3.6. Characteristic functions

Characteristic function of a random variable \(X\) is defined as

\[\varphi_X(t) := \mathbb{E} \left[ e^{i t X} \right] \ .\]

Characteristic function of a continuous random variable with proabibility density function \(f(x)\) thus reads

\[\varphi_X(t) := \mathbb{E} \left[ e^{i t X} \right] = \int_{x \in D_x} f(x) e^{i t x} \, dx \ ,\]

i.e. its the Fourier transform of its pdf.

Example 3.1 (Characteristic function of a multi-dimensional variable)

\[Z(\mathbf{Y})\]

\[\varphi_{Z(\mathbf{Y})} := \mathbb{E}\left[ e^{i t Z(\mathbf{Y})} \right] = \int_{\mathbf{y}} e^{i t Z(\mathbf{y})} f(\mathbf{y}) \, d \mathbf{y}\]

Example 3.2 (Characteristic function of a linear combination of independent variables)

\[Z(\mathbf{Y}) = a_1 Y_1 + \dots a_n Y_n \ ,\]

with

\[f(\mathbf{y}) = f(y_1, \dots, y_n) = f_1(y_1) \dots f_n(y_n) \ .\]

\[\begin{split}\begin{aligned} \varphi_{Z(\mathbf{Y})} & := \mathbb{E}\left[ e^{i t Z(\mathbf{Y})} \right] = \\ & = \int_{\mathbf{y}} e^{i t \left( \sum_k a_k y_k \right)} f(\mathbf{y}) \, d \mathbf{y} = \\ & = \int_{y_1} e^{i t a_1 y_1} f_1(y_1) \, d y_1 \, \dots \int_{y_n} e^{i t a_n y_n} f_n(y_n) \, d y_n = \\ & = \varphi_{Y_1}(a_1 t) \dots \varphi_{Y_n}(a_n t) \ . \end{aligned}\end{split}\]

Example 3.3 (Taylor expansion of characteristic function)

For “small” values of \(t\), an approximation of the characteristic function is provided by Taylor expansion around \(t=0\),

\[\begin{split}\begin{aligned} \int e^{ i y t } f(y) \, dy & = \int \left[ 1 + i y t - \dfrac{1}{2} (yt)^2 + o(t^2) \right] f(y) dy = && (1) \\ & = 1 + i \mu t - \dfrac{1}{2} t^2 \left( \sigma^2 + \mu^2 \right) + o(t^2) \end{aligned}\end{split}\]

as \((1)\) \(\sigma^2 = \mathbb{E}[(y - \mu)^2] = \mathbb{E}[y^2] - \mu^2\)

Example 3.4 (Characteristic function of a normal distribution \(\mathscr{N}(0,1)\))

\[f(x) = \dfrac{1}{\sqrt{2 \pi}} \exp\left( -\dfrac{x^2}{2} \right)\]

\[\begin{split}\begin{aligned} \int_{x=-\infty}^{+\infty} e^{ i x t } f(x) \, dx & = \dfrac{1}{\sqrt{2 \pi}} \int_{x=-\infty}^{+\infty} e^{ i x t - \frac{x^2}{2} } \, dx = && (1) \\ & = \dfrac{1}{\sqrt{2 \pi}} \int_{x=-\infty}^{+\infty} e^{ - \frac{(x-it)^2}{2} } \, dx \, e^{-\frac{t^2}{2}} = && (2) \\ & = \dfrac{1}{\sqrt{2 \pi}} \, \sqrt{2 \pi} \, e^{-\frac{t^2}{2}} = e^{-\frac{t^2}{2}} \\ \end{aligned}\end{split}\]

having \((1)\) completed the square \((x - it)^2 = x^2 - i 2 x t - t^2\), and evaluated the integral todo (it’s similar to the standard result \(\int_{-\infty}^{+\infty} e^{x^2} \, dx = \sqrt{2 \pi}\), but with complex variable. Link to math material, complex calculus).