Convolution of characteristic function

[alena_S]

Homework Statement

I am trying to figure out following problem.

Let A ⊂ R. Then we can define the characteristic function:
\begin{align}
\chi_A : R → \{0, 1\}, x = \begin{cases}
1 & \text{if } x \in A \\
0 & \text{else }
\end{cases}
\end{align}

Let a be bigger than 0. I am trying to find a following convolution:
\begin{align}
\chi_{[-a,a]} * \chi_{[-a,a]} * \chi_{[-a,a]}
\end{align}

Homework Equations

Convolution is given as
\begin{align}
f*g = \int f(x-y) g(y) dy
\end{align}

The Attempt at a Solution

I have started to do convolution of 1st 2 terms, my results are as follows(not sure about correctness)[/B]

\begin{align}
\phi * \phi (x) = \begin{cases}
0 & x \leq -a \\
x & \text{ if } -a \leq x \leq a\\
2a - x & \text{ if } a \leq x \leq 2a\\
0 & x \geq 2a
\end{cases}

\end{align}
but I am being stuck what should follow.

 

[S.G. Janssens]

I don't see a problem statement yet. What do you want to convolve $\chi_A$ with? Do you know more about $A$? Is the domain of the convolution integral the entire real line?

 

[alena_S]

sorry, there was a typo in post. I hope, its correct now

 

[Ray Vickson]

sorry, there was a typo in post. I hope, its correct now

No, it is NOT correct. The notation $\chi_A[-a,a]$ makes no sense. If you mean that $A = [-a,a]$, then you must write $\chi_{[-a,a]}$. If you mean something else, tell us what you mean.

 

[alena_S]

No, it is NOT correct. The notation $\chi_A[-a,a]$ makes no sense. If you mean that $A = [-a,a]$, then you must write $\chi_{[-a,a]}$. If you mean something else, tell us what you mean.

Indeed, I meant $\chi_{[-a,a]}$.

 

[Ray Vickson]

Indeed, I meant $\chi_{[-a,a]}$.

OK, so one way is to use Laplace transforms. Usually you see Laplace transforms defined for functions on $[0, \infty)$ but they CAN be defined for functions on $[-M,\infty)$ in the same way: for $f:[-M,\infty) \to \mathbb{R}$, the Laplace transform is
{\cal L}_f (s) = \int_{-M}^{\infty} f(x) e^{-st} \, dt,
and has the usual properties. In particular, the transform of a convolution is the product of the transforms. Then you can invert the result to get your function in terms of $x$.

Alternatively, you can apply the straight definition of convolution and compute the results directly by integration, doing first the function $f_2 = \chi_{[-a,a]} * \chi_{[-a,a]}$, then doing another integration to get $f_3 = \chi_{[-a,a]} * \chi_{[-a,a]} * \chi_{[-a,a]}$ as $f_3 = f_2 * \chi_{[-a,a]}$. Keeping track of integration limits and spitting the calculation up into cases is tricky but necessary.

By the way: you expression for $f_2$ in your eq. (4) is not correct: the lower and upper $x$-limits are wrong. Basically, your functions $\chi_{[-a,a]}$ are (up to a scale factor) the probability density functions of random variables uniformly distributed between $-a$ and $+a$. The two-fold convolution is the density function of a sum of two independent uniformly-distributed random variables, and so will be positive on values of $x$ ranging from $-2a$ to $+2a$. For the three-fold convolution that range will be from $-3a$ to $+3a$.

 

[alena_S]

OK, so one way is to use Laplace transforms. Usually you see Laplace transforms defined for functions on $[0, \infty)$ but they CAN be defined for functions on $[-M,\infty)$ in the same way: for $f:[-M,\infty) \to \mathbb{R}$, the Laplace transform is
{\cal L}_f (s) = \int_{-M}^{\infty} f(x) e^{-st} \, dt,
and has the usual properties. In particular, the transform of a convolution is the product of the transforms. Then you can invert the result to get your function in terms of $x$.

Alternatively, you can apply the straight definition of convolution and compute the results directly by integration, doing first the function $f_2 = \chi_{[-a,a]} * \chi_{[-a,a]}$, then doing another integration to get $f_3 = \chi_{[-a,a]} * \chi_{[-a,a]} * \chi_{[-a,a]}$ as $f_3 = f_2 * \chi_{[-a,a]}$. Keeping track of integration limits and spitting the calculation up into cases is tricky but necessary.

Is my attempt of f2 correct (eq 4 on beginning)? just to know, if should continue with such a result, or rather try it again?

