Conditional Transforms for Random Sums

Introduction

A sum with a fixed number of terms is transformed by multiplying the transforms of independent summands. When the number of terms is itself random, the result is a second layer of averaging: first average over the summands given the count, then average over the count. This leads to a compact formula, but only when the count is independent of the summand sequence.

The formula is useful across many models because it separates two ingredients: the distribution of the count and the common distribution of each term. No moment assumptions are needed for the characteristic-function version.

Definition and Assumptions

Definition: Let $N$ be a random variable taking values in $\{0,1,2,\ldots\}$, with $P(N<\infty)=1$. Let $X_1,X_2,\ldots$ be an iid sequence of real-valued random variables, and suppose $N$ is independent of the entire sequence $(X_1,X_2,\ldots)$. The random sum is $$ S_N=\sum_{j=1}^{N}X_j, $$ with the empty sum defined as $S_0=0$. Write $p_n=P(N=n)$, $\phi_X(t)=E[e^{itX_1}]$, and define the probability generating function of $N$ by $G_N(z)=\sum_{n=0}^{\infty}p_nz^n$, for $|z|\leq 1$.

The independence assumption means independence from the sequence as a whole, not just that $N$ has a particular marginal distribution. The iid assumption supplies the same characteristic function for each summand and makes the conditional transform given $N=n$ equal to $\phi_X(t)^n$.

Conditional Transform Formula

Theorem: Under the assumptions above, the characteristic function of the random sum is $$ \phi_{S_N}(t)=E[e^{itS_N}]=G_N\bigl(\phi_X(t)\bigr) =\sum_{n=0}^{\infty}p_n\phi_X(t)^n,\qquad t\in\mathbb R. $$

Proof. The events $\{N=n\}$, for $n\geq 0$, form a countable partition. On $\{N=n\}$, the random sum equals $S_n=\sum_{j=1}^{n}X_j$, where $S_0=0$. Independence of $N$ from the sequence gives, for each $n$, $$ E[e^{itS_N}\mid N=n]=E[e^{itS_n}] =\prod_{j=1}^{n}E[e^{itX_j}] =\phi_X(t)^n. $$ For $n=0$, the product is empty and equals $1$, as does $\phi_X(t)^0$. Averaging over the count therefore gives $$ E[e^{itS_N}] =\sum_{n=0}^{\infty}P(N=n)E[e^{itS_N}\mid N=n] =\sum_{n=0}^{\infty}p_n\phi_X(t)^n. $$ This sum converges absolutely because $|\phi_X(t)|\leq 1$ and $\sum_n p_n=1$. The final expression is $G_N(\phi_X(t))$, proving the result. $\square$

The same conditioning argument gives a distributional mixture, not just a transform identity. For every Borel set $A\subseteq\mathbb R$, $$ P(S_N\in A)=\sum_{n=0}^{\infty}p_nP(S_n\in A). $$ Indeed, partition $\{S_N\in A\}$ by the events $\{N=n\}$; independence implies $P(S_N\in A\mid N=n)=P(S_n\in A)$. The mixture formula makes clear that the case $N=0$ contributes mass $p_0$ at zero.

Worked Example: A Bernoulli Count and Normal Summands

Let $N\sim\operatorname{Bernoulli}(0.4)$, so $P(N=0)=0.6$ and $P(N=1)=0.4$. Independently, let the summands be iid $N(2,1)$. Find the characteristic function of $S_N$.

For a $N(\mu,\sigma^2)$ random variable, the characteristic function is $\exp(i\mu t-\sigma^2t^2/2)$. Thus $$ \phi_X(t)=\exp(2it-t^2/2),\qquad G_N(z)=0.6+0.4z. $$ Substitution into the conditional transform formula yields $$ \phi_{S_N}(t)=0.6+0.4\exp(2it-t^2/2). $$ The first term is the contribution from $N=0$, when the sum is exactly zero; the second is the contribution from $N=1$, when the sum has the summand distribution. In particular, at $t=0$ the expression is $0.6+0.4=1$, as every characteristic function must be.

Worked Example: A Poisson Count and Fair-Sign Summands

Let $N\sim\operatorname{Poisson}(2)$. Independently, let each $X_j$ equal $1$ or $-1$, each with probability $1/2$. Find the characteristic function of $S_N$.

For one summand, $$ \phi_X(t)=\tfrac12e^{it}+\tfrac12e^{-it}=\cos t. $$ The Poisson probability generating function is $G_N(z)=\exp(2(z-1))$. Therefore $$ \phi_{S_N}(t)=G_N(\cos t)=\exp\bigl(2(\cos t-1)\bigr). $$ At $t=0$, this is $\exp(0)=1$. Also, the expression is real and lies in $(0,1]$, since $\cos t\leq 1$. These checks are consistent with the fact that the sum is integer-valued and symmetric about zero.

Other Transforms

The same conditioning method applies to transforms other than characteristic functions, with suitable existence conditions. If $X_1\geq 0$ almost surely, define its Laplace transform by $L_X(s)=E[e^{-sX_1}]$ for $s\geq 0$. Since $0\leq L_X(s)\leq 1$, this expectation is finite. Conditioning on $N$ gives $$ E[e^{-sS_N}]=\sum_{n=0}^{\infty}p_nL_X(s)^n=G_N\bigl(L_X(s)\bigr). $$ Each term is nonnegative, so the conditional averaging and series are valid even without assumptions about moments of $N$ or $X_1$.

