# Can you Fourier transform probabilities?

If I have a rect function , and I convolute it with it's self, I get a triangle function. If I convolute with a rect function again, I get a bell-curve. I can continue, so long as I know how to convolute two functions.

Alternatively I can multiply the Fourier transform of the functions together.

When I learnt this a few years back on my Physics course, I began to wonder if this was possible, or had applications in probability, where the rect function can act as a function for uniform deviates. Specifically I thought of dice rolls, which can be expressed as a uniform deviate convoluted with a comb of Dirac function. If I wanted to know the distribution of say 3 six sided dice (or some arbitrarily complicated roll of dice), I could just multiply the appropriate Fourier transform, then transform it back.

The way I was taught about Fourier transforms (by my physics lecturer) though was that Frequency maps to Time and Distance maps to Angles when you transform.

What does the Fourier transform of probability map to? How can it help my endeavour to work out dice probability distributions?

• – Qiaochu Yuan May 9 '12 at 21:39
• @QiaochuYuan the article lead me to duality which is more baffling than my original question... – Pureferret May 9 '12 at 21:52
• The answer to a question is often more complicated than the question. You don't need Fourier transforms to work out dice distributions though because those are discrete; it suffices to use generating functions instead (see math.upenn.edu/~wilf/DownldGF.html for example). – Qiaochu Yuan May 9 '12 at 22:36
• Generating functions and characteristic functions are basically the same thing. – Alex R. May 10 '12 at 1:45
• @sam could you point me towards a book? – yiyi May 10 '12 at 4:10

A (real-valued) random variable is a measurable function $X:\Omega\to\mathbb R$, where $\Omega$ is a probability space. Let $\mu$ be the pushforward of the probability measure under $X$; equivalently, $\mu$ is the measure on $\mathbb R$ such that $\mu(-\infty,b)=P(X<b)$ for all $b\in\mathbb R$. The characteristic function of $X$ is the same as the Fourier transform of measure $\mu$. The interpretation of the axis $\mathbb R$ on which $\widehat \mu$ lives depends on the axis on which $\mu$ lives.
For example, if $X$ is the waiting time, then $\mu$ lives on the time axis, and its transform lives on the frequency axis. If waiting period for a bus tends to be a multiple of $20$ minutes, you will see a corresponding spike on the characteristic function of this random variable.
One problem with interpretation of $\widehat\mu$ in your dice example is lack of physical interpretation of the original data. What are the units for the score in the game of dice? How would you call the axis on which they are plotted? But even when the nature of $X$ is clear (height of a person, say), the Fourier transform isn't really physical. It gives a convenient way to study the distribution of sums of independent variables: $E(e^{i(X+Y)})=E(e^{iX}e^{iY})=E(e^{iX})E(e^{iY})$, and that's good enough.