Clipping, Censoring, and Point Masses

Why Clipping Creates a Mixed Distribution

A measurement may be recorded only within a permitted range. For example, a hypothetical instrument might report any value below a floor as the floor itself, and any value above a cap as the cap. This operation is called clipping. Even if the original measurement has a continuous distribution, the clipped value can equal either boundary with positive probability.

The result has both point masses and a continuous part. The point masses occur because many different original values are all mapped to the same endpoint. Between the endpoints, values are unchanged, so the original density remains there. This distinction matters whenever probabilities or expectations are calculated: integrating the interior density alone omits the endpoint probabilities.

Definition: Let $X$ have a continuous distribution with CDF $F_X$ and density $f_X$, and let $a<b$ be fixed constants. The variable clipped to $[a,b]$ is $$ Y=\min\bigl(\max(X,a),b\bigr). $$ Thus $Y=a$ when $X\le a$, $Y=X$ when $a<X<b$, and $Y=b$ when $X\ge b$. The endpoint probabilities are $P(Y=a)=F_X(a)$ and $P(Y=b)=1-F_X(b)$. On $a<y<b$, the density of $Y$ is $f_X(y)$.

The assumption that $X$ is continuous means it has no probability atoms of its own. Therefore, whether the clipping rule assigns equality at a boundary to the endpoint or to the interior does not change any probability. If $X$ were already mixed or discrete, its existing atoms would also have to be included.

The CDF, Atoms, and Interior Density

The cumulative distribution function gives a direct way to derive the clipped law. If $y<a$, the clipped variable cannot be at or below $y$. If $a\le y<b$, the event $Y\le y$ is exactly the event $X\le y$. At or above $b$, every possible value of $Y$ is at or below $y$. Hence

F_Y(y)= \begin{cases} 0, & y<a,\\ F_X(y), & a\le y<b,\\ 1, & y\ge b. \end{cases}

At $a$, the CDF jumps from 0 to $F_X(a)$, so the jump size is $P(Y=a)=F_X(a)$. At $b$, it jumps from $F_X(b)$ to 1, giving $P(Y=b)=1-F_X(b)$. Between these two points, differentiating $F_Y(y)=F_X(y)$ gives the interior density $f_X(y)$.

P(Y=a)+\int_a^b f_X(y)\,dy+P(Y=b) =F_X(a)+\bigl(F_X(b)-F_X(a)\bigr)+1-F_X(b)=1.

This total-probability check is useful: the interior integral is only the probability that $a<X<b$, not the full probability of $Y$. The clipped law is therefore a mixed distribution, consisting of two possible atoms and a continuous component. If the original support does not extend past one of the clipping boundaries, the corresponding atom may have probability zero.

Worked Example: Clipping a Uniform Measurement

A hypothetical sensor reading $X$ is uniformly distributed on $(-2,4)$. The recorded value is clipped to $[0,3]$. Find the endpoint probabilities, the interior density, and $E[Y]$. The method is to partition the outcomes into the lower endpoint, the interior, and the upper endpoint.

Lower endpoint. The uniform density is $1/(4-(-2))=1/6$. Thus $P(Y=0)=P(X\le0)=(0-(-2))/6=2/6=1/3$.

Upper endpoint. The probability of exceeding 3 is $P(Y=3)=P(X\ge3)=(4-3)/6=1/6$.

Interior. For $0<y<3$, the density is $f_Y(y)=1/6$. Its total probability is $\int_0^3(1/6)\,dy=3/6=1/2$. The probabilities add to $1/3+1/2+1/6=1$.

Expectation. Multiply each endpoint by its probability and integrate over the interior:

\begin{aligned} E[Y] &=0\left(\frac13\right)+\int_0^3 y\left(\frac16\right)\,dy+3\left(\frac16\right)\\ &=0+\left[\frac{y^2}{12}\right]_0^3+\frac12\\ &=\frac{9}{12}+\frac12=\frac34+\frac12=\frac54=1.25. \end{aligned}

The result is checked against the range $0\le Y\le3$, which the mean 1.25 satisfies. The common mistake avoided here is treating $1/6$ as a density over the whole clipped distribution; it describes only the interior, while the endpoints carry the remaining probability.

