29
$\begingroup$

The characteristic function of a random variable $X$ is given by $$\Phi_X(\omega) = \mathbb{E}e^{j\omega X}=\int_{-\infty}^\infty e^{j\omega x}f_X(x) dx.$$ One can easily capture the similarity between this integral and the Fourier transform.

For a standard normal random variable, the characteristic function can be found as follows: $$\Phi_X(\omega)= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}e^{j\omega x} dx = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(x^2-2j\omega x)}{2}\right)dx $$

I know that the answer must be $\Phi_X(\omega) = \exp(-\omega^2/2)$, but can you explain how to evaluate the integral with a complex number in the exponent?

$\endgroup$
2
  • 2
    $\begingroup$ You can take the derivative with respect to $\omega$, and integrate by parts to get a differential equation. $\endgroup$ Nov 27, 2011 at 13:32
  • 1
    $\begingroup$ Is it a good way to use the fact that "If $f$ and $g$ are real functions then $\int (f + i g) = \int f + i \int g$" math.stackexchange.com/q/85941/1281 ? $\endgroup$
    – Tim
    Nov 27, 2011 at 13:40

2 Answers 2

40
$\begingroup$

I will give two answers:

Do it without complex numbers, notice that

$$ \begin{eqnarray} \mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x \\ &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{-j \omega x} \mathrm{d} x \\ &=& 2 \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x \end{eqnarray} $$ Now, compute $\mathcal{F}^\prime(\omega)$, and integrate by parts: $$\begin{eqnarray} \mathcal{F}^\prime(\omega) &=& -\frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} x \sin(\omega x) \mathrm{d} x = \frac{2}{\sqrt{2\pi}} \int_0^\infty \sin(\omega x) \mathrm{d} \left( \mathrm{e}^{-\frac{x^2}{2}} \right) \\ &=& \frac{2}{\sqrt{2\pi}} \left. \mathrm{e}^{-\frac{x^2}{2}} \sin(\omega x) \right|_0^\infty - \frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} \omega \cos(\omega x) \mathrm{d} x \\ &=& - \omega \mathcal{F}(\omega) \end{eqnarray} $$ The solution to so obtained ODE, $\mathcal{F}^\prime(\omega) = - \omega \mathcal{F}(\omega)$ is $\mathcal{F}(\omega) = c \exp\left( - \frac{\omega^2}{2} \right)$, and the integration constant is seen to be one from normalization requirement $\mathcal{F}(0)=1$ of the Gaussian probability density.

Complex integration: As you have started, complete the square: $$ \left( -\frac{x^2}{2} + j \omega x \right) = \left( -\frac{x^2}{2} + j \omega x + \frac{\omega^2}{2} \right) - \frac{\omega^2}{2} = -\frac{1}{2} \left( x - j \omega \right)^2 - \frac{\omega^2}{2} $$ We then have: $$ \mathcal{F}(\omega) = \mathrm{e}^{-\frac{\omega^2}{2}} \cdot \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x $$ The integral $I = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x$ is indeed $1$. To see this, consider $$ \begin{eqnarray} I_L &=& \int_{-L}^L \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x = \int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\ &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\ &=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \mathcal{I}_L \end{eqnarray} $$ Here we denoted $\mathcal{I}_L = \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z$. Notice that $\lim\limits_{L \to \infty} \mathcal{I}_L = 1$.

Consider a complex contour $\mathcal{C}$, $ -L \to L \to L - j \omega \to -L - j \omega \to -L$: $$ \begin{eqnarray} I_L - \mathcal{I}_L &=&\left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) \\ &=& -\int_\mathcal{C} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \end{eqnarray} $$ The integral over $\mathcal{C}$ is zero, since the integrand is holomorphic. Therefore: $$ I-1 = \lim_{L \to \infty} (I_L-\mathcal{I}_L) = \lim_{L \to \infty} \left( - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \right) $$ And the limit above is easily seen to vanish. Indeed: $$ \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(-L - j \omega t)^2}{2}} \right| = \lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(L^2 + \omega^2 t^2)}{2}} \right| =0. $$

$\endgroup$
6
  • $\begingroup$ +1. Thanks! Why instead of directly computing $\int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x $ (btw, I don't know how to directly integrate it), you solved it indirectly by constructing a ODE? $\endgroup$
    – Tim
    Nov 27, 2011 at 14:14
  • $\begingroup$ @Tim It seemed like the simplest approach. The only two other approaches I am aware of are 1) to write $\cos(\omega x)$ into Taylor series and integrate term-wise, 2) use Mellin-convolution theorem. Neither is simpler. $\endgroup$
    – Sasha
    Nov 27, 2011 at 14:24
  • $\begingroup$ Sasha: the $+1$ term which appears at the third line in the computation of $I_L$ should read $1+o(1)$, I believe. $\endgroup$
    – Did
    Dec 10, 2011 at 21:35
  • $\begingroup$ @DidierPiau Welcome back and thank you for the feedback. I have edited the post to make the equation precise. $\endgroup$
    – Sasha
    Dec 10, 2011 at 22:17
  • $\begingroup$ how in the first method, is moving d/dw into the integral justified? $\endgroup$
    – DinkyDoe
    Nov 13, 2012 at 0:19
3
$\begingroup$

There is an alternative approach which is based on the relation between moment-generating function and characteristic function. More accurately, assume that $X$ is a random variable such that there exists $\delta\in(0,\infty]$ for which

$$M_X(t)\equiv Ee^{tX}<\infty\ , \ \forall t\in(-\delta,\delta)\,.$$

Then, it is known (e.g., see my answer to How to prove: Moment Generating Function Uniqueness Theorem) that

$$\phi_X(z)\equiv Ee^{zX}\ \ , \ \ \forall z\in \Omega\equiv\left\{x+iy\ ;\ x\in(-\delta,\delta)\right\}$$

is an analytic continuation of $M_X(\cdot)$ to the domain $\Omega$. This implies that for every $t\in\mathbb{R}$ $$M_X(t)=\phi_X(t)$$ and $$\psi_X(t)\equiv Ee^{itX}=\phi_X(it)\,.$$

Now, it is straightforward that for every $z\in\Omega$ it is possible to derive an expresssion for $\phi_X(iz)$ by computing a formula for $\phi_X(z)$ and then insert to this formula $iz$ instead of $z$. Thus, as a special case, in order to derive an expression for $\psi_X(t)$ at some point $t\in(-\delta,\delta)$, it is possible to compute $M_X(t)$ and then to insert $it$ instead of $t$ into the resulting expression.

In particular, if $X\sim N(0,1)$, then it is straightforward to show that for every $t\in\mathbb{R}$ $$M_X(t)=e^{\frac{t^2}{2}}<\infty$$ (that is $\delta=\infty$) and hence, replacing $t$ by $it$ implies that

$$\psi_X(t)= e^{-\frac{t^2}{2}}\ , \ \forall t\in\mathbb{R}\,.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.