Identifying Distributional Identities by Uniqueness

From a Transform Calculation to a Distributional Identity

A characteristic function is useful not only for describing a distribution, but also for proving that two random variables have the same distribution. The key step is to calculate their characteristic functions and compare them for every real argument. If the functions agree everywhere, the uniqueness theorem for characteristic functions identifies the distributions.

This method is especially effective for sums. When the summands are independent, their characteristic functions multiply. A product that simplifies to the characteristic function of a familiar distribution gives a concise route to an exact distributional identity. The argument establishes equality of laws; it does not assert that the random variables themselves are equal.

Definition: The characteristic function of a real-valued random variable \(X\) is \(\varphi_X(t)=E[e^{itX}]\), for \(t\in\mathbb{R}\). Random variables \(X\) and \(Y\) are equal in distribution, written \(X\overset{d}{=}Y\), if they have the same distribution function. Equivalently, they assign the same probability to every Borel subset of \(\mathbb{R}\).

Theorem (uniqueness of characteristic functions): If \(X\) and \(Y\) are real-valued random variables and \(\varphi_X(t)=\varphi_Y(t)\) for every \(t\in\mathbb{R}\), then \(X\overset{d}{=}Y\).

The uniqueness theorem is the final step, not a substitute for the calculation. To use it, first establish that the candidate distribution is correctly specified and then show that the characteristic function of the variable under study equals the candidate characteristic function at all real \(t\). The standard characteristic functions used below are

\varphi_X(t)=\exp\left(i\mu t-\frac{\sigma^2t^2}{2}\right) \quad\text{if }X\sim N(\mu,\sigma^2),\qquad \varphi_Z(t)=\exp\left(\lambda(e^{it}-1)\right) \quad\text{if }Z\sim\operatorname{Poisson}(\lambda).

Here \(\sigma^2>0\) and \(\lambda\geq 0\). These are characteristic functions, not moment generating functions; their arguments are real \(t\), and the exponential includes the imaginary unit \(i\).

The Identity Principle for Independent Sums

If \(X\) and \(Y\) are independent, then the characteristic function of their sum factors:

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

The factorization uses independence to separate the expectation of the product. Without independence, the multiplication step is not generally valid. For finitely many mutually independent variables, applying the same argument repeatedly yields the product of their characteristic functions.

Theorem (identity by characteristic-function comparison): Let \(S\) and \(T\) be real-valued random variables. If \(\varphi_S(t)=\varphi_T(t)\) for every \(t\in\mathbb{R}\), then \(S\overset{d}{=}T\). In particular, if \(X_1,\ldots,X_n\) are mutually independent and the product \(\prod_{j=1}^n\varphi_{X_j}(t)\) equals the characteristic function of a specified random variable \(T\) for every real \(t\), then \(\sum_{j=1}^n X_j\overset{d}{=}T\).

Proof. Mutual independence gives \(\varphi_{\sum_{j=1}^n X_j}(t)=\prod_{j=1}^n\varphi_{X_j}(t)\) for each real \(t\). By the stated equality, this is \(\varphi_T(t)\) for every \(t\). The uniqueness theorem therefore gives \(\sum_{j=1}^n X_j\overset{d}{=}T\). The first assertion is the same uniqueness theorem applied directly to \(S\) and \(T\). \(\square\)

Find the characteristic function of the target.
Choose a candidate distribution and write its characteristic function with all parameters specified.

Calculate the characteristic function of the variable of interest.
For an independent sum, multiply the summands' characteristic functions; include any signs or scale factors in their arguments.

Compare for every real argument.
Algebraic agreement for all \(t\in\mathbb{R}\), followed by uniqueness, proves equality in distribution.

Normal Sums

The characteristic function of a normal distribution has an exponent that is quadratic in \(t\). Multiplication adds these exponents, which is why independent normal sums remain normal. The following result handles differing means and variances.

Theorem: If \(X_1,\ldots,X_n\) are mutually independent and \(X_j\sim N(\mu_j,\sigma_j^2)\), with \(\sigma_j^2>0\), then \(\sum_{j=1}^n X_j\sim N\left(\sum_{j=1}^n\mu_j,\sum_{j=1}^n\sigma_j^2\right)\).

Proof. Independence and the normal characteristic-function formula give, for every real \(t\),

\begin{aligned} \varphi_{\sum_{j=1}^n X_j}(t) &=\prod_{j=1}^n \exp\left(i\mu_jt-\frac{\sigma_j^2t^2}{2}\right)\\ &=\exp\left(i t\sum_{j=1}^n\mu_j-\frac{t^2}{2}\sum_{j=1}^n\sigma_j^2\right). \end{aligned}

The final expression is the characteristic function of the normal distribution with mean \(\sum_{j=1}^n\mu_j\) and variance \(\sum_{j=1}^n\sigma_j^2\). Uniqueness gives the claimed identity in distribution. \(\square\)

Worked Example: Adding Two Independent Normals

Let \(X\sim N(2,4)\) and \(Y\sim N(-1,9)\) be independent. Identify the distribution of \(S=X+Y\) by comparing characteristic functions.

The variances are \(4\) and \(9\), so \(\varphi_X(t)=\exp(2it-4t^2/2)=\exp(2it-2t^2)\) and \(\varphi_Y(t)=\exp(-it-9t^2/2)\). Independence gives

\begin{aligned} \varphi_S(t) &=\varphi_X(t)\varphi_Y(t)\\ &=\exp(2it-2t^2)\exp(-it-9t^2/2)\\ &=\exp\left(it-\frac{13t^2}{2}\right). \end{aligned}

This is the characteristic function of \(N(1,13)\): the means sum to \(2+(-1)=1\), and the variances sum to \(4+9=13\). Thus \(X+Y\sim N(1,13)\). The variance is \(13\), not the square of \(13\), because the second normal parameter denotes variance.

