Can double convolutions be simplified using a change of variable?

[muzialis]

Let us write a convolution
$$\int_{0}^{t} A(t-\tau) \mathrm{d}x(\tau)$$ as
$$A \star \mathrm{d}x$$
I would like to write down the expression for the double convolution
$$A \star \mathrm{d}x \star \mathrm{d}x $$
Following the definition I obtain
$$ \int_{0}^{t} \int_{0} ^{t-\tau} A(t-\tau-s) \mathrm{d}x(s) \mathrm{d}x(\tau)$$
Can this be given a more compact form, especially in reference to the upper limit of integration in the inner integral?
I would like to perform the change of variable $$t-\tau = w$$ but unsure as to how to proceed, any hint would be the most appreciated, thanks

 

[Stephen Tashi]

Explain what you mean by:

Let us write a convolution
$$\int_{0}^{t} A(t-\tau) \mathrm{d}x(\tau)$$ as
$$A \star \mathrm{d}x$$

Perhaps you are using some notation known in engineering. However, in the usual mathematical terminology, one convolves two functions and if that's what your talking about, the notation doesn't make it clear if you are using are two functions.

For example if $A()$ is the function defined by $A(r) = 3r^2 + r + 1$ then what does $A * dx$ mean? In particular what would $dx(\tau)$ be?

 

[muzialis]

Stephen,

many thanks for your input.
I am not using an engineering notation, whatever yo mean by that, I am simply writing the convolution down as a Stijeltes convolution (using Stjeltes instead of Riemann integration), as the notation is more compact.
Under certain technical conditions which can definitely be assumed in the present case, one can write
$$\int_0^{t} A(t-\tau) \frac{\mathrm{d}g}{\mathrm{d}\tau}\mathrm{d}\tau$$
as
$$\int_0^{t} A(t-\tau) \mathrm{d}g(\tau)$$
(formally looking as a change of variable procedure)
Many thanks again, I hope this clarifies and you will help me further.

 

[Stephen Tashi]

I am simply writing the convolution down

Your notation needs some words to put it in context.

Which section of the wikipedia article on convolution http://en.wikipedia.org/wiki/Convolution refers to the type of convolution in your notation?

 

[muzialis]

Well, there is no mention in Wikipedia of Stjeltes convolutions (although they are very strongly related to the measure theory approach described in the Section "Measure"), would it help if I edited my question using Riemann integrals notation?

 

[Stephen Tashi]

Does you question amount to "How do we do a change of variable in a Stieltjes integral?" I'm not an expert on that topic, but perhaps we can find expositions about how to do it.

 

[muzialis]

Stephen, you are right in pointing out that such notation is confusing. Let me then rephrase my question clearly:
Let us denote a convolution
$$\int_0^{t} A(t-\tau) x(\tau) \mathrm{d}\tau$$
With the notation
$$A \star x$$
I would like to write down the expression for the double convolution
$$A\star x \star x$$
Following the definition I could write
$$\int_{0}^{t} \int_0^{t-\tau} A(t-\tau-s) x(s)x(\tau)\mathrm{d}s \mathrm{d}\tau$$

Or, exploiting the associativity property
$$\int_0^{t} A(t-\tau) \int_{0}^{\tau} x(\tau-s)x(s)\mathrm{d}s \mathrm{d}\tau$$

But I would like to write the double convolution using double integrals with the same limits of integration, how can I achieve that (especially in the first of the two expression I wrote down)?
Many thanks

 

[bpet]

Let us write a convolution
$$\int_{0}^{t} A(t-\tau) \mathrm{d}x(\tau)$$ as
$$A \star \mathrm{d}x$$
I would like to write down the expression for the double convolution
$$A \star \mathrm{d}x \star \mathrm{d}x $$
Following the definition I obtain
$$ \int_{0}^{t} \int_{0} ^{t-\tau} A(t-\tau-s) \mathrm{d}x(s) \mathrm{d}x(\tau)$$
Can this be given a more compact form, especially in reference to the upper limit of integration in the inner integral?
I would like to perform the change of variable $$t-\tau = w$$ but unsure as to how to proceed, any hint would be the most appreciated, thanks

If A(t) and x(t) have properties similar to probability CDFs, you could write the convolution integrals in symmetric form using indicator functions, eg $$\int\int I(t_1+t_2\le t)dA(t_1)dx(t_2)$$ - the three-way version should be fairly straight forward.