# Characteristic function and Fourier transform for a discrete random variable!

Let $$\phi_{x}(t)= E [ e^{itx}]$$ be the characteristic function

If X is a continuous random variable, then:

$$\phi_{x}(t)= E [ e^{itx}] = \int e^{itx} f(x)dx$$ (being $$f(x)$$ the probability density function of x)

If X is a discrete random variable, then:

$$\phi_{x}(t)= E [ e^{itx}] = \sum_{k} e^{itx_{k}} p(x_{k})$$ (being $$p(x)$$ the probability mass function of x)

To be general Fourier transform can be define as (Ref: https://www.johndcook.com/blog/fourier-theorems/):

$$F(s)=\frac{1}{A}\int_{-\infty}^{\infty} e^{iBst}f(t)dt$$

The choices that are found in practice are:

$$A=\sqrt{2\pi}, B=\pm 1;$$ $$A=1, B=\pm 2\pi;$$ $$A=1, B=\pm 1$$

Choosing A=1 and B=1 to relate it to the characteristic function, the inverse Fourier transform is:

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-ist}F(s)ds$$

In continuous case:

$$\phi_{x}(t) = \int e^{itx} f(x)dx \rightarrow f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-ixt}\phi_{x}(t)dt$$

That is the utility of the characteristic function, it allows me to know the probability function

But in the discrete case I get problems:

If discrete Fourier transform (DFT) is :

$$F(s)=\sum_{n=0}^{N-1} e^{-2\pi sn/N}f(n)$$

The inverse DFT is:

$$f(t)=\frac{1}{N}\sum_{n=0}^{N-1} e^{2\pi sn/N}F(s)$$

So, Would the "inverse" characteristic function be

$$\phi_{x}(t) = \sum_{k} e^{itx_{k}} p(x_{k}) \rightarrow P(x) = \frac{1}{2\pi N}\sum_{k} e^{-ixt_{k}}\phi_{x}(t_{n})$$ ?

So if FT can have different definitions for the continuous case. What happen with the discrete case? How many definitions are?

For the discrete case, you need to look at DTFT, not DFT. N-point DFT assumes that the underlying function is periodic, which is not the case for probability mass functions. You can then apply similar logic for inverse transform:

$$\phi_{x}(t)= E [ e^{itx}] = \sum_{k} e^{itx_{k}} p(x_{k})\rightarrow p(x)=\frac{1}{2\pi}\int_0^{2\pi}e^{-ixt}\phi_x(t)dt$$

For example, characteristic function for Bernouilli RV is $$\phi_x(t)=1-p+pe^{it}$$. Applying the formula yields the following:

\begin{align}p(x)&={1\over 2\pi}\int_0^{2\pi}e^{-ixt} (1-p+pe^{it})dt\\&=(1-p){1\over 2\pi}\underbrace{\int_0^{2\pi} e^{-itx}dt}_{2\pi\delta_x} + p\frac{1}{2\pi}\underbrace{\int_0^{2\pi}e^{-it(x-1)}dt}_{2\pi\delta_{x-1}}\\&=(1-p)\delta_x+p\delta_{x-1}\end{align}

where $$\delta_x$$ is the Kronecker delta function, i.e. it's $$1$$ if $$x=0$$, and $$0$$ otherwise.

This is from Fourier transform perspective. Probability theory has its own principled way of calculating inverses.

• +1 and can you provide more details may be with an example? Jul 7, 2020 at 9:42
• @HaitaoDu great advice. I've added an example for Bernoulli RV. Jul 7, 2020 at 10:46
• @gunes Why is the upper limit of the integral 2$\pi$? Also I know I can use Stieltjes integral as $F(b)-F(a)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{ e^{-itb}- e^{-ita}}{-it}f(t)dt$ But if $F_{x}(x)=P(X\leq x) = \sum_{x_{i} \leq x} P(X=x_{i}) = \sum_{x_{i} \leq x} p(x_{i})$ How do you integrate it and put inside the integral? Jul 7, 2020 at 16:52
• @gunus And also I think $\delta_{x}$ is Dirac delta no Kronecker delta Jul 7, 2020 at 18:35
• for a probability theory treatment, this source seems good: statweb.stanford.edu/~adembo/stat-310b/lnotes.pdf for fourier transform review, I strongly recommend Oppenheim Jul 7, 2020 at 19:36