Why Stability Matters
For many distributions, adding independent variables changes the form of the distribution. Stable distributions are special: when independent variables have a common stability index, their weighted sum remains in the same family. The characteristic function makes this property precise and turns the calculation into addition of exponent terms.
The normal and Cauchy distributions are especially useful examples. Normal sums combine variances, while Cauchy sums combine their usual scale parameters. Both facts are instances of one characteristic-function identity, provided the parameters are interpreted consistently.
Definition and Characteristic-Function Form
Stable distributions include asymmetric as well as symmetric examples. The focus here is the symmetric stable family, for which the characteristic function gives a particularly direct description. A characteristic function uniquely determines a distribution, so equality of the characteristic functions below establishes equality in distribution.
For every \(0<\alpha\leq 2\), this formula is a characteristic function. The case \(\mu=0\) is strictly stable. A nonzero \(\mu\) translates the distribution; it does not alter the symmetric stable form of its centered characteristic function. The exponent coefficient \(c\) and the conventional scale \(s\) are different parameterizations and should not be interchanged.
Weighted Sums: The Main Identity
The absolute value in \(|a_j|^\alpha\) is essential: a negative weight reverses a variable but cannot give a negative decay coefficient in a characteristic function. The location, on the other hand, is multiplied by the signed weight.
Proof. Independence gives the characteristic function of the sum as the product of the characteristic functions of its terms. For each \(j\), $$ E[e^{it a_jX_j}]=\varphi_{X_j}(a_jt) =\exp\!\left(i\mu_j a_jt-c_j|a_jt|^\alpha\right) =\exp\!\left(i\mu_j a_jt-c_j|a_j|^\alpha|t|^\alpha\right). $$ Multiplying these expressions adds their exponents: $$ \varphi_S(t) =\prod_{j=1}^m E[e^{it a_jX_j}] =\exp\!\left(it\sum_{j=1}^m a_j\mu_j-|t|^\alpha\sum_{j=1}^m c_j|a_j|^\alpha\right). $$ This is the characteristic function of a symmetric \(\alpha\)-stable distribution with the stated location and coefficient. If at least one weight is nonzero, the coefficient is positive, so the distribution is nondegenerate. \(\square\)
Scaling Sums of Independent Copies
When \(X_1,\ldots,X_n\) are independent copies of a symmetric \(\alpha\)-stable variable \(X\) with location \(\mu\) and coefficient \(c\), the weighted-sum identity gives a concise scaling rule.
Proof. Applying the weighted-sum result with every weight equal to one, the sum has characteristic function $$ \exp\!\left(in\mu t-nc|t|^\alpha\right). $$ The random variable \(n\mu+n^{1/\alpha}(X-\mu)\) has the same characteristic function: \(X-\mu\) has characteristic function \(\exp(-c|t|^\alpha)\), and scaling it by \(n^{1/\alpha}\) gives \(\exp(-nc|t|^\alpha)\). Adding \(n\mu\) multiplies this by \(\exp(in\mu t)\). The characteristic functions therefore agree, which proves equality in distribution. \(\square\)
For a centered variable, \(\mu=0\), this reduces to \(X_1+\cdots+X_n\overset{d}{=}n^{1/\alpha}X\). Thus the typical size of the sum scales as \(n^{1/\alpha}\), rather than necessarily as the square root of \(n\).
Worked Examples
Worked Example: A Weighted Sum of Normal Variables
Let \(X\sim N(1,4)\) and \(Y\sim N(-2,9)\) be independent. Find the distribution of \(S=2X-Y\), using the convention that the second argument of \(N\) is the variance.
A normal variable \(N(\mu,\sigma^2)\) has characteristic function \(\exp(i\mu t-\sigma^2t^2/2)\). Therefore it is symmetric \(2\)-stable with exponent coefficient \(c=\sigma^2/2\). For \(X\), \(c_X=4/2=2\); for \(Y\), \(c_Y=9/2=4.5\).
The location of \(S\) is \(2(1)-1(-2)=4\). Its exponent coefficient is $$ c_S=2|2|^2+4.5|-1|^2=8+4.5=12.5. $$ For a \(2\)-stable normal variable, the variance is \(2c_S=25\). Hence \(S\sim N(4,25)\). This also agrees with the independent-normal variance calculation \(2^2(4)+(-1)^2(9)=25\).
