Differences and Symmetrization via Characteristic Functions

Why Take the Difference of Two Copies?

Suppose $X'$ is an independent random variable with the same distribution as $X$. The difference $X-X'$ compares two independent outcomes from one distribution. Its distribution is always symmetric about zero: a positive difference and the corresponding negative difference have the same probability. The characteristic function makes this symmetrization especially transparent: it is the squared modulus of the original characteristic function.

This identity needs no moment assumptions and applies to discrete, continuous, and mixed distributions. It also gives a useful way to separate the magnitude of a characteristic function from its complex phase. The examples below show the identity in discrete and continuous settings, while keeping clear which facts follow directly from the construction and which require additional calculations.

The Symmetrization Identity

Definition: Let $X$ be a real-valued random variable with characteristic function $\phi_X(t)=E[e^{itX}]$, for $t\in\mathbb{R}$. Let $X'$ be independent of $X$ and have the same distribution as $X$. The random variable $D=X-X'$ is called the symmetrization of $X$, and its characteristic function is $\phi_D(t)=E[e^{itD}]$.

Independence lets us factor the expectation of the product $e^{itX}e^{-itX'}$. The fact that $X'$ has the same distribution as $X$ then relates the second factor to the complex conjugate of $\phi_X(t)$.

Theorem: If $X$ and $X'$ are independent and identically distributed real-valued random variables, then for every $t\in\mathbb{R}$, $$ \phi_{X-X'}(t)=|\phi_X(t)|^2. $$ In particular, $\phi_{X-X'}(t)$ is real and nonnegative for every $t$. Moreover, $X-X'$ and $-(X-X')$ have the same distribution.

Proof of the characteristic-function identity. Since $X$ and $X'$ are independent, the expectation factors:

\begin{aligned} \phi_{X-X'}(t) &=E[e^{it(X-X')}]\\ &=E[e^{itX}e^{-itX'}]\\ &=E[e^{itX}]E[e^{-itX'}]. \end{aligned}

The variables $X$ and $X'$ have the same distribution, so $E[e^{-itX'}]=E[e^{-itX}]$. Complex conjugation commutes with expectation for bounded complex random variables, and $e^{-itX}$ is the complex conjugate of $e^{itX}$. Therefore $E[e^{-itX}]=\overline{\phi_X(t)}$. Hence

\phi_{X-X'}(t)=\phi_X(t)\overline{\phi_X(t)}=|\phi_X(t)|^2.

This expression is real and nonnegative because it is a squared modulus.

Proof of symmetry. Independence and identical distribution imply that the pair $(X,X')$ has the same joint distribution as $(X',X)$. Swapping the two coordinates changes $X-X'$ into $X'-X=-(X-X')$. Thus, for every Borel set $A\subseteq\mathbb{R}$,

P(X-X'\in A)=P(X'-X\in A)=P(-(X-X')\in A).

This is precisely the statement that $X-X'$ and $-(X-X')$ have the same distribution. Neither part of the proof requires $X$ to have a density or any finite moments. $\square$

The squared modulus retains the size of $\phi_X(t)$ but removes its complex phase. As a result, the symmetrized characteristic function cannot, by itself, retain all the location information in the original distribution. In particular, adding a constant to both independent copies has no effect on their difference.

Worked Examples

Worked Example: The Difference of Two Bernoulli Variables

Let $X$ and $X'$ be independent Bernoulli random variables with success probability $p=1/3$. Find the distribution and characteristic function of $D=X-X'$.

Each variable takes the values 0 and 1. The difference is 1 when $X=1$ and $X'=0$, and it is $-1$ when $X=0$ and $X'=1$. Independence gives

\begin{aligned} P(D=1)&=P(X=1)P(X'=0)=\frac13\cdot\frac23=\frac29,\\ P(D=-1)&=P(X=0)P(X'=1)=\frac23\cdot\frac13=\frac29,\\ P(D=0)&=P(X=X')=\frac13\cdot\frac13+\frac23\cdot\frac23=\frac59. \end{aligned}

The probabilities sum to $2/9+5/9+2/9=1$, and the probabilities at 1 and $-1$ agree, as symmetry predicts. Directly from this distribution,

\phi_D(t)=\frac29e^{it}+\frac59+\frac29e^{-it} =\frac59+\frac49\cos t.

For comparison, one Bernoulli variable has characteristic function $\phi_X(t)=2/3+(1/3)e^{it}$. The symmetrization identity gives

|\phi_X(t)|^2 =\left(\frac23+\frac13e^{it}\right)\left(\frac23+\frac13e^{-it}\right) =\frac59+\frac49\cos t,

in agreement with the direct calculation.

Worked Example: The Difference of Two Poisson Variables

Let $X$ and $X'$ be independent Poisson random variables, each with mean 1. Derive the characteristic function of $D=X-X'$, and calculate exact expressions for $P(D=0)$ and $P(D=1)$.

For a Poisson random variable with mean 1, its characteristic function follows by summing its probability mass function:

\phi_X(t)=\sum_{j=0}^{\infty}e^{itj}\frac{e^{-1}}{j!} =e^{-1}\exp(e^{it}) =\exp(e^{it}-1).

By the symmetrization identity,

\phi_D(t)=|\phi_X(t)|^2 =\exp(e^{it}-1)\exp(e^{-it}-1) =\exp\!\bigl(2\cos t-2\bigr).

