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Characteristic Function — Definition, Formula & Examples

A characteristic function is a complex-valued function that completely describes the probability distribution of a random variable. It is essentially the expected value of eitXe^{itX}, where XX is the random variable and tt is a real number.

For a random variable XX, the characteristic function φX(t)\varphi_X(t) is defined as φX(t)=E[eitX]\varphi_X(t) = E[e^{itX}] for all tRt \in \mathbb{R}, where ii is the imaginary unit. Unlike moment-generating functions, the characteristic function always exists and uniquely determines the distribution of XX.

Key Formula

φX(t)=E[eitX]=eitxfX(x)dx\varphi_X(t) = E\bigl[e^{itX}\bigr] = \int_{-\infty}^{\infty} e^{itx}\, f_X(x)\, dx
Where:
  • φX(t)\varphi_X(t) = Characteristic function of random variable X evaluated at real number t
  • ii = Imaginary unit, where i² = −1
  • tt = Real-valued parameter
  • fX(x)f_X(x) = Probability density function of X (continuous case)

How It Works

You evaluate the characteristic function by computing the expectation of eitXe^{itX} over the distribution of XX. For discrete random variables, this becomes a weighted sum; for continuous ones, an integral involving the density function. Two distributions are identical if and only if their characteristic functions are equal. This makes the characteristic function a powerful tool for proving convergence results, such as the Central Limit Theorem. Derivatives of φX(t)\varphi_X(t) evaluated at t=0t=0 yield the moments of XX: specifically, E[Xn]=1inφX(n)(0)E[X^n] = \frac{1}{i^n}\varphi_X^{(n)}(0).

Worked Example

Problem: Find the characteristic function of a Bernoulli random variable X with success probability p = 0.5.
Set up the expectation: X takes value 1 with probability 0.5 and value 0 with probability 0.5. Apply the definition directly.
φX(t)=E[eitX]=eit0P(X=0)+eit1P(X=1)\varphi_X(t) = E[e^{itX}] = e^{it \cdot 0}\cdot P(X=0) + e^{it \cdot 1}\cdot P(X=1)
Substitute the values: Since e⁰ = 1, and each outcome has probability 0.5:
φX(t)=10.5+eit0.5=0.5+0.5eit\varphi_X(t) = 1 \cdot 0.5 + e^{it} \cdot 0.5 = 0.5 + 0.5\,e^{it}
Simplify: Factor out 0.5 for a cleaner form.
φX(t)=0.5(1+eit)\varphi_X(t) = 0.5\bigl(1 + e^{it}\bigr)
Answer: The characteristic function is φX(t)=0.5(1+eit)\varphi_X(t) = 0.5(1 + e^{it}).

Why It Matters

Characteristic functions are essential in proving the Central Limit Theorem, one of the most important results in probability and statistics. In mathematical finance and signal processing, they provide a way to work with distributions that may not have well-defined moment-generating functions, such as the Cauchy distribution.

Common Mistakes

Mistake: Confusing the characteristic function with the moment-generating function (MGF).
Correction: The MGF uses E[etX]E[e^{tX}] with a real parameter and may not exist for all distributions. The characteristic function uses E[eitX]E[e^{itX}] with a complex exponential and always exists.