Differences of Geometric Counts and Discrete Laplace Laws

Why a Difference of Counts Has a Simple Form

Differences of counts arise whenever two independent processes compete or are compared: for example, the difference between events recorded in two equal observation periods. For Poisson counts, this difference has the Skellam distribution. A related calculation for geometric counts leads to a different, particularly simple discrete law: its probability decreases geometrically with the absolute value of the difference.

The calculation depends on a precise convention for the geometric distribution. We use counts of failures before the first success, so the possible values begin at zero. If instead a geometric random variable counts trials through the first success, its support begins at one; subtracting two such counts gives the same difference law, but only after translating both variables consistently.

Definition: For $0<p<1$, a geometric count $X$ with success probability $p$ has support $\{0,1,2,\ldots\}$ and mass function $$ P(X=j)=p(1-p)^j,\qquad j=0,1,2,\ldots. $$ Write $q=1-p$. If $X$ and $Y$ are independent geometric counts with this same success probability, their difference is $D=X-Y$.

The task is to find $P(D=k)$ for every integer $k$, then identify the distribution and its characteristic function. Write $D=k$ as $X=Y+k$ when $k\geq 0$. This representation turns the probability into a sum over all possible values of $Y$.

Deriving the Mass Function

Theorem: Let $X$ and $Y$ be independent geometric counts with success probability $p$, under the support convention above. For every $k\in\mathbb Z$, $$ P(D=k)=\frac{p}{2-p}(1-p)^{|k|}. $$ This is a symmetric discrete Laplace distribution on the integers.

For $k\geq 0$, independence gives

\begin{aligned} P(D=k) &=\sum_{j=0}^{\infty}P(X=j+k,Y=j)\\ &=\sum_{j=0}^{\infty}p^2q^{j+k}q^j\\ &=p^2q^k\sum_{j=0}^{\infty}q^{2j}\\ &=\frac{p^2q^k}{1-q^2} =\frac{p}{2-p}q^k. \end{aligned}

The geometric series converges because $0<q<1$, and $1-q^2=(1-q)(1+q)=p(2-p)$. For $k<0$, exchange the roles of $X$ and $Y$: the pair $(Y,X)$ has the same joint distribution as $(X,Y)$, so $D$ and $-D$ have the same distribution. Thus $P(D=k)=P(D=-k)$, and the formula for nonnegative $k$ gives the claimed expression for all integers. $\square$

To see directly that the displayed masses sum to one, set $c=p/(2-p)$. Then

\sum_{k\in\mathbb Z}cq^{|k|} =c\left(1+2\sum_{k=1}^{\infty}q^k\right) =c\left(1+\frac{2q}{1-q}\right) =\frac{p}{2-p}\frac{1+q}{1-q} =1.

Here $1+q=2-p$ and $1-q=p$. The masses are nonnegative, and this calculation verifies their normalization independently of their derivation as probabilities.

Worked Example: Computing Individual Probabilities

Suppose $X$ and $Y$ are independent geometric counts with success probability $p=1/3$. Find $P(D=0)$, $P(D=2)$, and $P(D=-2)$.

Here $q=1-p=2/3$, and the coefficient in the mass function is $$ \frac{p}{2-p}=\frac{1/3}{5/3}=\frac15. $$ Therefore $$ P(D=0)=\frac15,\qquad P(D=2)=\frac15\left(\frac23\right)^2=\frac4{45}. $$ Symmetry gives $P(D=-2)=P(D=2)=4/45$. The equal probabilities for $2$ and $-2$ reflect the fact that the two independent counts have identical distributions.

Characteristic Function

For an integer-valued random variable $D$, its characteristic function is $\varphi_D(t)=E[e^{itD}]$, for real $t$, where $i^2=-1$. Independence lets us calculate the characteristic function of the difference from the characteristic functions of the two counts. The characteristic function of one geometric count is a convergent geometric series:

\varphi_X(t) =\sum_{j=0}^{\infty}p q^j e^{itj} =\frac{p}{1-qe^{it}}.

The series converges absolutely since $|qe^{it}|=q<1$. Since $D=X-Y$, independence gives

Result: The characteristic function of the difference is $$ \varphi_D(t) =\varphi_X(t)\varphi_Y(-t) =\frac{p^2}{(1-qe^{it})(1-qe^{-it})} =\frac{p^2}{1-2q\cos t+q^2}, \qquad t\in\mathbb R. $$

The denominator is $|1-qe^{it}|^2$, which is positive because $q<1$. In particular, the expression is real and nonnegative, and $\varphi_D(0)=p^2/(1-q)^2=1$. This is consistent with a characteristic function. The same result follows by summing the derived mass function:

\begin{aligned} \sum_{k\in\mathbb Z}cq^{|k|}e^{itk} &=c\left(1+\sum_{k=1}^{\infty}q^ke^{itk} +\sum_{k=1}^{\infty}q^ke^{-itk}\right)\\ &=c\left(1+\frac{qe^{it}}{1-qe^{it}} +\frac{qe^{-it}}{1-qe^{-it}}\right)\\ &=\frac{c(1-q^2)}{1-2q\cos t+q^2} =\frac{p^2}{1-2q\cos t+q^2}. \end{aligned}

The sums converge absolutely, so the geometric-series evaluations are valid. Also $c(1-q^2)=p^2$, as $c=p/(2-p)$ and $1-q^2=p(2-p)$. This confirms the characteristic function directly from the probability mass function.

