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I have a mathematical problem that leads me to a particular necessity. I need to calculate the convolution of a function for itself for a certain amount of times.

So consider a generic function $f : \mathbb{R} \mapsto \mathbb{R}$ and consider these hypothesis:

  • $f$ is continuos in $\mathbb{R}$.
  • $f$ is bound, so: $\exists A \in \mathbb{R} : |f(x)| \leq A, \forall x \in \mathbb{R}$.
  • $f$ is integral-defined, so its area is a real number: $\exists \int_a^bf(x)\mathrm{d}x < \infty, \forall a,b \in \mathbb{R}$. Which implies that such a function at ifinite tends to zero.

Probability mass functions: Such functions fit the constraints given before. So it might get easier for you to consider $f$ also like the pmf of some continuos r.v.

Consider the convolution operation: $a(x) \ast b(x) = c(x)$. I name the variable always $x$.

Consider now the following function:

$$ F^{(n)}(x) = f(x) \ast f(x) \ast \dots \ast f(x), \text{for n times} $$

I want to evaluate $F^{(\infty)}(x)$. And I would like to know whether there is a generic final result given a function like $f$.

My trials

I tried a little in Mathematica using the Gaussian distribution. What happens is that, as $n$ increases, the bell stretches and its peak always gets lower and lower until the function almost lies all over the x axis. It seems like $F^{(\infty)}(x)$ tends to $y=0$ function...

Trials in Mathematica

As $n$ increases, the curves gets lower and lower.

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    $\begingroup$ Maybe work out this result: The convolution of a Gaussian density with mean zero and variance $a$ times a Gaussian density with mean zero and variance $b$ is a Gaussian density with mean zero and variance ?? $\endgroup$ – GEdgar Apr 12 '13 at 19:04
  • $\begingroup$ Variance increases everytime... and it seems not reaching a stable value... $\endgroup$ – Andry Apr 12 '13 at 19:06
  • $\begingroup$ Your three conditions don't guarantee that the convolutions exist. For example, let $f \equiv 1$. In your third condition, do you mean $\int_{-\infty}^\infty |f(x)|\,dx < \infty$? $\endgroup$ – Stefan Smith Apr 13 '13 at 0:46
  • $\begingroup$ Yes! I thought it was the same right? $\endgroup$ – Andry Apr 13 '13 at 7:54
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    $\begingroup$ @Audry: Maybe my guess was wrong. I tried $f(x) = 100\exp(-|x|/10)$ and took a couple of convolutions using Maple and $F^{(n)}(0)$ seems to get bigger and bigger. Note that this $f$ decays more slowly than a Gaussian. $\endgroup$ – Stefan Smith Apr 13 '13 at 17:50
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While I don't know about the general case of your $F^{(\infty)}(x)$ for any such generic function $f$, it seems that in some cases you can find explicit closed forms for $n$-fold convolutions for some functions $f$. Maybe then you can study the limit as $n\rightarrow\infty$ for your purposes?

For example, I found useful information in Grindstead and Snell's Introduction to Probability which has a page about this question for the case of a finite number of convolutions (chapter 7, pg 300 in my copy) and lists the equation

$$ f_{S_n}(x)=\frac{1}{\sqrt{2\pi n}}e^{-x^2/2n} $$

representing $n$ convolutions of a Gaussian, with $S_n=X_1+X_2+\dots+X_n$ where $X_1,X_2,\dots,X_n$ are independent Gaussian random variables with mean 0 and variance 1. $S_n$ is defined by their $n$-fold convolution as $f_{S_n}=(f_{X_1}*f_{X_2}*\dots*f_{X_n})(x)$. So for your Gaussian experiment, maybe this can provide some insight about the convergence to zero.

The book shows similar examples for uniform and exponential random variables. They note that this can be done for certain cases, implying that it is not generally applicable. They also provide the following reference for additional information (which I have not studied): J. B. Uspensky, Introduction to Mathematical Probability (New York: McGraw-Hill, 1937), p. 277.

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I had a similar question for years. Only recently I was able to solve. So here you go.

As you have mentioned, you can assume $f$ as a pdf of a random variable multiplied by a scaling factor, since it satisfies all the required properties you've mentioned.

So following the approach, let me first consider a function $f(x)$, which is a pdf of a random variable $X$. Also consider a sequence of $n$ random variables, $X_1 , X_2 , X_3 , \dots , X_n $ that are iid ( Independent and Identically Distributed RVs ) with a pdf $f(x)$.

Now Central Limit Theorem says that \begin{equation} Y = \frac{1}{\sqrt n} \sum\limits_{i=1}^{n} X_i \end{equation} converges in probability to a normal distribution as $n$ approaches $\infty$. But by sum property of random variable, the pdf of the random variable $Y$ is simply $\frac{1}{\sqrt n} ( f(x)*f(x)*\dots f(x)) $.

This means that in your case $F^{\infty}(x)$ converges to $\sqrt n a^n \mathcal{N} (\mu,\sigma)$, which tends to $0$ as $n$ tends to $\infty$, if $|a| \leq 1 $, where $a$ is the scaling factor required to normalize the area under the curve to $1$ for the equivalent pdf. This is the reason why your function becomes flatter and flatter with increasing $n$. Now try the same experiment after normalizing the function with $ \sqrt n a^n$, you must get a smooth bell curve.Hope it helps.

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  • $\begingroup$ You can use the Fourier transform to estimate $\underbrace{f \ast \ldots \ast f}_n$ and prove the CLT this way $\endgroup$ – reuns Aug 25 '17 at 22:19
  • $\begingroup$ Yes. But since I already have CLT, I am using it to solve this problem rather than deriving it from scratch. $\endgroup$ – Bragadeesh Sep 2 '17 at 14:32
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Well for your trials with Gaussians you're only rescaling them. My guess would be that it will tend to an infinitely stretched Gaussian regardless of the initial function. It least it is equivalent to having a sum of $n \to \infty$ random variables with the same probability density function $f$, which should tend to normal distribution stretched by the factor of $n$.

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  • $\begingroup$ I had sort of the same feeling... Would you be able to link me to a proof or something? Thank you $\endgroup$ – Andry Feb 14 '15 at 19:03
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You can use the cumulants of the original distribution and make the inverse Fourier Transform. Being m_1 the mean and k_n the cumulants of f, after N self convolutions, the resulting mean is Nm_1 and the cumulants Nk_n. If f have all cumulants well defined, the result tends to a gaussian with mean Nm_1 and variance Nk_2 (actually the central limit theorem).

Note: in your mathematical experiment, (which have 0 mean and unit variance), the result is a gaussian with variance N: it is more dispersed. It is the reason you see it as if it tends to zero, the area, which is conserved, is expanded over a large interval, so the maximum lowers... If you expand the x-axis you will recover a gaussian...

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