Independent Sums and Products of Characteristic Functions

Why Independence Simplifies a Sum

A sum of random variables can be difficult to describe by listing all possible outcomes, particularly when there are many summands. Characteristic functions provide a compact alternative: for independent summands, the characteristic function of the sum is the product of the individual characteristic functions. The same relationship appears in the distribution itself through convolution.

Throughout, the characteristic function of a real-valued random variable \(X\) is \(\varphi_X(t)=E[e^{itX}]\), for \(t\in\mathbb R\). Because \(|e^{itX}|=1\), this expectation always exists. Independence is the key condition in the product rule; identical distributions are not required.

Definition: If \(X\) and \(Y\) are independent real-valued random variables with probability laws \(\mu\) and \(\nu\), their convolution law, denoted \(\mu*\nu\), is the probability law defined on Borel sets \(A\subseteq\mathbb R\) by

(\mu*\nu)(A)=\int_{\mathbb R}\int_{\mathbb R}\mathbf 1_A(x+y)\,\nu(dy)\,\mu(dx).

It is the law of \(X+Y\). Equivalently, convolution is the distribution obtained by adding independent random variables with laws \(\mu\) and \(\nu\).

The Product Rule

Theorem: If \(X\) and \(Y\) are independent real-valued random variables, then, for every \(t\in\mathbb R\),

\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t).

More generally, if \(X_1,\ldots,X_n\) are mutually independent, then

\varphi_{\sum_{j=1}^n X_j}(t)=\prod_{j=1}^n\varphi_{X_j}(t).

Here, mutual independence means that for any Borel sets \(B_1,\ldots,B_n\), the probability that all \(X_j\) belong to their respective sets is the product of the marginal probabilities. In particular, this is stronger than requiring each pair to be independent.

To justify the product rule, we first use a consequence of independence. If \(g\) and \(h\) are bounded Borel-measurable complex-valued functions, then \(E[g(X)h(Y)]=E[g(X)]E[h(Y)]\). For simple functions, this follows by writing \(g(X)=\sum_r a_r\mathbf 1_{\{X\in A_r\}}\) and \(h(Y)=\sum_s b_s\mathbf 1_{\{Y\in B_s\}}\), and applying independence to each pair of events. Every bounded Borel-measurable complex-valued function can be approximated uniformly by simple Borel-measurable functions. Uniform approximation allows the expectations of the functions and their products to converge, proving the stated factorization for bounded \(g\) and \(h\).

Proof of the theorem: For \(g(x)=e^{itx}\) and \(h(y)=e^{ity}\), the factorization just established applies because both functions are bounded and Borel-measurable. Since \(e^{it(X+Y)}=e^{itX}e^{itY}\),

\varphi_{X+Y}(t) =E[e^{itX}e^{itY}] =E[e^{itX}]E[e^{itY}] =\varphi_X(t)\varphi_Y(t).

For \(n\) mutually independent variables, apply the same factorization to the bounded functions \(e^{itX_j}\). Their product is \(e^{it\sum_{j=1}^nX_j}\), so its expectation is \(\prod_{j=1}^n E[e^{itX_j}]\). This proves both claims. \(\square\)

The same independence condition identifies the law of a sum. Independence implies that the joint law of \((X,Y)\) is the product measure \(\mu\times\nu\): both measures assign the same probabilities to rectangles \(B\times C\), and rectangles determine a probability measure on the product Borel sets. Therefore, for any Borel set \(A\),

P(X+Y\in A) =\int_{\mathbb R}\int_{\mathbb R}\mathbf 1_A(x+y)\,\nu(dy)\,\mu(dx) =(\mu*\nu)(A).

In particular, the distribution function of the sum can be written as

F_{X+Y}(z)=\int_{\mathbb R}F_Y(z-x)\,\mu(dx).

Indeed, for a fixed \(x\), the inner probability that \(x+Y\le z\) is \(F_Y(z-x)\). This expression and the product rule describe the same independent sum in two ways: one through its distribution and one through its characteristic function.

Worked Sums and Convolutions

Worked Example: Two Independent Bernoulli Summands

Let \(X\sim\operatorname{Bernoulli}(0.3)\) and \(Y\sim\operatorname{Bernoulli}(0.6)\) be independent. A Bernoulli random variable with success probability \(p\) equals \(1\) with probability \(p\) and \(0\) with probability \(1-p\). Calculate the characteristic function and distribution of \(S=X+Y\).

For \(X\), the two possible values give \(\varphi_X(t)=0.7+0.3e^{it}\); likewise, \(\varphi_Y(t)=0.4+0.6e^{it}\). Independence gives

\varphi_S(t) =(0.7+0.3e^{it})(0.4+0.6e^{it}) =0.28+0.54e^{it}+0.18e^{2it}.

We can check the distribution directly. The event \(S=0\) requires two failures, so \(P(S=0)=0.7(0.4)=0.28\). Exactly one success can occur as \(X=1,Y=0\) or \(X=0,Y=1\), giving \(P(S=1)=0.3(0.4)+0.7(0.6)=0.12+0.42=0.54\). Finally, \(P(S=2)=0.3(0.6)=0.18\). The probabilities sum to \(0.28+0.54+0.18=1\), matching the coefficients in the characteristic function.

