0

If we have a piecewise function defined as

$f(x) = \begin{cases} 1, & \text{0 $\le$ $x$ $\le$ 1} \\ 0, & \text{otherwise} \end{cases}$

Explain how the convolution of $f$ with itself for $n$ times i.e. $(f*f*f.......*f)$ $n$ times might begin to look like bell curve of the central limit theorem. I have no idea how to proceed. Any help would be appreicated.

meta_finance
  • 327
  • 2
  • 12
  • Have you computed a few of these? It's pretty instructive to compute $ff$ and $ff*f$. More than that and things rapidly get tedious, but those few would be good to try. – Ian Jul 23 '16 at 22:04
  • I tried computing them, but failed to recognize how will they converge to a normal distribution. – meta_finance Jul 23 '16 at 22:06
  • 2
    Well, as for why they do, you should already know that when you add together two independent continuous random variables, the pdf of the sum is the convolution of the pdfs. So you already somehow know that the convergence is occurring. I think the point of the exercise is to get you to think about what convolution actually does in such a way that you can see the curve form. $fff$ already looks a bit like a truncated bell curve. – Ian Jul 23 '16 at 22:09
  • Anyway, take a look at the Wikipedia article about convolution, there is a description in terms of a "moving window" that you might find instructive (especially in this case of an indicator function of an interval). Basically the point here is to understand that near the middle, the convolution is big, sufficiently far away it is zero, and the region where it is nonzero grows as you do more and more convolutions. Getting the actual shape of a normal distribution per se sounds like a more delicate matter than this assignment wants you to do. – Ian Jul 23 '16 at 22:09

1 Answers1

2

As I have no idea of your background, I will do as @Ian has done : try to give you an intuitive feeling of what is "behind the curtain".

$f$ is the pdf of a uniformly distributed Random Variable $X$ which is simulated by a call to a (pseudo) random number generator "rand" (or "Random" or ... depending on the language you use).

Have you seen that the pdf of a sum of independant identically distributed (i.i.d.) random variables is the convolution of their pdfs ?

If you have, say, four independent variables $X_k, (k=1 \cdots 4)$, have you tried to plot an histogram, of 100,000 sums $X_1+X_2$ (100,000 times rand+rand) ? (This is the first thing I ask to my students to do). You should observe that, though not gaussian-shaped, it is no longer uniformly distributed on $(0,2)$. You may even observe that it looks like a tent).

Then proceed to sums $X_1+X_2+X_3$ which begin to take a bell-curve shape on $(0,3)$, and so on...

Jean Marie
  • 81,803
  • If we are doing convolutions on the function itself, then how is it i.i.d? – meta_finance Jul 23 '16 at 22:30
  • Convolution is a mathematical operation that ignores independence issues... – Jean Marie Jul 23 '16 at 22:31
  • While doing $fff$, how does the limit becomes $(0,3)$? Wouldn't it be the same $(0,2)$? @JeanMarie – meta_finance Jul 24 '16 at 02:31
  • 1
    @Rishris Think intuitively: $fff$ is the pdf of a sum of 3 uniform $(0,1)$ pdfs. So its support is on $(0,3)$. You may be confusing this with the domain of integration; you can indeed set this up to be just an integral over $(0,1)$. For example, $(fff)(x)=\int_0^1 (f*f)(x-y) f(y) dy$. But that's another matter. – Ian Jul 24 '16 at 17:45
  • 1
    @Rishris All $X_k$ are such that $0<X_k<1$, then $0<X_1+X_2+X_3<3$ (The difficulty you find is common to many beginners in probability : trying to figure out on the $y$ axis something that in fact has to be be looked on the $x$-axis). – Jean Marie Jul 24 '16 at 19:43
  • @ JeanMarie @Ian Yes, it's a bit confusing. But I am getting the hang of what you are trying to say. Thank you! – meta_finance Jul 24 '16 at 20:16