Worked Example: Simplifying a Characteristic Function

Let $p=1/3$, so $q=2/3$. Substitute these values into the characteristic function: $$ \varphi_D(t) =\frac{(1/3)^2}{1-\frac43\cos t+\frac49} =\frac{1/9}{(13-12\cos t)/9} =\frac{1}{13-12\cos t}. $$ At $t=0$, this equals $1/(13-12)=1$, as required. At $t=\pi$, it equals $1/(13+12)=1/25$. The function is even in $t$, matching the symmetry of $D$.

Moments and Tail Probabilities

Symmetry gives $E[D]=0$ when the expectation exists; here it does, since the mass decreases geometrically. The variance can be found either from the difference of two independent geometric counts or directly from the discrete Laplace mass function. The latter route also illustrates why the absolute-value form is useful.

Result: For $D$ above, $$ E[D]=0,\qquad \operatorname{Var}(D)=\frac{2q}{p^2}. $$

To verify the variance directly, use symmetry and the power-series identity $\sum_{k=1}^{\infty}k^2q^k=q(1+q)/(1-q)^3$, valid for $0<q<1$. The identity follows by differentiating the geometric series twice, or by applying $q\,d/dq$ twice to $\sum_{k=0}^{\infty}q^k=1/(1-q)$; the differentiated series converge for $q$ in this range. Thus

\begin{aligned} E[D^2] &=2c\sum_{k=1}^{\infty}k^2q^k\\ &=2\frac{p}{2-p}\frac{q(1+q)}{(1-q)^3}\\ &=\frac{2q}{p^2}, \end{aligned}

because $1-q=p$ and $1+q=2-p$. Since $E[D]=0$, this second moment is the variance. As a check, a geometric count has variance $q/p^2$, so independence gives $\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)=2q/p^2$.

Worked Example: A Two-Sided Tail and the Variance

Take $p=2/5$, so $q=3/5$ and $c=p/(2-p)=1/4$. Find $P(|D|\geq 2)$ and $\operatorname{Var}(D)$.

For any integer $m\geq1$, summing the two equal tails gives $$ P(|D|\geq m)=2c\sum_{k=m}^{\infty}q^k =\frac{2c q^m}{1-q} =\frac{2q^m}{2-p}. $$ For $m=2$, this is $$ P(|D|\geq2)=\frac{2(3/5)^2}{8/5} =\frac{18}{25}\frac58 =\frac9{20}. $$ The variance is $$ \operatorname{Var}(D)=\frac{2q}{p^2} =\frac{2(3/5)}{(2/5)^2} =\frac{6/5}{4/25} =\frac{15}{2}. $$ The tail probability is less than one, as it must be, and the variance is positive. It need not be an integer even though $D$ takes integer values.

Parameterizations and Common Pitfalls

A common parameterization of the symmetric discrete Laplace law is $$ P(D=k)=\frac{1-r}{1+r}r^{|k|},\qquad k\in\mathbb Z,\quad 0<r<1. $$ The difference-of-geometric-counts law has this form with $r=q=1-p$, because $(1-r)/(1+r)=p/(2-p)$. This parameterization is useful for recognizing the law, but the parameter $r$ is not the success probability $p$ of the original geometric counts.

A frequent error is to write $P(D=k)=p^2q^{|k|}$, using only the first pair of outcomes that produces the difference. For $k\geq0$, every pair $(X,Y)=(j+k,j)$ contributes, not just $(k,0)$. Omitting the sum over $j$ loses the factor $1/(1-q^2)$. Another error is to forget that $D$ can be negative; a formula derived for $k\geq0$ must be extended using symmetry, or derived separately for negative $k$.

If $p=1$, each count is zero with probability one, so $D=0$ with probability one. This degenerate case is the natural endpoint of the formula. The case $p=0$ is not included: there is no finite first success, and the geometric count distribution is not a probability distribution on the nonnegative integers.

Key takeaway: For independent, identically distributed geometric counts of failures before success, the difference has mass $p(1-p)^{|k|}/(2-p)$. Its characteristic function is $p^2/[1-2(1-p)\cos t+(1-p)^2]$, and its variance is $2(1-p)/p^2$.

Check Your Understanding

Use the convention that a geometric count records failures before the first success.

For independent geometric counts with success probability $p$, explain why $P(D=k)=P(D=-k)$.
Write the mass function of $D$ when $p=1/2$, and calculate $P(D=1)$.
State the characteristic function of $D$ in terms of $p$ and $q=1-p$. Why is its denominator positive for every real $t$?
For $p=1/4$, calculate $P(|D|\geq1)$ using the tail formula.
What factor is omitted if one computes $P(D=k)$, for $k\geq0$, using only the pair $(X,Y)=(k,0)$?