Expectation and Other Functions of the Clipped Variable

For any function $g$ for which the expectation exists, the same partition gives a general formula. The values of $g(Y)$ at the endpoints must be weighted by their atom probabilities; the integral handles only the values strictly between them.

E[g(Y)] =g(a)F_X(a)+\int_a^b g(y)f_X(y)\,dy+g(b)\bigl(1-F_X(b)\bigr).

Taking $g(y)=y$ gives the mean. Taking $g(y)=y^2$ gives the second raw moment, from which the variance can be found using $\operatorname{Var}(Y)=E[Y^2]-(E[Y])^2$. The formula follows by splitting the expectation over the disjoint events $X\le a$, $a<X<b$, and $X\ge b$. On the first event $Y=a$; on the second $Y=X$; on the third $Y=b$. The probabilities and density integral in the formula are exactly the probabilities of those three parts.

Find the lower atom.
Calculate $F_X(a)$, the probability that the original value is at or below the lower clipping point.

Keep the interior law.
For $a<y<b$, use the original density $f_X(y)$. Integrate only over this interval for interior probabilities and expectation contributions.

Find the upper atom and check.
Calculate $1-F_X(b)$, then verify that the two endpoint masses and the interior probability sum to 1.

Worked Example: Exponential Values Clipped to an Interval

Let $X$ have an exponential distribution with rate $0.5$ per unit, so $f_X(x)=0.5e^{-0.5x}$ for $x\ge0$ and $F_X(x)=1-e^{-0.5x}$ for $x\ge0$. Clip $X$ to $[1,3]$. Find both atoms, the interior probability, and $E[Y]$. The rate is per unit of $X$, and all calculations use that same unit.

Lower atom. Since values up to 1 are mapped to 1, $P(Y=1)=F_X(1)=1-e^{-0.5}\approx0.3935$.

Upper atom. Values at or above 3 are mapped to 3, so $P(Y=3)=1-F_X(3)=e^{-1.5}\approx0.2231$.

Interior probability. The probability of $1<X<3$ is $F_X(3)-F_X(1)=e^{-0.5}-e^{-1.5}\approx0.3834$. The three probabilities sum to $0.3935+0.3834+0.2231=1.0000$, to four decimal places.

Expectation. The endpoint contributions are $1(1-e^{-0.5})$ and $3e^{-1.5}$. The interior contribution is

\begin{aligned} \int_1^3 y(0.5e^{-0.5y})\,dy &=\left[-(y+2)e^{-0.5y}\right]_1^3\\ &=3e^{-0.5}-5e^{-1.5}. \end{aligned}

Combining all three contributions,

\begin{aligned} E[Y] &=(1-e^{-0.5})+(3e^{-0.5}-5e^{-1.5})+3e^{-1.5}\\ &=1+2e^{-0.5}-2e^{-1.5}\\ &\approx1+2(0.6065307)-2(0.2231302)=1.7668. \end{aligned}

The check $1\le E[Y]\le3$ is satisfied. A common integration error is to evaluate the interior contribution as $2e^{-0.5}-4e^{-1.5}$. The antiderivative at $y=1$ is $-3e^{-0.5}$, and at $y=3$ it is $-5e^{-1.5}$; subtracting the former from the latter gives $3e^{-0.5}-5e^{-1.5}$. Omitting either endpoint contribution would also give the wrong mean.

When Only One Endpoint Has an Atom

Clipping does not guarantee positive probability at both boundaries. If the original support lies entirely below the upper clipping point, then $P(X\ge b)=0$, so there is no upper atom. The clipped variable can still have a lower atom if the original distribution extends below $a$. The CDF and expectation formulas remain valid; a zero endpoint probability simply contributes zero.

Worked Example: An Unreached Upper Clipping Point

Let $X$ be uniform on $(0,5)$, with density $1/5$, and clip it to $[2,7]$. Find the distribution’s endpoint masses, $P(Y\le4)$, and $E[Y]$. The upper clipping point is above the support of $X$, so check rather than assume that it has an atom.

Lower endpoint. The event $Y=2$ is $X\le2$, giving $P(Y=2)=2/5=0.4$.

Upper endpoint. Since $X<5$ with probability 1 and $5<7$, $P(X\ge7)=0$, so $P(Y=7)=0$. On $2<y<5$, the density is $1/5$; on $5\le y<7$, it is zero. The interior probability is $\int_2^5(1/5)\,dy=3/5$, and $0.4+0.6+0=1$.

