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I have to compute the convolution of $ f(t) = \frac{1}{\pi}\frac{1}{t^2 + 1} $ with itself 4 times, i.e. $$ f \star f \star f \star f $$

I slightly doubt that doing it in steps, i.e. taking $f \star f$ first and then taking the convolution of the result with f and then do the same once more, is the correct way of computing it..

Could anyone help me out with a hint of what method I should use?

Thank you!

user42751
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1 Answers1

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You can use the fact that a convolution in the time domain is equivalent to a multiplication in the Fourier domain.

Thus if you carry out a Fourier transform of $f(t)$ as defined below:-
$$F(\omega)=FT(f(t))=\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$ then $$F'(\omega)=FT(f(t)\ast f(t) )=F(\omega)F(\omega)$$ Therfore, to calculate the result of $f(t)\ast f(t)\ast f(t)\ast f(t)$ you would calculate the Fourier transform of $f(t)=\frac{1}{\pi}\frac{1}{1+t^2}$ similar to what is shown here for example, then multiply this result with itself three times, and finally apply the inverse Fourier transform.

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Alijah Ahmed
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  • The particularly nice property in this particular case is that the Fourier transform is of the form $a\exp(-b|\omega|)$ whose fourth power is $a^4\exp(-4b|\omega|)$ and this makes the inverse Fourier transform quite easy to apply. From a probability perspective, the convolution gives the density of the sum of $4$ independent standard Cauchy random variables (more generally $n$), and it is well-known that the density of such a sum is the density of $4X$ (more generally, $nX$) where $X$ is a standard Cauchy random variable. – Dilip Sarwate Jun 01 '14 at 13:29
  • @Dilip Sarwate: The nice form of the Fourier transform and the inverse of the products are indeed specific to the case in question. Your observations on the probability perspective are interesting, especially the result that the sum of $n$ independent Cauchy random variables has a Cauchy density of random variable $nX$. I find this quite a remarkable result, as it means that if we observe a process with Cauchy distributed error, increasing the number of measurements will not yield a more accurate result. – Alijah Ahmed Jun 01 '14 at 14:59
  • @Dilip Sarwate: Regarding the extra number of measurements for a system with Cauchy distributed noise not resulting in a more accurate estimate, I should have qualified the statement on the basis that the estimator uses an averaging process. – Alijah Ahmed Jun 01 '14 at 15:51