For moment-generating functions, write $M_X(t)=E[e^{tX_1}]$. If $M_X(t)<\infty$, then conditioning gives the extended nonnegative expectation $$ E[e^{tS_N}]=\sum_{n=0}^{\infty}p_nM_X(t)^n. $$ This may be infinite. Thus the composition is a finite moment-generating function only at values of $t$ for which the displayed series is finite. Unlike characteristic functions, moment-generating functions need not exist in a neighborhood of zero for every distribution.

Worked Example: A Geometric Count and Exponential Summands

Suppose $P(N=n)=2^{-(n+1)}$ for $n=0,1,\ldots$, and independently $X_j\sim\operatorname{Exponential}(2)$, where $2$ is the rate. Find the Laplace transform of $S_N$, and evaluate it at $s=2$.

The count probabilities sum to one because $\sum_{n=0}^{\infty}2^{-(n+1)}=1$. Its generating function is $$ G_N(z)=\sum_{n=0}^{\infty}2^{-(n+1)}z^n =\frac{1/2}{1-z/2} =\frac{1}{2-z},\qquad |z|\leq 1. $$ An exponential random variable with rate $2$ has density $2e^{-2x}$ for $x\geq0$, so direct integration gives $$ L_X(s)=\int_0^\infty e^{-sx}2e^{-2x}\,dx =\frac{2}{2+s},\qquad s\geq0. $$ Consequently, $$ E[e^{-sS_N}]=G_N\bigl(L_X(s)\bigr) =\frac{1}{2-\frac{2}{2+s}} =\frac{s+2}{2s+2}. $$ At $s=2$, the transform is $4/6=2/3$. At $s=0$, it is $2/2=1$, which checks the normalization. As $s$ tends to infinity, the transform tends to $1/2=P(N=0)$, consistent with the atom at zero from the empty sum.

Why Independence Is Essential

The formula $G_N(\phi_X(t))$ cannot generally be obtained from the marginal distributions of $N$ and $X_1$ when the count depends on the summands. In that situation, the conditional distribution of the included terms can change with $N$; the step $E[e^{itS_N}\mid N=n]=\phi_X(t)^n$ may fail.

Worked Example: Dependence Changes the Answer

Let $X_1,X_2,\ldots$ be iid fair signs, so $P(X_1=1)=P(X_1=-1)=1/2$. Define $N=1$ when $X_1=1$, and $N=0$ when $X_1=-1$. This count has $P(N=0)=P(N=1)=1/2$, but it is not independent of the summand sequence.

By the empty-sum convention, $S_N=1$ when $X_1=1$, and $S_N=0$ when $X_1=-1$. Hence the actual characteristic function is $$ E[e^{itS_N}]=\tfrac12e^{it}+\tfrac12. $$ If one incorrectly used the independent-count formula with the marginal distributions, then $G_N(z)=\tfrac12+\tfrac12z$ and $\phi_X(t)=\cos t$, giving $$ G_N(\phi_X(t))=\tfrac12+\tfrac12\cos t. $$ These expressions are not generally equal. For example, at $t=\pi$, the actual characteristic function is $0$, while the incorrect expression is $0$ as well; this value does not expose the difference. At $t=\pi/2$, however, the actual value is $(1+i)/2$, whereas the incorrect value is $1/2$. The discrepancy demonstrates why independence is a hypothesis, not a technical decoration.

Common Pitfall and Practical Use

A common mistake is to condition on $N=n$ and immediately write $\phi_X(t)^n$ without checking that conditioning on the count leaves the summands’ joint law unchanged. This is justified when $N$ is independent of the entire iid sequence. It is not justified merely because the $X_j$ have identical marginal distributions or because the count has a familiar distribution.

When the assumptions do hold, the formula is efficient: determine the count generating function, determine the one-term transform, and compose them. It also clarifies which features of a model are controlled by the count and which by the summand law. For integer-valued nonnegative summands, the same composition applies to probability generating functions whenever those are defined on the relevant arguments.

Check the structure.
Confirm that $N$ is finite and nonnegative integer-valued, and independent of the whole iid sequence.

Transform one summand.
Find $\phi_X(t)$, or use a Laplace or moment-generating transform with its required existence conditions.

Transform the count.
Evaluate $G_N(z)$ at the one-summand transform, checking that the argument is in the domain where the generating series is valid.

Key takeaway: For a finite random count independent of an iid summand sequence, conditioning gives $\phi_{S_N}(t)=G_N(\phi_X(t))$. Dependence between the count and the summands can invalidate this composition, even when their marginal distributions are unchanged.

Check Your Understanding

Try these questions before moving on.

Why is the empty sum convention necessary when $P(N=0)>0$, and what value does its characteristic function contribute?
State the independence assumption needed to derive $\phi_{S_N}(t)=G_N(\phi_X(t))$.
If $N$ is Poisson with mean $\lambda$ and is independent of iid summands with characteristic function $\phi_X(t)$, write the characteristic function of $S_N$.
For nonnegative summands, explain why the Laplace-transform composition is finite for every $s\geq0$.
In the fair-sign dependence example, calculate the actual and incorrectly composed characteristic functions at $t=\pi/2$.