Geometric-Count Random Sums

Why a Geometric Count Changes a Sum

A sum with a fixed number of independent terms has a distribution built from repeated addition. If the number of terms is itself random, the resulting law also averages over the possible counts. When the count is geometric, those count values have a simple sequence of weights, but the resulting distribution need not be geometric or belong to the same family as the summands.

The convention for the geometric distribution matters here. We use a count of failures before the first success, so the count can be zero. Consequently, the random sum has a positive probability of being exactly zero from taking no terms at all. That feature must be included even when each individual summand is continuous.

Definition: Let $N$ be independent of an iid sequence $X_1,X_2,\ldots$, and suppose $0<p\leq 1$. Write $q=1-p$ and define the geometric count by $$ P(N=n)=p q^n,\qquad n=0,1,2,\ldots. $$ The geometric-count random sum is $S=\sum_{i=1}^{N}X_i$, with the empty sum defined as $S=0$ when $N=0$. For $n\geq 1$, write $F_n$ for the distribution of $X_1+\cdots+X_n$, and let $F_0$ be the point mass at zero.

The weights sum to one because $\sum_{n=0}^{\infty}p q^n=1$ when $0\leq q<1$. Also, $P(N>m)=q^{m+1}$, which tends to zero as $m$ tends to infinity. Thus $N$ is finite with probability one, and the sum is well-defined. Some texts define a geometric random variable on $1,2,\ldots$, counting the trial on which success occurs. That is a different count convention; the formulas below use the number of summands, which starts at zero.

Expanding the Distribution Over Count Values

For any event concerning $S$, separate the possibilities $N=0,N=1,\ldots$. Independence ensures that, given $N=n$, the summands still have their original joint distribution. The following formula makes this expansion precise.

Theorem: For every Borel set $A\subseteq\mathbb{R}$, $$ P(S\in A)=\sum_{n=0}^{\infty}p q^n F_n(A), $$ where $F_n(A)=P(X_1+\cdots+X_n\in A)$ for $n\geq 1$, and $F_0(A)=\mathbf{1}_{\{0\in A\}}$. In particular, for every real $x$, $$ P(S\leq x)=\sum_{n=0}^{\infty}p q^n F_n(x). $$

Proof. The events $\{N=n\}$, $n\geq0$, are disjoint and their union has probability one. Therefore, by countable additivity, $$ P(S\in A)=\sum_{n=0}^{\infty}P(S\in A,\ N=n) =\sum_{n=0}^{\infty}P(S\in A\mid N=n)P(N=n). $$ When $n\geq1$, on $\{N=n\}$ the random sum equals $X_1+\cdots+X_n$. Independence of $N$ from the sequence implies $P(S\in A\mid N=n)=F_n(A)$. When $n=0$, $S=0$ with certainty, so the conditional probability is $F_0(A)$. Substituting $P(N=n)=p q^n$ gives the stated expansion. Taking $A=(-\infty,x]$ gives the distribution-function formula. $\square$

This result is a mixture over fixed-count sums: first choose a count $n$ with weight $p q^n$, then use the distribution of the sum of $n$ iid terms. For discrete summands, the formula gives a probability mass function by expanding each $F_n$. For absolutely continuous summands with density $f$, the sum of $n\geq1$ terms has the $n$-fold convolution density $f^{*n}$. In that case the random sum has an atom at zero and a continuous component with density $$ g(x)=\sum_{n=1}^{\infty}p q^n f^{*n}(x), $$ where the density formula is understood up to sets of Lebesgue measure zero.

To justify the density statement, let $B$ be any Borel set. The probability that $S$ lies in $B$ through a positive count is $$ \sum_{n=1}^{\infty}p q^n\int_B f^{*n}(x)\,dx =\int_B\sum_{n=1}^{\infty}p q^n f^{*n}(x)\,dx. $$ The interchange is valid by Tonelli’s theorem because every summand is nonnegative. The remaining count value $n=0$ contributes mass $p$ at zero. In particular, if every $X_i$ has an absolutely continuous distribution, then $P(S=0)=p$: the empty sum contributes $p$, while each positive-count sum has probability zero of equaling any fixed point.

Worked Examples

Worked Example: A Geometric Count of Bernoulli Terms

Let $P(N=n)=0.4(0.6)^n$, $n\geq0$, and let the independent summands be Bernoulli with success probability $a=0.5$. Find the full probability mass function of $S=\sum_{i=1}^N X_i$.

Given $N=n$, the sum is binomial with parameters $n$ and $0.5$. For $k\geq0$, only counts $n\geq k$ can contribute, so $$ P(S=k)=\sum_{n=k}^{\infty}0.4(0.6)^n {n\choose k}(0.5)^k(0.5)^{n-k}. $$ Set $m=n-k$. Then $$ P(S=k)=0.4(0.3)^k\sum_{m=0}^{\infty}{m+k\choose k}(0.3)^m. $$ For $|r|<1$, the power-series identity $\sum_{m=0}^{\infty}{m+k\choose k}r^m=(1-r)^{-k-1}$ follows by differentiating the geometric series $k$ times and dividing by $k!$. Applying it at $r=0.3$ gives $$ P(S=k)=\frac{0.4}{0.7}\left(\frac{0.3}{0.7}\right)^k =\frac{4}{7}\left(\frac{3}{7}\right)^k,\qquad k=0,1,2,\ldots. $$ For example, $P(S=0)=4/7$, and $P(S=1)=12/49$. This is a valid mass function: its terms are nonnegative and their sum is $(4/7)/(1-3/7)=1$. The event $N=0$ contributes probability $0.4$ to $S=0$; positive counts with all Bernoulli terms zero contribute the rest.

