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