Truncated Laws as Conditional Distributions

Truncation as Conditioning

A truncated distribution describes values from an original random variable after values outside a specified retained set are excluded. The central idea is simple but essential: the law of a value that is retained is the original law conditioned on retention. Dividing by the probability of retention renormalizes the remaining probability to one.

This definition describes the distribution of an observed value when the observation rule retains it exactly when it belongs to the specified set. It does not, by itself, describe other selection mechanisms that might change the relative chances of retaining different values within that set.

Definition: Let $X$ be a real-valued random variable and let $R$ be a Borel set such that $q=P(X\in R)>0$. The law of $X$ truncated to $R$ is the conditional distribution of $X$ given $X\in R$. For each Borel set $B$, $$ P_R(B)=P(X\in B\mid X\in R) =\frac{P(X\in B\cap R)}{P(X\in R)}. $$ Here $P_R$ denotes the truncated probability law.

The requirement $P(X\in R)>0$ is not a technical decoration: the conditional probability in the definition divides by this quantity. If retention has probability zero, this ordinary event-conditioning definition does not specify a truncated law. The set $R$ must also be measurable, so that the probabilities in the definition are defined.

The CDF and the Normalizing Constant

Write $F$ for the original CDF and $F_R$ for the CDF after truncation to $R$. By definition, a CDF records the probability that the variable is at most $x$. Apply the conditional-law definition to the event $\{X\le x\}$:

F_R(x)=P(X\le x\mid X\in R) =\frac{P(X\le x,\ X\in R)}{P(X\in R)}.

The numerator keeps only probability that is both at or below $x$ and retained. The denominator is the total probability retained, so that the restricted law is rescaled to have total probability one.

Theorem: If $P(X\in R)=q>0$, then $P_R(B)=P(X\in B\cap R)/q$ defines a probability law. Its CDF is $F_R(x)=P(X\le x,\ X\in R)/q$.

Proof. For every Borel set $B$, $P_R(B)\ge 0$, since it is a nonnegative probability divided by $q>0$. Also,

P_R(\mathbb{R})=\frac{P(X\in \mathbb{R}\cap R)}{q} =\frac{P(X\in R)}{q}=1.

If $B_1,B_2,\ldots$ are pairwise disjoint Borel sets, then the sets $B_i\cap R$ are pairwise disjoint. Countable additivity of the original probability gives

P_R\left(\bigcup_{i=1}^{\infty}B_i\right) =\frac{P\left(\bigcup_{i=1}^{\infty}(B_i\cap R)\right)}{q} =\frac{\sum_{i=1}^{\infty}P(B_i\cap R)}{q} =\sum_{i=1}^{\infty}P_R(B_i).

Thus $P_R$ is a probability law. Substituting $B=(-\infty,x]$ into its definition gives the stated CDF. $\square$

Mass Functions and Densities After Truncation

For a discrete $X$ with mass function $f$, the probability of retaining the value $x$ is zero unless $x\in R$. If it is retained, its original mass is divided by the total retention probability. For a continuous $X$, the same restriction-and-renormalization principle applies to its density.

Theorem: If $X$ is discrete with mass function $f$, its truncated mass function is $$ f_R(x)=\frac{\mathbf{1}_{\{x\in R\}}f(x)}{P(X\in R)}. $$ If $X$ has density $f$ with respect to Lebesgue measure, its truncated density is $$ f_R(x)=\frac{\mathbf{1}_{\{x\in R\}}f(x)}{P(X\in R)} $$ for almost every $x$. In both formulas, the denominator is positive.

Proof for the discrete case. For each possible value $x$, apply the conditional-law definition to the event $\{X=x\}$. If $x\in R$, then $\{X=x\}\cap\{X\in R\}=\{X=x\}$, so the conditional mass is $f(x)/q$. If $x\notin R$, the intersection is empty and the conditional mass is zero. These two cases give the displayed formula. Its masses sum to one:

\sum_x f_R(x) =\frac{\sum_{x\in R}f(x)}{q} =\frac{P(X\in R)}{q}=1.

Proof for the density case. For any Borel set $B$, the conditional-law definition and the density representation of the original law give

P_R(B) =\frac{P(X\in B\cap R)}{q} =\frac{1}{q}\int_{B\cap R} f(x)\,dx =\int_B \frac{\mathbf{1}_{\{x\in R\}}f(x)}{q}\,dx.

The final integrand therefore is a density for the truncated law. Its integral over the real line is $\int_R f(x)\,dx/q=P(X\in R)/q=1$. A density is determined only up to changes on sets of Lebesgue measure zero, which is why the formula is asserted almost everywhere. $\square$

These formulas do not mean that every truncated law has a density. A mixed law, for example, can retain both an atom and a continuous part. The conditional-law definition applies without change to such a distribution, but its parts must be described by their appropriate masses and densities.

Worked Example: A Poisson Law Retained at Two Values

Let $X\sim\operatorname{Poisson}(1)$, so $P(X=k)=e^{-1}/k!$ for $k=0,1,2,\ldots$. Retain observations only when $X\in R=\{0,2\}$. Find the retention probability and the truncated mass function.