Worked Example: Poisson Summands

Let $P(N=n)=0.5(0.5)^n$, and suppose the independent summands have a Poisson distribution with parameter $2$. Calculate $P(S=0)$ and $P(S=1)$.

Given $N=n$, the sum of $n$ independent Poisson variables with parameter $2$ is Poisson with parameter $2n$; for $n=0$, it is the point mass at zero. Thus $$ P(S=0)=\sum_{n=0}^{\infty}0.5(0.5)^n e^{-2n} =0.5\sum_{n=0}^{\infty}r^n =\frac{0.5}{1-r}, $$ where $r=0.5e^{-2}\approx0.06767$. Therefore $P(S=0)\approx0.5363$. This includes the count-zero probability $0.5$ and the additional chance of getting zero from positive numbers of zero-valued Poisson terms.

For $S=1$, the $n=0$ term contributes zero, and for $n\geq1$ a Poisson variable with parameter $2n$ equals one with probability $2n e^{-2n}$. Hence $$ P(S=1)=\sum_{n=1}^{\infty}0.5(0.5)^n 2n e^{-2n} =\sum_{n=1}^{\infty}n r^n =\frac{r}{(1-r)^2} \approx0.0779. $$ The last identity follows by differentiating $\sum_{n=0}^{\infty}r^n=(1-r)^{-1}$ and multiplying by $r$. These calculations illustrate how count-value expansion can give selected probabilities even when a simple named distribution for the entire random sum is not evident.

Worked Example: Exponential Summands

Let $P(N=n)=0.25(0.75)^n$, and let the independent summands have exponential density $f(x)=2e^{-2x}$ for $x>0$, and zero otherwise. Find the atom at zero and the density for positive values.

The count-zero event gives $P(S=0)=0.25$. For $n\geq1$, the sum of $n$ such exponential variables has a gamma density $$ f^{*n}(x)=\frac{2^n x^{n-1}e^{-2x}}{(n-1)!},\qquad x>0. $$ The continuous part of the random-sum law therefore has density $$ g(x)=\sum_{n=1}^{\infty}0.25(0.75)^n\frac{2^n x^{n-1}e^{-2x}}{(n-1)!},\qquad x>0. $$ Taking out factors that do not depend on the summation index and setting $m=n-1$, $$ g(x)=0.25(1.5)e^{-2x} \sum_{m=0}^{\infty}\frac{(1.5x)^m}{m!} =0.375e^{-2x}e^{1.5x} =0.375e^{-0.5x},\qquad x>0. $$ The density integrates to $0.375/(0.5)=0.75$, exactly the probability $P(N\geq1)=0.75$. Together with the atom $0.25$, the total probability is one. For $x\geq0$, this also gives $P(S>x)=0.75e^{-0.5x}$.

Interpreting and Checking the Result

The geometric count supplies the weights, but the summand law determines the fixed-count distributions being mixed. As the examples show, the final law may be purely discrete, or it may have both a point mass and a continuous component. A density alone cannot describe the whole exponential example, because integrating its density gives only $0.75$, not one; the missing probability is the atom at zero.

A useful check is to track the count-zero term separately. For any summand distribution, $P(N=0)=p$ contributes $p$ to $P(S=0)$. For discrete summands, positive counts can contribute further probability at zero. For absolutely continuous summands, positive counts contribute no atom at zero, though the density may place substantial probability near zero.

Common Mistake

A frequent error is to start the expansion at $n=1$, or to use weights $p q^{n-1}$ while treating $n$ as the number of summands. That silently changes the model to a count on $1,2,\ldots$, omitting the possibility of an empty sum. With the convention here, the $n=0$ term has weight $p$, and every calculation must include it. A separate but related check is not to call the continuous component a probability density for the entire law when there is an atom: its integral is the probability of a positive count, $q$, not one.

Key takeaway: Expand a geometric-count random sum by conditioning on $N=n$, weighting the fixed-$n$ sum law by $p q^n$, and including the $n=0$ point mass. For continuous summands, sum the positive-count convolution densities and keep the atom at zero separate.

Check Your Understanding

Use the convention $P(N=n)=p(1-p)^n$ for $n\geq0$.

Why must the random sum have at least probability $p$ at zero when the count is independent of the summands?
Write the count expansion for $P(S\leq x)$ and identify its $n=0$ term.
If the summands are absolutely continuous, why does the density of the random sum generally integrate to $1-p$, rather than one?
For Bernoulli summands with success probability $a$, explain why only count values $n\geq k$ contribute to $P(S=k)$.
In the exponential example, verify that the atom and continuous component together have total probability one.

What You'll Learn