Worked Example: Repeated Independent Rademacher Variables

Let \(X_1,\ldots,X_4\) be independent, each taking the values \(1\) and \(-1\) with probability \(1/2\). Find the characteristic function and distribution of \(S=X_1+X_2+X_3+X_4\).

For one summand, \(\varphi_{X_j}(t)=\tfrac12e^{it}+\tfrac12e^{-it}=\cos t\). The repeated-sum rule therefore gives

\varphi_S(t)=(\cos t)^4.

If exactly \(k\) of the four variables equal \(1\), the remaining \(4-k\) equal \(-1\), so the sum is \(k-(4-k)=2k-4\). Each of the \(2^4=16\) sign patterns has probability \(1/16\), and there are \(\binom{4}{k}\) patterns with \(k\) positive signs. Thus the possible sums and probabilities are

\begin{array}{c|ccccc} s&-4&-2&0&2&4\\ P(S=s)&1/16&4/16&6/16&4/16&1/16 \end{array}

The probabilities total \((1+4+6+4+1)/16=1\). This is the fourfold convolution of the common two-point law: each new independent summand convolves the current sum law with that same law.

Worked Example: A Weighted Sum of Independent Bernoulli Variables

Let \(X\sim\operatorname{Bernoulli}(1/2)\) and \(Y\sim\operatorname{Bernoulli}(1/4)\) be independent, and set \(T=X+2Y\). Find \(\varphi_T\) and the distribution of \(T\).

The summands in \(T\) are \(X\) and \(2Y\). Since \(e^{it(2Y)}=e^{i(2t)Y}\), their characteristic functions are \(\varphi_X(t)=\tfrac12+\tfrac12e^{it}\) and \(\varphi_{2Y}(t)=\varphi_Y(2t)=\tfrac34+\tfrac14e^{2it}\). Independence yields

\varphi_T(t) =\left(\frac12+\frac12e^{it}\right) \left(\frac34+\frac14e^{2it}\right) =\frac38+\frac38e^{it}+\frac18e^{2it}+\frac18e^{3it}.

The four outcomes \((X,Y)=(0,0),(1,0),(0,1),(1,1)\) give \(T=0,1,2,3\), respectively. Their probabilities are \(\tfrac12\tfrac34=\tfrac38\), \(\tfrac12\tfrac34=\tfrac38\), \(\tfrac12\tfrac14=\tfrac18\), and \(\tfrac12\tfrac14=\tfrac18\). These sum to \(1\), and give exactly the coefficients displayed in \(\varphi_T\). The different support spacing here illustrates that the convolution adds the values of the actual summands, including their weights.

Convolution, Repetition, and a Common Pitfall

For probability laws \(\mu_1,\ldots,\mu_n\), repeated convolution is defined by applying the two-law operation successively. Its association does not affect the resulting law: adding \(x,y,z\) gives the same value regardless of grouping. If \(X_1,\ldots,X_n\) are independent with respective laws \(\mu_1,\ldots,\mu_n\), their sum has law \(\mu_1*\cdots*\mu_n\), and its characteristic function is the product of the individual characteristic functions.

If the summands are independent and identically distributed with common law \(\mu\) and characteristic function \(\varphi\), then the \(n\)-term sum has law \(\mu^{*n}\), the \(n\)-fold convolution of \(\mu\), and characteristic function \(\varphi(t)^n\). Identical distribution is what makes every factor the same; independence alone gives a product, but the factors need not match.

Mistake: “The characteristic function of a sum is always the product of the marginal characteristic functions.” This argument silently assumes independence.
Correction: The product rule follows from independence. Without it, \(E[e^{itX}e^{itY}]\) need not equal \(E[e^{itX}]E[e^{itY}]\). For example, if \(Y=X\) and \(X\) is equally likely to be \(1\) or \(-1\), then \(\varphi_{X+Y}(t)=\cos(2t)\), while \(\varphi_X(t)\varphi_Y(t)=\cos^2(t)\); these differ at \(t=\pi/2\).
Mistake: Using \(\varphi_X(t)^n\) for a sum of independent but differently distributed variables.
Correction: Use \(\prod_{j=1}^n\varphi_{X_j}(t)\) unless all summands have the same distribution.

These identities are useful because a convolution can be awkward to calculate directly, while multiplying characteristic functions is often simple. The product formula gives a compact description of independent sums and, for repeated identically distributed summands, exposes the repeated structure immediately. It does not by itself establish that two distributions are equal from matching characteristic functions; that converse requires a separate uniqueness result.

Check Your Understanding

Use independence carefully in each answer.

State the condition under which \(\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)\).
If \(X_1,\ldots,X_n\) are independent with characteristic functions \(\varphi_1,\ldots,\varphi_n\), write the characteristic function of their sum.
What is the convolution law of two independent random variables intended to describe?
If \(X_1,\ldots,X_n\) are independent and identically distributed with characteristic function \(\varphi\), what is the characteristic function of their sum?
Why does the product rule not follow merely from knowing the two marginal distributions of \(X\) and \(Y\)?

Independent Sums and Products of Characteristic Functions

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