The retention probability is the sum of the two original masses:

q=P(X\in\{0,2\}) =P(X=0)+P(X=2) =e^{-1}+\frac{e^{-1}}{2} =\frac{3}{2e}.

For $k=0$, the conditional mass is

P_R(X=0)=\frac{P(X=0)}{q} =\frac{e^{-1}}{3/(2e)}=\frac{2}{3}.

For $k=2$, it is

P_R(X=2)=\frac{P(X=2)}{q} =\frac{e^{-1}/2}{3/(2e)}=\frac{1}{3}.

Every other value is outside $R$, so its truncated mass is zero. The retained probabilities sum to $2/3+1/3=1$, as required. Notice that the truncated law is supported only at 0 and 2, although the original Poisson law assigns positive mass to every nonnegative integer.

Worked Example: A Continuous Law Retained on Two Pieces

Let $X\sim\operatorname{Uniform}(0,4)$, with density $f(x)=1/4$ for $0<x<4$. Retain values in $R=[0,1]\cup[3,4]$. Find the retention probability, density, and CDF of the truncated law.

The retained set has total length 2 within an interval of length 4, so

q=P(X\in R)=\frac{2}{4}=\frac{1}{2}.

On $R$, the original density is $1/4$. Dividing by $q=1/2$ gives

f_R(x)= \begin{cases} 1/2, & 0<x<1 \text{ or } 3<x<4,\\ 0, & \text{otherwise}. \end{cases}

For the CDF, accumulate the retained density from left to right. Before 0, no retained probability has accumulated. Between 0 and 1, the accumulated probability is $x/2$. Between 1 and 3, it remains $1/2$, because this gap is not retained. Between 3 and 4, the accumulated probability is $1/2+(x-3)/2=(x-2)/2$. Thus

F_R(x)= \begin{cases} 0, & x<0,\\ x/2, & 0\le x\le 1,\\ 1/2, & 1<x<3,\\ (x-2)/2, & 3\le x\le 4,\\ 1, & x>4. \end{cases}

At $x=1$ the CDF is $1/2$, and it stays there until the retained interval begins again at 3. At $x=4$ it reaches 1. The formula displays directly how a gap in the retained set appears as a flat section of the CDF.

Worked Example: Truncating a Mixed Distribution

Suppose $X$ has probability $1/3$ at 0 and, with probability $2/3$, is uniformly distributed on $[0,2]$. Retain values in $R=[0,1]$. Find the conditional atom at 0 and the continuous density on $(0,1)$.

The atom at 0 is retained in full. The continuous component contributes probability $(2/3)(1/2)=1/3$ to $[0,1]$, because half of the uniform interval is retained. Therefore

q=P(X\in[0,1])=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}.

The conditional probability of the atom is

P_R(X=0)=\frac{P(X=0)}{q} =\frac{1/3}{2/3}=\frac{1}{2}.

The original continuous density on $(0,2)$ is $(2/3)(1/2)=1/3$: the mixture weight $2/3$ multiplies the uniform density $1/2$. On the retained portion $(0,1)$, dividing this density by $q$ gives $f_R(x)=(1/3)/(2/3)=1/2$. Its continuous mass is $\int_0^1(1/2)\,dx=1/2$. Together, the atom and continuous part have mass $1/2+1/2=1$. A density alone would fail to describe the atom at 0.

Why the Normalization Matters

Truncation changes a distribution, not merely its support. Restricting a mass function or density to $R$ discards probability outside $R$; dividing by $P(X\in R)$ reallocates the remaining probability so that the conditional law totals one. The examples also show that the form of the retained law depends on the original law: the retained Poisson masses are rescaled, the uniform density becomes supported on separated intervals, and the mixed law keeps both an atom and a continuous part.

A common mistake is to set the truncated density equal to $f(x)$ on $R$ and zero elsewhere, without dividing by the retention probability. In the uniform example, that incorrect function would have integral $1/2$, not 1, so it cannot be a probability density. The correction is to divide by $q=P(X\in R)$; the resulting integral is $q/q=1$.

A second mistake is to assume that conditioning removes every feature outside a density formula. In a mixed distribution, an atom has positive probability at a single point and is not represented by an ordinary density with respect to length. The conditional-law definition avoids this problem: it applies to the full probability measure, including both atoms and continuous portions.

Key takeaway: For a measurable retained set $R$ with positive probability, truncation is conditioning on $X\in R$. Intersect with $R$, then divide by $P(X\in R)$; use the resulting conditional law to obtain the CDF, masses, density, or all components of a mixed distribution.

Check Your Understanding

Try these questions using the conditional-law definition.

Why must $P(X\in R)>0$ for the truncation formula to define an ordinary conditional distribution?
A discrete variable has masses $0.2$, $0.5$, and $0.3$ at $0$, $1$, and $2$, respectively. If only values in $\{0,2\}$ are retained, what are the conditional masses?
For a continuous variable with density $f$, explain why restricting $f$ to $R$ without a normalizing factor will generally not produce a density.
How does the truncated CDF $F_R(x)$ express the requirement that a value be both retained and at most $x$?
Why is an ordinary density insufficient to describe every truncated mixed distribution?

Truncated Laws as Conditional Distributions

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