## Pathological PDFs. eg: ratio of normals... including Cauchy.

Chiro,

I made an out of context remark regarding Marsaglia. He is quite justified in going to the trouble of using a 50 digit number -- My statement that his approximations were "crude" was based on the general notion of approximation vs. exact integrals...-- I didn't mean to give an impression of his work being garbage, and after re-reading my quick remark; I just realized you might have taken my comment that way.

Until I actually give my own approximation, I have no idea whether or not I can trump his.

Cheers.
--Andrew.

 Quote by Stephen Tashi Reputable math texts don't bother to mention all the things that are NOT true about a mathematical topic. Engineering texts may have their own conventions about integrals, but "reputable" math texts that talk about integrals of real valued functions are referring to Riemann or Lebesgue integration, not Cauchy principal value integration. If they wish to treat the Cauchy principal value type integration, they would feel obligated to mention it explicitly.
I don't expect that Engineering texts are different. In the derivation I gave, I specifically asked regarding the appropriateness of the way I did the limit; My point is that Cauchy principal value did come up in a problem regarding Fourier transforms, and we were to use it under a certain set of conditions.

If you go back to the earlier thread from which this branched out, and you look for Chiro' giving me a suggestion on computing the integral -- That's where Fourier came into the discussion...

 Marsiglia provides results for the case when the denominator is conditioned by a restriction that prevents it from being zero.
Yes, and his particular restriction seem arbitrary.

 You aren't going to get around the fact that by the standard definition of the mean of a distribution, the mean value of the ratio of gaussian distributions doesn't exist.
I realize that; But I also need to understand how real means which do exist -- for samples -- are related to none for the distribution as a whole.

 Talking about what happens in numerical approximations and simulations doesn't weasel around it. If numerical approximations and simulations don't reproduce a proven theoretical result,
What is the theoretical result? eg: Infinite is not the THEORETICAL result even though Marsaglia mentions integrals going infinite (he is perhaps numerically speaking, or half integral speaking?)

All distributions using Gaussian randoms implicitly come from a limiting approach based on samples. The Gaussian was discovered by asking the question, if a measurement (discrete) is repeated what is the probability of the error.

To answer the question; Gauss took a limit for infinite repetitions, and the bell curve was arrived at.

There is, then, in each distribution (& Cauchy) -- an implicit limit going from the discrete/finite to the continuous.

If the continuous integrals for the mean had always diverged toward positive infinity or negative infinity, there would be no question of what theory said/predicted... But the moment theory can arrive at infinity minus infinity -- that isn't a result of any kind. If that is true, then an estimate of 0 is as good as any other estimate -- they are all *equally* bad.

BUT: In a real experiment, there is going to be a sampling granularity -- and an actual finite value for the mean.

When I hear the words that the integral does not converge, then I understand The QUESTION was somehow asked wrong -- and the boundary conditions need to be looked at to decide why.

 it shows the numerical approximations and simualtions are inadequate, not that the theoretical result is wrong. The legalistic nature of mathematics is price one must pay for keeping statements unambiguous and not proving claims that are false.
Bravo -- I'll clap my hands. I think that's wonderful.

Simulations are a way to look for problems, and inconsistencies -- because they indicate the results a real measurement would report. The problem may be resolved in many different ways -- without changing the theory itself.

But, I need *quantitative* models of how simulation of Cauchy distribution using discrete samples is going to be different from the continuous case.

 A problem with integrating the ratio of gaussians is not merely that the denominator might be exactly zero. There no bound to the ratio even if we exclude the possiblity that the denominator is exactly zero.
I agree that removing the zero point is insufficient. Your analysis, here, however isn't clear to me yet. Since Cauchy is a limit of a ratio of binomials taken to n-->infinity; The correlation between the character of the discrete distribution -- and the continuous one changed it's properties somewhere -- and that bothers me.

 If you do a numerical approximation method that establishes a bound for the ratio, it's the numerical approximation that's wrong. If you observe that denominators very close to zero never showed up in a simulation then it didn't explore all the possibilties of a gaussian denominator.
Of course! I think you have gotten the wrong impression about my numerical simulations. My apology belatedly...

I think (intuition) what we're missing is the idea of a confidence interval. How likely an experiment will "hit" the div/zero discontinuity close enough to disturb an otherwise tranquil mean...

