Truncating Discrete Distributions

Why Discrete Truncation Needs Care

When observations are available only if they fall in a specified range, the distribution of the observed values is not the original distribution. For a discrete random variable, the observed probabilities are the original masses on the retained values, divided by the total probability of retaining an observation. That renormalization is essential: simply deleting the excluded masses leaves probabilities that do not sum to one.

Cutoffs also require attention to the integer support. For example, if $X$ is integer-valued, then $X>2.4$ means $X\geq 3$, while $X\geq 2.4$ also means $X\geq 3$. But $X\geq 2$ includes the mass at 2. The difference between including and excluding a boundary value can change both the normalizing probability and the resulting moments.

Definition: Let $X$ be a discrete random variable with probability mass function $f(k)=P(X=k)$, and let $A$ be a set of possible values of $X$ with $P(X\in A)>0$. The distribution of $X$ truncated to $A$ is the conditional distribution given $X\in A$. Its mass function is $$ f_A(k)=P(X=k\mid X\in A)= \begin{cases} \dfrac{f(k)}{P(X\in A)}, & k\in A,\\[6pt] 0, & k\notin A. \end{cases} $$

Write $c=P(X\in A)$ for the retention probability. The retained masses sum to one because $\sum_{k\in A} f_A(k)=\sum_{k\in A}f(k)/c=c/c=1$. For integer bounds $L\leq U$, the retained set may be $A=\{L,L+1,\ldots,U\}$, $A=\{k:k\geq L\}$, or another specified set. In every case, the same rule applies: retain the original masses and divide each by $c$.

Conditional Probabilities and Moments

The renormalization rule also gives formulas for probabilities and moments. In the following result, sums are over the support of $X$; terms outside $A$ contribute zero.

Theorem: Let $A$ be an event of the form $X\in B$, where $B$ is a set of possible values and $P(A)=c>0$. For any set $D$, $$ P(X\in D\mid A)=\frac{\sum_{k\in D\cap B}f(k)}{c}. $$ If $E[|X|^r]<\infty$ for a positive integer $r$, then $$ E[X^r\mid A]=\frac{\sum_{k\in B} k^r f(k)}{c}. $$ In particular, if $E[X^2]<\infty$, then $$ \operatorname{Var}(X\mid A) =\frac{\sum_{k\in B} k^2f(k)}{c} -\left(\frac{\sum_{k\in B} kf(k)}{c}\right)^2. $$

Proof. Since $A=\{X\in B\}$, the event $\{X\in D\}\cap A$ consists exactly of the values $k\in D\cap B$. Thus $P(X\in D,A)=\sum_{k\in D\cap B}f(k)$. Dividing by $P(A)=c$ gives the conditional probability formula.

For the moment formula, the conditional mass function is $f(k)/c$ on $B$ and zero elsewhere. Therefore, by the definition of expectation under this mass function, $E[X^r\mid A]=\sum_{k\in B}k^r f(k)/c$. The assumed absolute moment ensures this sum is finite. Applying this result for $r=1$ and $r=2$, then using variance equal to the second moment minus the square of the mean, gives the variance formula. $\square$

Worked Example: A Binomial Distribution Truncated to an Interval

Let $X\sim\operatorname{Binomial}(4,0.5)$, and retain observations with $1\leq X\leq 3$. Find the conditional probability mass function, mean, and variance.

The original probabilities at $0,1,2,3,4$ are respectively $1/16,4/16,6/16,4/16,1/16$. The retention probability is $c=(4+6+4)/16=14/16=7/8$. Dividing each retained mass by $7/8$ gives

P(X=1\mid 1\leq X\leq 3)=\frac{2}{7},\quad P(X=2\mid 1\leq X\leq 3)=\frac{3}{7},\quad P(X=3\mid 1\leq X\leq 3)=\frac{2}{7}.

The masses sum to $2/7+3/7+2/7=1$, as required. The conditional mean is $1(2/7)+2(3/7)+3(2/7)=14/7=2$. By symmetry about 2, or by direct calculation,

\operatorname{Var}(X\mid 1\leq X\leq 3) =(1-2)^2\frac{2}{7}+(2-2)^2\frac{3}{7}+(3-2)^2\frac{2}{7} =\frac{4}{7}.

The conditional distribution is not the original binomial distribution: the endpoint masses at 0 and 4 have been removed, and the retained masses have been rescaled.

Truncated Geometric Distributions

Use the geometric convention that $X$ counts failures before the first success, so $X$ takes values $0,1,2,\ldots$ and, for success probability $p$ with $0<p<1$, $P(X=k)=p(1-p)^k$. Write $q=1-p$. A lower-tail truncation $X\geq m$, for an integer $m\geq0$, has retention probability $\sum_{k=m}^{\infty}pq^k=q^m$. The resulting conditional mass at $m+j$, for $j\geq0$, is

P(X=m+j\mid X\geq m) =\frac{pq^{m+j}}{q^m}=pq^j.

Thus the retained value has the distribution of $m$ plus a geometric random variable with the original success probability. This identity follows directly from the masses; it is not a claim that truncation leaves the value of $X$ unchanged.

Worked Example: Geometric Distribution Truncated Below at 3

Suppose $X$ counts failures before a success with $p=0.25$, so $q=0.75$. Only values $X\geq3$ are retained. Find the retention probability, a conditional tail probability, the conditional mean, and the conditional variance.