 

[Ray Vickson]

Is my attempt of f2 correct (eq 4 on beginning)? just to know, if should continue with such a result, or rather try it again?

Read my edited message # 6, as it deals exactly with that issue. I posted the edited version before your new message appeared on my screen; I often encounter such PF delays.

 

[alena_S]

Read my edited message # 6, as it deals exactly with that issue. I posted the edited version before your new message appeared on my screen; I often encounter such PF delays.

So, should it be as:
for f2:

\begin{align}
\phi * \phi (x) = \begin{cases}
0 & x \leq -2a \\
x & \text{ if } -2a \leq x \leq a\\
2a - x & \text{ if } a \leq x \leq 2a\\
0 & x \geq 2a
\end{cases}
\end{align}

And for f3, do we need 7 different intervals, or did I get it wrong?

 

[Ray Vickson]

So, should it be as:
for f2:

\begin{align}
\phi * \phi (x) = \begin{cases}
0 & x \leq -2a \\
x & \text{ if } -2a \leq x \leq a\\
2a - x & \text{ if } a \leq x \leq 2a\\
0 & x \geq 2a
\end{cases}
\end{align}

And for f3, do we need 7 different intervals, or did I get it wrong?

Still wrong: the graph of $y = f_2(x)$ should be an isosceles triangle.

Basically, you are taking the convolution of even functions, so ought to get back an even function. After all: if $f_2(x) = \int_{-\infty}^{\infty} f(y) f(x-y) \, dy$, then without actually doing the integral you can see that $f \;\text{even} \longrightarrow f_2 \; \text{even}$, just by changing variables and doing some manipulations. It becomes intuitively obvious if you use the "probabilistic" interpretation of the problem.

 

[alena_S]

Does this look more correct?

\begin{align}
\phi * \phi (x) = \begin{cases}
0 & x > 2a \ \ \text{or} \ \ x < -2a \\
x + 2a & \text{ if } x \leq [-2a, 0]\\
2a - x & \text{ if } x \in [0,2a]\\
\end{cases}
\end{align}

 

[Ray Vickson]

Does this look more correct?

\begin{align}
\phi * \phi (x) = \begin{cases}
0 & x > 2a \ \ \text{or} \ \ x < -2a \\
x + 2a & \text{ if } x \leq [-2a, 0]\\
2a - x & \text{ if } x \in [0,2a]\\
\end{cases}
\end{align}

Not "more" correct; just plain correct.

I hope you got the result by actual calculation, not by "guessing", because you will not easily guess the formula for $f_3$.

 

[alena_S]

Not "more" correct; just plain correct.

I hope you got the result by actual calculation, not by "guessing", because you will not easily guess the formula for $f_3$.

After some steps, I have obtained following for f3:
$$\chi \ast \chi \ast \chi(x) = \begin{cases} 0, & x \notin [-2a,2a], \\ \int 1_{[-a,2a+x]}(y) \cdot (x+2a) \, dy, & x \in [-2a,-a], \\ \int 1_{[x,a]}(y) \cdot (x+2a) + 1_{[-a,x]}(y) \cdot (2a-x) \, dy, & x \in [-a,a] \\ \int 1_{[-2a+x,a]}(y) \cdot (2a-x) \,dy, & x \in [a,2a]. \end{cases}$$,
which is after evaluation of integrals results into

$$\chi \ast \chi \ast \chi(x) = \begin{cases}
0, & x \notin [-2a,2a], \\
(x+2a) & x \in [-2a,-a], \\
4a & x \in [-a,a] \\
(2a-x) & x \in [a,2a].
\end{cases}$$

 

[alena_S]

Not "more" correct; just plain correct.

I hope you got the result by actual calculation, not by "guessing", because you will not easily guess the formula for $f_3$.

Is this what I am supposed to get?

 

[Ray Vickson]

Is this what I am supposed to get?

No, nothing like it. Go back to square 1.

 

[alena_S]

No, nothing like it. Go back to square 1.

Evaluation of integrals is wrong? or all result?

 

[Ray Vickson]

Evaluation of integrals is wrong? or all result?

Result is wrong, so integrals must be evaluated or written incorrectly as well. These are my last words on the topic.

 

[vela]

Evaluation of integrals is wrong? or all result?

It would help immensely if, rather than simply posting your final result, you showed your work and explained your reasoning. We can't see your paper, so we don't know what you did. And trying to guess what you did gets old really fast. Even if you get the right result, you could still be making mistakes.