Requested probability. For $2\le4<7$, $P(Y\le4)=P(X\le4)=4/5=0.8$.

Expectation. Only the lower atom and the interval from 2 to 5 contribute:

\begin{aligned} E[Y] &=2\left(\frac25\right)+\int_2^5 y\left(\frac15\right)\,dy+7(0)\\ &=\frac45+\left[\frac{y^2}{10}\right]_2^5\\ &=\frac45+\frac{25-4}{10} =\frac45+\frac{21}{10}=2.9. \end{aligned}

The mean is between the minimum and maximum possible clipped values, 2 and 5, as expected. Assigning a positive mass to 7 just because 7 is a clipping boundary would be a mistake: an endpoint atom exists only when the original variable has positive probability beyond that boundary.

Clipping, Censoring, and Common Traps

Clipping describes a transformation of a random variable: values outside the interval are replaced by the nearest endpoint. In a censored-data setting, a value beyond a threshold may instead be recorded as “at least the threshold” or “at most the threshold,” with an indication that the exact value is unknown. If a censored observation is numerically represented by the threshold, its recorded number has the same endpoint accumulation as clipping. But the interpretation is different: the threshold is not necessarily the underlying measurement.

For distribution calculations, keep the model explicit: specify the original $X$, its distribution and parameters, and the clipping interval. Then use the transformed variable $Y$, not the original variable, for probabilities and moments of the recorded result.

Using only a density. The interior density integrates to $P(a<X<b)$, which may be less than 1. Include both endpoint atoms when calculating total probabilities or expectations.
Putting an atom at the wrong boundary. $P(Y=a)=P(X\le a)$, while $P(Y=b)=P(X\ge b)$. The lower and upper tails must not be interchanged.
Using the original CDF everywhere. $F_Y(y)=F_X(y)$ only for $a\le y<b$. Outside that range, the clipped CDF is 0 or 1.
Assuming both atoms are positive. A clipping boundary outside the original support may have zero probability.

These calculations are aligned with the SOA VEE Mathematical Statistics syllabus: the key skill is to derive the transformed law from events, identify its support and atoms, and account for every part of the distribution when calculating a quantity such as a mean.

Check Your Understanding

For each problem, identify the endpoint atoms and use the clipped distribution rather than treating its interior density as a complete probability law.

$X$ is uniform on $(-1,5)$ and is clipped to $[0,4]$. Find $P(Y=0)$, $P(Y=4)$, the interior density, and $E[Y]$. (Answer: $P(Y=0)=1/6$, $P(Y=4)=1/6$, $f_Y(y)=1/6$ for $0<y<4$, and $E[Y]=2$.)
$X$ is exponential with rate $0.25$ per unit and is clipped to $[0,2]$. Find $P(Y=2)$ and $P(Y<1)$. Give probabilities to four decimals. (Answer: $P(Y=2)=e^{-0.5}\approx0.6065$, and $P(Y<1)=1-e^{-0.25}\approx0.2212$.)
$X\sim N(0,4)$, where the second parameter is the variance, and $X$ is clipped to $[-2,3]$. Find the two endpoint masses to four decimals. (Answer: $P(Y=-2)=\Phi(-1)\approx0.1587$, and $P(Y=3)=1-\Phi(1.5)\approx0.0668$.)
$X$ is uniform on $(1,9)$ and is clipped to $[3,7]$. Find both endpoint masses, the interior density, and $E[Y]$. (Answer: $P(Y=3)=1/4$, $P(Y=7)=1/4$, $f_Y(y)=1/8$ for $3<y<7$, and $E[Y]=5$.)
$X$ is exponential with rate 1 per unit and is clipped to $[0.5,2]$. Find both endpoint masses, the interior probability, and $E[Y]$ to four decimals. (Answer: $P(Y=0.5)=1-e^{-0.5}\approx0.3935$, $P(Y=2)=e^{-2}\approx0.1353$, $P(0.5<Y<2)=e^{-0.5}-e^{-2}\approx0.4712$, and $E[Y]=0.5+e^{-0.5}-e^{-2}\approx0.9712$.)

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