The probability of retention is $P(X\geq3)=\sum_{k=3}^{\infty}0.25(0.75)^k=(0.75)^3=0.421875$. For $j\geq0$, dividing the mass at $3+j$ by this probability gives $0.25(0.75)^j$. In particular,

P(X\geq5\mid X\geq3) =\frac{P(X\geq5)}{P(X\geq3)} =\frac{(0.75)^5}{(0.75)^3} =(0.75)^2=0.5625.

The conditional value is 3 plus a geometric variable with $p=0.25$. To calculate its moments, the probability generating function of that geometric variable is $G(t)=p/(1-qt)$. Differentiating gives $E[X]=G'(1)=q/p$ and $E[X(X-1)]=G''(1)=2q^2/p^2$. Hence its variance is $E[X(X-1)]+E[X]-E[X]^2=q/p^2$. Adding the constant 3 shifts the mean but does not change the variance. Therefore,

E[X\mid X\geq3]=3+\frac{0.75}{0.25}=6,\qquad \operatorname{Var}(X\mid X\geq3)=\frac{0.75}{(0.25)^2}=12.

Truncated Poisson Distributions

For a Poisson random variable, conditional moments on a retained set can be calculated by summing the retained masses. For a tail event, a useful shortcut comes from shifting the index in the Poisson mass function. If $X\sim\operatorname{Poisson}(\lambda)$, with $\lambda>0$, and $r$ is a nonnegative integer, then

E[X(X-1)\cdots(X-r+1)\mathbf{1}_{\{X\geq m\}}] =\lambda^r P(X\geq m-r),\qquad m\geq r.

Here $\mathbf{1}_{\{X\geq m\}}$ equals 1 when $X\geq m$ and 0 otherwise. To verify the identity, write the expectation as a sum from $k=m$ to infinity. For $k\geq r$, $k(k-1)\cdots(k-r+1)e^{-\lambda}\lambda^k/k! =\lambda^r e^{-\lambda}\lambda^{k-r}/(k-r)!$. Setting $j=k-r$ changes the sum to $\lambda^r\sum_{j=m-r}^{\infty}P(X=j)$, which is the stated expression. $\square$

The cases $r=1$ and $r=2$ yield the first and second factorial moments on a tail. Since $X^2=X(X-1)+X$, these give the ordinary first and second moments needed for the conditional variance.

Worked Example: Poisson Distribution Truncated to Values at Least 2

Let $X\sim\operatorname{Poisson}(3)$, and retain observations with $X\geq2$. Calculate the retention probability, the conditional probability of $X=2$, the conditional mean, and the conditional variance.

The excluded values are 0 and 1, so the retention probability is $c=1-e^{-3}-3e^{-3}=1-4e^{-3}$, approximately $0.800852$. The conditional mass at 2 is

P(X=2\mid X\geq2) =\frac{P(X=2)}{c} =\frac{e^{-3}3^2/2}{1-4e^{-3}} \approx\frac{0.224042}{0.800852} \approx0.2798.

For the mean, the $r=1$ identity gives $E[X\mathbf{1}_{\{X\geq2\}}]=3P(X\geq1)=3(1-e^{-3})$. Thus

E[X\mid X\geq2] =\frac{3(1-e^{-3})}{1-4e^{-3}} \approx3.5595.

For the second factorial moment, the $r=2$ identity gives $E[X(X-1)\mathbf{1}_{\{X\geq2\}}]=9P(X\geq0)=9$. Therefore the conditional second ordinary moment is $9/c+E[X\mid X\geq2]$, and the variance is

\operatorname{Var}(X\mid X\geq2) =\frac{9}{c}+E[X\mid X\geq2]-\big(E[X\mid X\geq2]\big)^2 \approx\frac{9}{0.800852}+3.5595-(3.5595)^2 \approx2.1274.

The conditional mean exceeds 3 because the values 0 and 1, which pull the untruncated mean downward, are no longer observed.

Common Mistakes and Practical Checks

A frequent error is to report the original mass $f(k)$ as the probability of a retained value. After truncation, that mass must be divided by the retention probability. A second error is to mishandle a strict or non-integer cutoff. For integer-valued $X$, $X>a$ is equivalent to $X\geq\lfloor a\rfloor+1$; $X\geq a$ is equivalent to $X\geq\lceil a\rceil$. Determine the integer values first, then sum their masses.

A useful check is to add the conditional masses over the retained set; the result must be one. For moments, calculate the conditional first and second moments using the same retention probability, then form the variance. Do not truncate the mean or variance of the original distribution directly: in general, neither moment can be adjusted correctly without accounting for the removed masses.

Truncation describes which values are observed: values outside the retained set are absent from the observed distribution. It does not place those observations at a boundary or replace them by a reported cutoff value. Keeping that distinction clear helps prevent confusing a conditional distribution with a distribution formed by recording boundary values.

Key takeaway: To truncate a discrete distribution, identify the retained integer values, compute their total probability $c$, and divide each retained mass by $c$. Conditional moments are the original moment sums restricted to those values and divided by the same $c$.

Check Your Understanding

For each question, identify the retained set before calculating.

If $X$ is integer-valued and the retained observations satisfy $X>4$, what is the smallest retained value?
A discrete distribution assigns masses $0.2, 0.5, 0.3$ to $0,1,2$, respectively. Find the truncated mass function when only $X=1$ or $X=2$ is retained.
For a geometric variable counting failures before success, explain why $P(X\geq m)=q^m$.
For a Poisson variable with mean $\lambda$, use the factorial-moment identity to express $E[X\mid X\geq2]$ in terms of tail probabilities.
Why is the variance after truncation generally not the original variance with the excluded values simply omitted from its calculation?