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Which distributions are their own Fourier transform besides the normal distribution and the generalized arcsine distribution?

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Suppose the Fourier transform of $x(t)$ is $X(f)$ where $$X(f) = \int_{-\infty}^{\infty} x(t) \exp(-i2\pi f t) \mathrm dt$$ where $i = \sqrt{-1}$. The inverse transform is $$x(t) = \int_{-\infty}^{\infty} X(f) \exp(i2\pi f t) \mathrm df$$

Some properties of the Fourier transform are as follows:

  • The Fourier transform of $X(t)$ is $x(-f)$

  • If $x(t)$ is a real-valued even function of $t$, then $X(f)$ is a real-valued even function of $f$.

Thus, if $x(t)$ is a real-valued even function of $t$, then the Fourier transform of the real-valued even function $X(t)$ is $x(f)$

Now suppose that $x(t)$ is an even probability density function (so that $x(t) \geq 0$ for all $t$) with the additional property that $x(0) = 1$. Suppose also that its Fourier transform $X(f)$ has the property that $X(f) \geq 0$ for all $f$. Then, since $$x(0) = 1 = \int_{-\infty}^{\infty} X(f) \mathrm df$$ $X(f)$ is a even non-negative real-valued function of $f$ with area $1$, that is, $X(f)$ is also a probability density function with the property that $X(0) = 1$. One example of such a pair of functions is the normal distribution cited by OP Neil G $$x_1(t) = \exp(-\pi t^2), ~~ X_1(f) = \exp(-\pi f^2)$$ and another example is $$x_2(t) = (1 - |t|)\mathbf 1_{[-1,1]}, ~~ X_2(f) = \operatorname{sinc}^2(f) = \begin{cases}\displaystyle \left(\frac{\sin(\pi f)}{\pi f}\right)^2,&f\neq 0,\\ 1,&f=0.\end{cases}$$

Now note that $\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)$ is a mixture density whose Fourier transform is $\frac{1}{2} X_2(f) + \frac{1}{2}x_2(f)$ which is the same mixture density.

Thus, if $x(t)$ is a density function whose Fourier transform $X(f)$ is a density function, then the mixture density function $\frac{1}{2} x(t) + \frac{1}{2}X(t)$ is its own Fourier transform.

Finally, given two densities that are their own Fourier transforms, e.g. $x_1(t)$ and $\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)$, any mixture density $$\alpha x_1(t) + (1-\alpha)\left[\frac{1}{2} x_2(t) + \frac{1}{2}X_2(t)\right]$$ where $\alpha \in [0,1]$ is a density function that is its own Fourier transform.

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  • 7
    $\begingroup$ (+1) This is quite clever. It should be noted that to guarantee a valid transform pair, we need an integrability condition on $X(f)$. Namely, $\int_{-\infty}^\infty X(f) \,\mathrm{d}f < \infty$ will guarantee that the stated inversion will recover the appropriate density. In a sense you employ such a condition later on. (I've already assumed the nonnegativity constraint on $X(f)$ has been imposed, so it needs no modulus.) $\endgroup$ – cardinal Jan 16 '12 at 23:37

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