(Don't answer that! it's in English -- and I need to quantify it)

 I don't understand your resistance to stating your results in terms of conditional gaussian distributions.
You're just moving through the problem more quickly than I can possibly do;
I am looking at your equations, and I don't find them objectionable -- but I'm not sure they give the same result that I have... (I'll comment later).

The question, to me, is not about whether or not we condition the distributions -- the question is; how, and on what basis, and why.

As I said in the OP -- I'll have to explain some of this later in the thread -- (I'm working on it, hard...)

When I wrote the OP -- I was expecting corrections inside the derivation, your approach is unexpected, and that's difficult for me to adjust to. I'm trying.

 Quote by Stephen Tashi Marsiglia provides results for the case when the denominator is conditioned by a restriction that prevents it from being zero. The standard interpretation of that would be that he replaces the distribution in the denominator by another distribution, a "conditional" distribution. I see nowhere in the paper that he claims that the moments exist without that replacement.
Yes, but earlier you made a strong impression on me that a distribution may not be called the same if it were modified in any way. Thus, the title of Marsaglias's paper (for this reason) appeared to me to be puporting something to be a Gaussian -- which is actually not a Gaussian, and computing the mean of a ratio of ????.

In contrast, if I now take this new view you are giving:
I implicitly gave the information about how the Gaussian's were conditioned in my derivation, even if I didn't know I was doing it.

At most it's a notational issue... and I asked people for help about formalizing the derivation; intentionally attempting to uncover issues like this one, directly.

Hindsight, of course... Hopefully, tomorrow will be better.
 Stephen, I think this will be a simple idea... The Cauchy has a CDF of ${1 \over 2} + {1 \over \pi} tan ^{ -1 } ( x )$; When estimating a mean, I might think like this: (assume purely positive numbers). The mean must be between the lowest value in the sample and the highest; If I assume the highest value takes the whole weight of the average (n samples times highest value / n), Then the mean must be lower or equal to that value. Hence, I can ask -- what is the greatest distance from 0 that the mean could be? 90% of the time, the largest sample will be no higher than: 6.31376; hence the weight of that is 5.1... ( 0.9 * 6.313.. ) 99% of the time, the largest sample will be no more than 63.65675; hence the weight of that is 5.72... ( 63.65675 * 0.09 ) 99.9% of the time -- 636.61925 and weight = 5.72... (636 * 0.009) So, I can expect with some confidence -- that a mean of no more than 18 will be computed at least a certain definite percentage of the time; depending on the number of samples taken. Hence, it doesn't seem right to say the mean is totally unbounded -- but there must be some kind of relationship between sample size and a typical mean. If there were no rhyme nor reason to the mean, all values would be equally likely. But that's not the case...
 Basically, it looks what you are doing is talking about a different distribution with each level of 'confidence'. One suggestion I have is that for the X/Y problem, you should modify Y so that you exclude a region of a neighbourhood around 0 (i.e. you censor this region where P(-e < X < e) = 0 for some epsilon e) and then recompute the density function. The idea of using a pure Gaussian for the denominator, even for something like NIST is absolutely stupid and if they want to use X/Y without any modification, then they are going to deal with the case of no moments existing. Since all you are doing is effectively changing the distribution for each level of confidence, you are probably IMO better off in just creating the distribution you intended and then calculating the mean in the way that it is calculated rather than trying to fudge the calculation of the mean for a distribution where it does not exist. This way, you'll keep to the definitions (which are there for a reason because they work both theoretically and practically) and you will be able to clarify your assumptions by the nature of the definition of the actual distribution (for example censor the region around 0 is due to getting rid of dividing by numbers close to zero). (Also remember if you do censoring, you have to normalize the distribution to make sure it integrates to 1).

 Quote by chiro Basically, it looks what you are doing is talking about a different distribution with each level of 'confidence'. One suggestion I have is that for the X/Y problem, you should modify Y so that you exclude a region of a neighbourhood around 0 (i.e. you censor this region where P(-e < X < e) = 0 for some epsilon e) and then recompute the density function. The idea of using a pure Gaussian for the denominator, even for something like NIST is absolutely stupid and if they want to use X/Y without any modification, then they are going to deal with the case of no moments existing.
Believe me -- I'd love to throw it out.
I'm not a glutton for punishment...