Worked Example: A Weighted Sum of Cauchy Variables
Let \(X\) and \(Y\) be independent Cauchy variables with locations \(0\) and \(3\), and conventional scales \(2\) and \(1/2\), respectively. Find the distribution of \(2X-3Y\).
A Cauchy variable with location \(\mu\) and scale \(\gamma>0\) has characteristic function \(\exp(i\mu t-\gamma|t|)\). Thus \(\alpha=1\), and its conventional scale \(\gamma\) equals the exponent coefficient \(c\). The resulting location is \(2(0)-3(3)=-9\), and the coefficient is $$ c_S=2|2|^1+0.5|-3|^1=4+1.5=5.5. $$ Therefore \(2X-3Y\) is Cauchy with location \(-9\) and scale \(5.5\). The negative weight changes the location contribution’s sign but contributes a positive amount to the scale.
Worked Example: Combining Three Stable Variables with Index \(3/2\)
Suppose \(X_1,X_2,X_3\) are independent symmetric \(3/2\)-stable variables with locations \(1,-1,2\) and exponent coefficients \(0.4,0.8,1.2\), respectively. Find the characteristic function of \(S=2X_1-X_2+0.5X_3\).
The location is $$ 2(1)-1(-1)+0.5(2)=4. $$ The combined exponent coefficient is $$ 0.4|2|^{3/2}+0.8|-1|^{3/2}+1.2|0.5|^{3/2} =0.8\sqrt{2}+0.8+0.3\sqrt{2} =0.8+1.1\sqrt{2}. $$ This is approximately \(2.356\), which is positive as required. Consequently, $$ \varphi_S(t)=\exp\!\left(4it-(0.8+1.1\sqrt{2})|t|^{3/2}\right). $$ The sum remains symmetric \(3/2\)-stable; its conventional scale is \((0.8+1.1\sqrt{2})^{2/3}\).
What the Special Cases Reveal
The stability index determines the power applied to the weights and the growth rate of sums. For \(\alpha=2\), the formula combines normal variances: since variance is \(2c\), the resulting variance is \(\sum_j 2c_j a_j^2\). For \(\alpha=1\), it combines Cauchy scales by \(\sum_j c_j|a_j|\). For other indices in \(0<\alpha<2\), the same power-law formula applies to the exponent coefficients.
For centered independent copies, division by \(n^{1/\alpha}\) leaves the same distribution: \(n^{-1/\alpha}\sum_{j=1}^nX_j\overset{d}{=}X\). This differs from the familiar square-root scaling of finite-variance normal sums except when \(\alpha=2\). In particular, Cauchy variables have \(\alpha=1\), so their sum scales linearly in \(n\), not by \(\sqrt{n}\).
Common Mistake: Raising the Coefficient to the Stability Index
A frequent parameterization error is to define the characteristic function as \(\exp(-c|t|^\alpha)\), then claim that the combined coefficient is \(\sum_j c_j^\alpha|a_j|^\alpha\). That is incorrect: in this parameterization, the coefficient combines as \(\sum_j c_j|a_j|^\alpha\). The power \(\alpha\) applies to the absolute weight, not to \(c_j\).
For example, if \(\alpha=2\), weights are both one, and \(c_1=2,c_2=3\), the combined coefficient is \(2+3=5\), not \(2^2+3^2=13\). If instead the characteristic function is parameterized as \(\exp(-(s|t|)^\alpha)\), then \(c=s^\alpha\), and it is the \(\alpha\)th powers of the conventional scales that combine: \(s_{\mathrm{combined}}^\alpha=\sum_j s_j^\alpha|a_j|^\alpha\).
Check Your Understanding
Use the characteristic-function parameterization in this tutorial.
- For independent centered symmetric \(\alpha\)-stable variables with the same coefficient \(c\), what coefficient has their sum when there are \(n\) variables?
- If \(X\) has characteristic function \(\exp(i\mu t-c|t|^\alpha)\), state the distributional scaling identity for a sum of \(n\) independent copies.
- Two independent Cauchy variables have scales \(2\) and \(5\). What is the scale of their sum?
- For a normal variable \(N(\mu,\sigma^2)\), what is its exponent coefficient \(c\) when \(\alpha=2\)?
- Explain why a negative weight changes a location contribution’s sign but not its contribution to the exponent coefficient.