Characteristic function with an indicator function

We know that $$\varphi(t) = \cos{t}$$ is a characteristic function of Rademacher random variable. Is the function $$\phi(t) = \cos{t} \cdot 1_{|t| \le \pi /2}$$ still a characteristic function of some random variable?

The only thing we have to check is if $$\phi$$ is positively-defined. I tried looking for some examples contradicting it, but with no success, so I suspect it is a characteristic function. When I write the condition of positive-definiteness of $$\phi$$, it looks very similar to the condition for $$\varphi$$, but with more zeros, but I cannot deduce anything out of it.

• Can you find the inverse Fourier transform of $\phi(t)$ and check if it is a probability mass distribution? Commented Nov 16, 2022 at 16:24
• @OliverDíaz but it is continuous. Commented Nov 18, 2022 at 17:52
• @Barbara, I jumped the gun to fast, Meant to write positive definite ( see Bochner-Herglotz theorem). Commented Nov 18, 2022 at 20:39

If your given function $$\phi (t) = {\Bbb E}(e^{ i t X})$$ is the characteristic function of a random variable $$X$$ whose probability density is $$f(x)dx$$, then $$\phi(t)$$ is the Fourier transform of $$f(x)$$ and the extremely useful Fourier inversion formula allows us to recover the function $$f(x)$$ by the general identity $$f(x) = \frac{1}{2\pi} \int_{t=-\infty}^{t=\infty} e^{- i t x} \phi(t) dt$$ (See footnote regarding sign conventions.)

In your example this simplifies to $$f(x) =\frac{1}{2\pi} \int_{t=-\pi/2}^{t=\pi/2} \ (\cos t ) \ e^{-i t x} dt$$ which turns out to be $$f(x)=\frac{ \cos(\pi x/2)}{ \pi ( 1- x^2)}$$

Plotting the graph of $$f(x)$$, we see that it fails to be always non-negative; thus $$f(x)$$ is NOT a probability density.

P.S. Wikipedia and many other sources use a slightly different convention for the Fourier transform (engineer's version) that shuffles the constant $$2\pi$$ into other places. The version used above is what is used in statistics and in theoretical physics. See e.g equations 3 and 4 in reference below.

Physicist's Fourier formulas

P.P.S. In response to a comment that raises a subtle issue I would add one clarification.

There is no a priori assumption that $$X$$ has a density. Since $$\phi$$ has compact support there is no doubt that it is the Fourier transform of a smooth function $$f(x)$$. And the uniqueness theorem for Fourier transforms of generalized functions implies that no finite measure $$X$$ could exist that has the same characteristic function as this $$f(x)$$. Thus ultimately the pivotal question is whether or not this function $$f(x)$$ is non-nonnegative.

The other characteristic function mentioned, $$\cos t$$ did not decay at $$\infty$$ so by the contrapositive of the Riemann-Lebesgue Lemma it could not be the characteristic function of any absolutely integrable function. Instead it is of course the Fourier transform of a pair of Dirac delta functions (point masses) which are of course generalized functions.

Suppose $$\phi$$ is the Fourier transform of a finite measure $$\mu$$ on $$(\mathbb{R},\mathscr{B}(\mathbb{R}))$$. By Lévy inversion \begin{aligned}\frac{\mu(3)+\mu(5)}{2}+\mu((3,5))&=\int_{[-\pi/2,\pi/2]}\frac{e^{i5\xi}-e^{i3\xi}}{i\xi}\cos(\xi)d\xi\approx -0.18\end{aligned} But $$\mu(B)\geq 0$$ for all Borel $$B$$, so $$\phi$$ is not the CF of a finite measure $$\mu$$.