For the probability at zero, both variables must take the same nonnegative integer value:

\begin{aligned} P(D=0) &=\sum_{j=0}^{\infty}P(X=j)P(X'=j)\\ &=e^{-2}\sum_{j=0}^{\infty}\frac{1}{(j!)^2}. \end{aligned}

For the probability at 1, the possible pairs are $(X,X')=(j+1,j)$, where $j=0,1,2,\ldots$. Thus

P(D=1) =\sum_{j=0}^{\infty}P(X=j+1)P(X'=j) =e^{-2}\sum_{j=0}^{\infty}\frac{1}{(j+1)!\,j!}.

Swapping $X$ and $X'$ shows that $P(D=-1)=P(D=1)$. These exact sums illustrate the difference distribution without needing to assign it a special-function name.

Worked Example: The Difference of Two Uniform Variables

Let $X$ and $X'$ be independent, each uniformly distributed on $[0,1]$. Derive the density of $D=X-X'$, then verify its characteristic function using both the density and the symmetrization identity.

The joint density of $(X,X')$ is 1 on the unit square and 0 elsewhere. For a fixed difference $z$, the density of $D$ is obtained by integrating $f_X(x)f_{X'}(x-z)$ over values of $x$ for which both $x$ and $x-z$ lie in $[0,1]$. The allowable interval has length $1-|z|$ when $|z|\leq 1$, and is empty when $|z|>1$. Therefore

f_D(z)= \begin{cases} 1-|z|,& |z|\leq 1,\\ 0,& |z|>1. \end{cases}

The density is symmetric, and its total integral is $2\int_0^1(1-z)\,dz=1$. From the density, for $t\ne 0$,

\begin{aligned} \phi_D(t) &=2\int_0^1(1-z)\cos(tz)\,dz\\ &=2\left(\frac{1-\cos t}{t^2}\right) =\frac{2(1-\cos t)}{t^2}. \end{aligned}

For the middle equality, integration by parts gives $\int_0^1(1-z)\cos(tz)\,dz=(1-\cos t)/t^2$. At $t=0$, the characteristic function is 1, either from its definition or by integrating the density.

The characteristic function of one uniform variable is

\phi_X(t)=\int_0^1e^{itx}\,dx =\frac{e^{it}-1}{it},\quad t\ne 0, \qquad \phi_X(0)=1.

Consequently,

|\phi_X(t)|^2 =\frac{|e^{it}-1|^2}{t^2} =\frac{2(1-\cos t)}{t^2},\quad t\ne 0,

which agrees with the transform of the triangular density. At $t=0$, both expressions have value 1.

A Moment Consequence

The characteristic-function identity does not require moments. When a second moment does exist, the difference also gives a simple variance relation.

Result: If $X$ and $X'$ are independent and identically distributed with $E[X^2]<\infty$, then $E[X-X']=0$ and $$ \operatorname{Var}(X-X')=2\operatorname{Var}(X). $$

Proof. The finite second-moment assumption ensures that the means and variances exist. Since $X$ and $X'$ have the same mean,

$$ E[X-X']=E[X]-E[X']=0. $$

Independence and finite second moments give $E[XX']=E[X]E[X']$, so $\operatorname{Cov}(X,X')=0$. Expanding the variance of the difference therefore yields

\begin{aligned} \operatorname{Var}(X-X') &=\operatorname{Var}(X)+\operatorname{Var}(X')-2\operatorname{Cov}(X,X')\\ &=\operatorname{Var}(X)+\operatorname{Var}(X) =2\operatorname{Var}(X). \end{aligned}

This consequence is consistent with the examples: the Bernoulli difference has variance $4/9$, twice the variance $2/9$ of a Bernoulli variable with success probability $1/3$. $\square$

Common Pitfall and Uses

A frequent error is to use the same-distribution condition but overlook the independence condition. Equal marginal distributions alone do not justify multiplying characteristic functions. For instance, if $X'=X$, then $X'$ has the same distribution as $X$, but the two variables are dependent and $X-X'=0$ with probability 1. Its characteristic function is 1, whereas $|\phi_X(t)|^2$ need not be 1.

The identity is useful because it turns an independent comparison into an expression involving only one characteristic function. It immediately shows that the symmetrized characteristic function is real and nonnegative, even when the original characteristic function is complex. The difference also removes any shared location shift: if a constant $c$ is added to each copy, $(X+c)-(X'+c)=X-X'$. These features make symmetrization a natural tool when the behavior of a distribution around its center matters more than its absolute location.

Keep the assumptions and conclusions distinct: independence and identical distribution produce the squared-modulus identity and symmetry; finite second moments are needed only for the variance formula. A real, nonnegative characteristic function is the conclusion for the symmetrized variable, not an assumption on the original variable.

Check Your Understanding

Use the independence and identical-distribution assumptions carefully in answering these questions.

For independent copies $X$ and $X'$, show in one line why $E[e^{-itX'}]=\overline{\phi_X(t)}$.
If $X$ is Bernoulli with success probability $p$, what are the three possible values of $X-X'$, and what are their probabilities?
Why is the difference of two independent identically distributed random variables symmetric even if the original distribution is not symmetric?
What fails in the symmetrization identity if $X'=X$ rather than an independent copy?
Which additional assumption is needed to conclude $\operatorname{Var}(X-X')=2\operatorname{Var}(X)$?