &=1-e^{-1.5(1.25-0.5)}\\ &=1-e^{-1.125}\\ &\approx1-0.324652\\ &\approx0.675348. \end{aligned}

The density at that point is

f_W(1.25)=1.5e^{-1.125}\approx1.5(0.324652)\approx0.486979.

The density is a probability per unit of lifetime, not a point probability. The probability of the single exact value $W=1.25$ is zero; in contrast, the exact boundary value $W=2$ has probability approximately $0.105399$.

As a check, the continuous part has probability $1-e^{-2.25}\approx0.894601$, and $0.894601+0.105399=1$. The mean is

\begin{aligned} E[W\mid X>0.5] &=0.5+\frac{1-e^{-2.25}}{1.5}\\ &=0.5+\frac{0.894601}{1.5}\\ &\approx0.5+0.596401\\ &\approx1.096401. \end{aligned}

Again, the mean lies between the entry and censoring times, $0.5$ and $2$.

Common Mistake: Forgetting What the Event Includes

A frequent error is to use the continuous density formula for an event that includes the censoring endpoint. The density describes probability accumulated over intervals below $b$; it does not include the mass concentrated at $b$. In the first worked example, subtracting the two exponential terms gives the probability of $1.1<W<1.4$, approximately $0.369402$. It does not give the probability of $1.1<W\leq1.4$, because that event also includes the boundary mass $e^{-0.8}\approx0.449329$.

A second error is to condition on $X>a$ in one part of a calculation but then use an unconditional probability elsewhere. The denominator $P(X>a)$ is required when starting from probabilities for $X$; after deriving the conditional density $\lambda e^{-\lambda(x-a)}$, the entry conditioning is already built in. Applying the unconditional density without that normalization produces the wrong distribution for retained units.

Key takeaway: Entry conditions the lifetime distribution on $X>a$; censoring then maps every retained lifetime at or beyond $b$ to $b$. The observed law has density $\lambda e^{-\lambda(w-a)}$ on $(a,b)$, an atom $e^{-\lambda(b-a)}$ at $b$, and mean $a+(1-e^{-\lambda(b-a)})/\lambda$.

Check Your Understanding

Use $X\sim\operatorname{Exp}(\lambda)$, with $\lambda>0$, entry threshold $a$, and censoring time $b>a$.

Why is the density of a retained lifetime $\lambda e^{-\lambda(x-a)}$ rather than $\lambda e^{-\lambda x}$? State its support.
Derive $P(W=b\mid X>a)$ by conditioning on $X>a$, and explain why this probability is an atom.
For $a=0$, $b=2$, and $\lambda=1$, find $P(W<2)$ and $P(W=2)$.
For the same parameters, calculate $E[W]$ and verify that it lies between $0$ and $2$.
Explain why the probability of an interval ending at $b$ may differ depending on whether the endpoint is included.