But consider a realistic case; There is a wall 2000(1) mm away. Divide that distance by a *SLOPPY* 1 meter scale 1000(1) mm long. What are the results?

Well, the result is obviously about 2 (on mean).
However, this still factors into a form which Marsaglia treats as having a Cauchy.
( I forget exactly what he said about that -- my eyes glazed over...)

N(2000,1) / N(1000,1) =
(2000 + N(0,1)) / (1000 + N(0,1)) =
=2000/(1000+N(0,1)) + N(0,1)/(1000+N(0,1))

My solution so far, is just for the first part -- and even that is supposedly invalid because I used a cauchy principle value...

Yet the second part is the only thing really Cauchy distribution like in the problem, so I assume that's a Cauchy with the mode offset in one direction or the other....

But in any event, it's not possible to avoid having that second part in the equation -- and if it's mean can be anything -- then the sum of the two means, can be anything -- and well, the *very* practical problem has just become theoretically impossible when ACTUALLY using the theory and not faking it.

I find that quite perplexing. Gosh! the odds of hitting the zero point are 1 in 1000 sigmas. It *aint* gonna happen... BUT -- it's still a Cauchy?!!!!!

Now, let's talk about the choice of "theory". There is no reason both of these lengths couldn't be repetitively measured over and over -- so a Gaussian is the most appropriate distribution.

But once we do a ratio, we are going to have a Cauchy and a 1/Gaussian.

WOW.

 Since all you are doing is effectively changing the distribution for each level of confidence, you are probably IMO better off in just creating the distribution you intended and then calculating the mean in the way that it is calculated rather than trying to fudge the calculation of the mean for a distribution where it does not exist.
I have a formula that simulates well -- if I just knew what the distribution was that made that shape, I'd be done. Funny, I have an answer looking for a distribution...

 This way, you'll keep to the definitions (which are there for a reason because they work both theoretically and practically) and you will be able to clarify your assumptions by the nature of the definition of the actual distribution (for example censor the region around 0 is due to getting rid of dividing by numbers close to zero). (Also remember if you do censoring, you have to normalize the distribution to make sure it integrates to 1).
Ummm ... I'd like to see that practical part...

But,
Yes, that sounds possible -- although I need to start with the confidence interval: eg: I need someone to be able to tell me they want my result 99.9% certain, and then I need to compute how much of the zero divide to censor... THEN, I can do it.

I'm going to sleep on it tonight... Dunno....
 For the practical part, the first thing to focus on is getting the distribution for 1/X where X is censored and then look at Y/X after you get the censored distribution for 1/X. It's best if you leave the 1/X distribution in terms of the e mentioned above so that later you can see how this e effects the calculation of the mean of Y/X: this solves your problem of analyticity and you can use this to compare how many standard deviations you need to get a mean of a particular value, but looking at how the epsilon affects the final calculation of E[Y/X] where Y is Gaussian and 1/X is the transformation of the inverse of your censored distribution. You can simulate this extremely easily by using a method to simulate from a censored distribution (an MCMC approach will do this) and then simply simulating from the Gaussian giving a simulation for Y/X. The assumption of censoring is one that can quantified in the context of more general assumptions in the domain (i.e. engineering) by considering the nature of what is being calculated (i.e. scales of things, what these things are) in relation to the epsilon used in the censoring process. I think that the above suggestion will help you not only derive a distribution and ultimately a mean using censorship around 0 for the denominator RV, but also to actually quantify the characteristics and how the epsilon changes the value of not only the mean, but also the other moments as well.

Recognitions:
 Quote by andrewr Hence, it doesn't seem right to say the mean is totally unbounded -- but there must be some kind of relationship between sample size and a typical mean. If there were no rhyme nor reason to the mean, all values would be equally likely. But that's not the case...

As a very general observation (general enough to apply to the whole of mathematical society, not particularly to yourself), there is always a mental contest between formal mathematics vs the philosophy of mathematics that I would called Mathematical Platonism. The Wikipedia deistinguishes quite a number of species of Mathematical Platonism, but to me, the common element in this philosophy is the belief that things with mathematicl definitions have a reality that exists apart from the definition. In your particular case, you believe that the concept of "mean value of a distribution" has a reality beyond the formal definition, so you allow yourself to reason about this reality and reach conclusions based on your private vision of it. I think almost everybody does this to some degree.

