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A distribution has the characteristic function

$$\phi(t) = (1-t^2/2)\exp(-t^2/4),\ -\infty \lt t \lt \infty$$

Show that the distribution is absolutely continuous and write the density function of the distribution.

Attempt:

$$\int_{-\infty}^{\infty}|(1-t^2/2)\exp(-t^2/4)|dt =(-2/t)(1-t^2/2)\exp(-t^2/4)-2\exp(-t^2/4)|_{-\infty}^{0}$$

Similar result for $[0,\infty]$ since $t$ is squared.

I'm not quite sure I did the integration right, but if I can show that the absolute value of $\phi(t)$ is less than $\infty$, then the function is absolutely continuous.

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  • $\begingroup$ Use that $p(t)/\exp(t^2)\to 0$ for $t\to \pm\infty$ for any polynomial $p$. This ensures that both tails are integrable. $\endgroup$ Commented Jun 24, 2014 at 9:11

1 Answer 1

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Density functions are found with the inverse Fourier transform. The density function of the distribution, if such a density exists, will be given by

$$f(t) = \frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}\phi(x) dx = \frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}\left((1-x^2/2)e^{-x^2/4}\right) dx .$$

This integral can be split into two, each of which has an integrand of the form

$$\exp(-Q_t(x))x^{2k}$$

where $Q_t$ is a quadratic form with negative leading term and $k$ is a non-negative integer. This makes each integrand a Schwartz (rapidly decreasing) function, assuring its integrability for any $t$. The integrability proves it is continuous; the rapid decrease proves it is absolutely continuous. The integrals are readily performed by completing the square in the exponential, reducing them to multiples of even moments of the Gaussian distribution. The result is

$$f(t) = \frac{2}{\sqrt{\pi}} t^2 e^{-t^2}.$$

The continuity of $f$ confirms the earlier conclusion of absolute continuity of the distribution.

Plot of f

The square of this (symmetric) variable has a Gamma$(3/2, 1)$ distribution.


Alternatively, one might recognize that

$$\phi(t) = -2\left(-\frac{1}{2} + \frac{t^2}{4}\right)e^{-t^2/4} = (-i)^2\frac{d^2}{dt^2}2e^{-t^2/4}$$

is proportional to the second derivative of the Gaussian $e^{-t^2/4}$, implying (since the operator $-id/dt$ on characteristic functions is equivalent to multiplication of distribution functions by the variable) that the density $f(x)$ exists and is proportional to $x^2$ times the density whose c.f. is $2e^{-t^2/4}$. That is immediately recognizable as a Gaussian (Normal) distribution with density proportional to $e^{-x^2}$. At this point all one has to do is work out the normalizing constant of $2/\sqrt{\pi}$ via integration or by computing the variance of a Normal distribution with standard deviation $\sqrt{1/2}$.

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