Relevance of an orphaned equation

[nomadreid]

I came across an equation in a disconnected piece of my notes, and I remember that it was an instance of something, but I can't remember what. (I see how to do the integration, that is not what I am worried about.) Does anyone recognize either the function or the integration as an example of something?
(This would be a very weak maths question, so I didn't put it in the mathematics rubric.)

It appears to be discontinuous along the line x=y, so it would be strange as, say, a probability distribution, no?

 

[Charles Link]

I question the minus sign on the second expression. Otherwise by symmetry, wouldn't the double integral result in zero as the result?

 

[Charles Link]

I just tried integrating it over the section $y>x$, and I got that it diverges. It looks to me like this thing may be some kind of limit for the line $y=x$, but it doesn't seem to make much sense. Otherwise, perhaps I erred in my computation of the integral.

 

[nomadreid]

Thanks, Charles Link. Here is my integration; I would be grateful if you find my error

 

[Charles Link]

Both sections of the integral diverge. You can match up each little area of $y>x$ with one that is opposite for $y<x$, so they cancel each other, but you can not compute either section by itself and get a finite answer. I'm not an expert on this topic, but IMO that is poor mathematics and that type of cancellation doesn't work. It is basically the step where you cancel the blue -1 with the red +1.

What you did is interesting, but it doesn't work. :)

 

[Charles Link]

and a follow-on: I couldn't find that discussion when I searched for it, but it reminds me of a previous discussion on PF a few years ago where several were discussing the (il)legitimacy of the Lorentzian distribution where the mean is not zero, even though you can balance the integral on both sides.

 

[nomadreid]

Thanks, Charles Link. Perhaps I did take an incorrect integral from a PF discussion without noticing that it was invalid, and then tried to justify it by juggling the integration. Well, that is still a worthwhile attempt, in that it will lead me to look again at the calculation to figure out the errors. (So I quite like getting corrected, and thus am grateful to those who help me in this way.) So, thanks again!

 

[Charles Link]

The Lorentzian distribution is $f(x)=\frac{1}{\pi (1+x^2)}$. Try computing the mean and you will see what I am referring to. The mean is undefined, rather than being zero. The function is symmetric about $x=0$, but when you try computing the mean, the integral $\int \frac{x}{1+x^2}\, dx$
is similar to what you have above. The part $x<0$ diverges, and the part $x>0$ diverges.

Edit: Note that the anti-symmetric function that you have (plus the positive values (=1) on the line $y=x$) gives the result (=1) that you got only with some handwaving that is not completely valid mathematically.

 

[Charles Link]

@nomadreid Please see the Edit in post 8 above.
Edit: I'm reading the OP again though, and it looks like you have the negative values on the line $y=x$. If you work through your post 4 more carefully, you might find you were not consistent with your limits of integration.

 

[fresh_42]

The minus sign has been my problem, too. I haven't checked the integral, though. It only looked like the density function of an exponential distribution in two variables if there weren't the negative values. I also thought about Lie groups, but again, one ended up with determinant $1$ which is fine, but the other one with determinant $-1$ which is not so nice.

 

[Charles Link]

@fresh_42 The minus sign is in there to make the function anti-symmetric about $y=x$, but I think it still needs some tricky handwaving to work the line $y=x$. It looks like from the statement of the problem in the OP that there are negative values on the line $y=x$, so I don't agree with the positive result of $+ 1$ in the OP, even if you allow for some handwaving.

 

[fresh_42]

@fresh_42 The minus sign is in there to make the function anti-symmetric about $y=x$, but I think it still needs some tricky handwaving to work the line $y=x$. It looks like from the statement of the problem in the OP that there are negative values on the line $y=x$, so I don't agree with the positive result of $+ 1$ in the OP, even if you allow for some handwaving.

Yes, but it closed both doors I considered: probabilities and multiplicative groups. My internet search led me to fancy things like memory loss (exponential pdf) so I thought about something conditional if there only wasn't the minus.

 

[DrGreg]

The function $f$ does not have a well-defined integral. This means that when you attempt to decompose the 2D integral into two nested 1D integrations, you can get different answers depending on how you choose the decomposition.

I believe the decomposition given in post #1 does indeed give an answer of $+1$. However, if you were to integrate $y$ first and $x$ second, you'd get an answer of $-1$, and if you integrated along lines in another direction you could get zero along each line, or $\pm\infty$ along different lines in yet another direction.

You can tell $f$ isn't integrable because $|f|$ isn't either, and there isn't any way to decompose the integration of $|f|$ to get a finite answer. (See Absolutely integrable function.)

This all goes to show that playing with $\infty$ is dangerous.

 

[Charles Link]

If you do the $dy$ integral on the lower right from $y=0$ to $y=x-\Delta$, that later results in an uncancelled term after a Taylor expansion of $-\int\limits_{0}^{+\infty} \Delta \, dx$. It appears you could assign almost any value you wish to this integral. Thereby I think the problem statement in the OP isn't entirely valid. Even with some handwaving it remains not well-defined.

 

[nomadreid]

Thanks to all contributors -- from these comments, it is clear that the problem was ill-posed, and I can consider it dead. Several people said essentially that it might (or might not) be a valid distribution if it were not for the minus sign in front of exp(x-y), and there were also objections to the infinite range . So, would a redefinition of f as f(x,y)= exp(-|x-y|) for finite ranges for x and y be of any interest in any context?