Worked Example: The Average of Four Independent Normals

Suppose \(X_1,\ldots,X_4\) are independent with \(X_j\sim N(5,4)\) for each \(j\). Find the distribution of \(\overline X=(X_1+\cdots+X_4)/4\).

For a scaled variable, \(\varphi_{aX}(t)=\varphi_X(at)\), since \(E[e^{it(aX)}]=E[e^{i(at)X}]\). Therefore independence gives

\begin{aligned} \varphi_{\overline X}(t) &=\prod_{j=1}^4\varphi_{X_j}(t/4)\\ &=\left[\exp\left(5i\frac{t}{4}-\frac{4(t/4)^2}{2}\right)\right]^4\\ &=\exp\left(5it-\frac{t^2}{2}\right). \end{aligned}

This is the characteristic function of \(N(5,1)\). Hence \(\overline X\sim N(5,1)\). The calculation shows directly that averaging leaves the mean at \(5\) and divides the variance \(4\) by \(4\), giving variance \(1\).

Independent Poisson Sums

For Poisson variables, the exponent in the characteristic function is linear in the rate parameter. Consequently, the rates add when independent Poisson variables are summed. The identification again follows by an all-\(t\) comparison and uniqueness.

Theorem: If \(Z_1,\ldots,Z_n\) are mutually independent and \(Z_j\sim\operatorname{Poisson}(\lambda_j)\), where each \(\lambda_j\geq0\), then \(\sum_{j=1}^n Z_j\sim\operatorname{Poisson}\left(\sum_{j=1}^n\lambda_j\right)\).

Proof. For every real \(t\), use independence and the Poisson characteristic function:

\begin{aligned} \varphi_{\sum_{j=1}^n Z_j}(t) &=\prod_{j=1}^n\exp\left(\lambda_j(e^{it}-1)\right)\\ &=\exp\left(\left(\sum_{j=1}^n\lambda_j\right)(e^{it}-1)\right). \end{aligned}

This is the characteristic function of a Poisson variable with rate \(\sum_{j=1}^n\lambda_j\). By uniqueness, the sum has that distribution. If all rates are zero, each variable is zero with probability one, and the same conclusion holds for the Poisson distribution of rate zero. \(\square\)

Worked Example: Adding Independent Poisson Counts

A laboratory records two independent event counts during the same observation period. Suppose \(Z_1\sim\operatorname{Poisson}(2.5)\) and \(Z_2\sim\operatorname{Poisson}(4)\). Identify the distribution of the total count \(T=Z_1+Z_2\).

Their characteristic functions are \(\varphi_{Z_1}(t)=\exp(2.5(e^{it}-1))\) and \(\varphi_{Z_2}(t)=\exp(4(e^{it}-1))\). By independence,

\begin{aligned} \varphi_T(t) &=\exp(2.5(e^{it}-1))\exp(4(e^{it}-1))\\ &=\exp(6.5(e^{it}-1)). \end{aligned}

The result is the characteristic function of \(\operatorname{Poisson}(6.5)\), so \(T\sim\operatorname{Poisson}(6.5)\). The numerical rate is the sum \(2.5+4=6.5\); no further assumptions about the count values are needed beyond the stated distributions and independence.

Why the Conditions Matter

A characteristic-function identity proves equality of distributions, not equality of outcomes on a particular probability space. Two random variables can have the same distribution without being equal, and variables with specified marginal distributions need not have the distribution obtained by multiplying their characteristic functions. That multiplication requires independence.

For example, let \(X\sim N(0,1)\) and set \(Y=X\). Both variables have standard normal marginal distributions, but they are not independent. Their sum is \(X+Y=2X\), so \(X+Y\sim N(0,4)\). Treating them as independent would instead multiply their marginal characteristic functions and incorrectly obtain \(N(0,2)\). The error is using \(\varphi_X(t)\varphi_Y(t)\) without justification; the correct expression in this dependent example is \(\varphi_{X+Y}(t)=\varphi_X(2t)\).

Another insufficient argument is checking agreement only at \(t=0\). Every characteristic function satisfies \(\varphi_X(0)=1\), so this single value cannot distinguish distributions. Uniqueness requires equality for every real \(t\). In practice, keep track of the parameterization as well: for \(N(\mu,\sigma^2)\), the coefficient of \(-t^2/2\) is the variance \(\sigma^2\), while in the Poisson characteristic function the coefficient of \(e^{it}-1\) is the rate.

Key takeaway: To establish a distributional identity, calculate both characteristic functions, verify equality for every real argument, and invoke uniqueness. For a sum, factor the characteristic function only when independence has been established.

Check Your Understanding

Use characteristic-function comparison and uniqueness to answer these questions.

If independent \(X\sim N(1,2)\) and \(Y\sim N(3,5)\), what are the mean and variance of \(X+Y\)?
If \(X_1,\ldots,X_5\) are independent \(N(0,9)\) variables, what is the distribution of their average?
Independent \(Z_1\sim\operatorname{Poisson}(1.2)\) and \(Z_2\sim\operatorname{Poisson}(3.8)\) are added. What is the rate of their total count?
Why does knowing that \(X\) and \(Y\) have normal marginal distributions not by itself justify multiplying their characteristic functions to find the law of \(X+Y\)?
Why does agreement of two characteristic functions at \(t=0\) not establish equality in distribution?