Sometimes Mathematical Platonism leads nowhere. For example, if you look at threads on the forum that are inviitations to Mathematical Platonists, such as "Is multiplication repeated addition?", "Is dy/dx a ratio?", you find that many of the posts with a Platonic slant are opinionated and unimaginative. But sometimes you do find Platonic outlooks that are very helpful intuitive ways to think about mathematical ideas.

Physicists and engineers often take the Platonic view of mathematics and I suspsect the reason that physics and engineering are able to cruise along with the Platonists on board is that most concepts they deal with don't depend on a legalistic and precise application of logic. On the other hand, mathematics gets into a mess if it tries to develop results based on Platonic arguments. There are simply too many different contradictory private concepts of things like "limits", "infinity", "probability" etc. among human beings. The only way to arrive at definite results is to have formal definitions and develop arguments based on those definitions, not people's private visions of what things are.

I don't want to discourage you from Platonic reasoning. I just want to make the point that whatever conclusions you reach by that reasoning have to reconciled with the formal mathematical definitions and presented in those terms in order for them to be accepted as mathematics.

----

As to the observations about the sample mean:

The sample mean of the Cauchy distribution is a statistic that does have a distribution. For a sample size of 1, it is obviously just the Cauchy distribution. It's an interesting question what the distribution is for larger size samples. The Central Limit Theorem (that the mean of an independent samples of size n is approximately normally distributed for large n) doesn't apply to the Cauchy distribution since that theorem requires that the distribution being sampled have a (finite) variance.

(This sets me wondering about such things as: Are their distributions whose k-th moment doesn't exist, but the k-th moment of the sample distribution (for sample size > 1) exists? Are their distributions whose k-th moment doesn't exist, but such that the kth-moment of the limiting distribution of their sampling distribution (as sample size approches infinity) exists. If no kind soul happens to tell me, I may start a thread with such questions someday.)

The non-existence of the mean of Cauchy distribution involves (according the formal definition) the non-existence of a integral that is done using the distribution. Thus the entire distribution is considered when doing the integration. The fact that a particular large value of a Cauchy random variable is unlikely in a sample doesn't mean that you can leave that value out when you do the integral. The problem with the (formal) existence of the integral depends on how you define integrals (Riemann vs Lebesgue - either way, therei's a problem). Again, it's the people who frequent the Calculus & Analysis section can probably give an authoritative answer about that.

I'm sure you've studied integrals invovling infinite limits and various theorems about when they exist or not. Some function die-off quick enough so that the integral from 0 to infinity exists, other's die-off but not quickly enough. That's the type of thing involved in the integral for the mean of the Cauchy. In the integral for the mean of the ratio of Gaussians, the problem is that the integrand is unbounded.

I don't think you should give-up on using the Cauchy principal part in your calculations. I merely suggest that you rephrase the claim about what you are calculating. My (Platonic) view is that your are calculating a limit of the means of distributions that are "conditioned" by setting them equal to zero on parts of the real line. (A density f(x) defined on a the real line can be used to define another density g(x) that leaves out intervals of the real line. On the part that is not left out, define the modified density to be g(x) = f(x)(/ 1 - P) where P is the probability of the left-out part. On the part that is left out, you define g(x) = 0.)

 Quote by chiro For the practical part, the first thing to focus on is getting the distribution for 1/X where X is censored and then look at Y/X after you get the censored distribution for 1/X.
I am unable to solve that problem.
Introducing a censoring window causes the angle of integration to change from 0 to 2π, to a set of bounds that depend on r. I am unable to solve the new integral itself.

Not only that, but in trying to solve for the Cauchy distribution itself (often done implicitly when solving for the first moment); I noticed that the Cauchy distribution requires the same integral I did before divided by r; which means that it, too, has a point which is infinite. Hence, the Cauchy itself requires the use of a Cauchy principle value to derive.... If changing the limits with which one integrates any such integral is capable of changing the results, I would have to assume the Cauchy distribution is itself invalid unless there is a way to justify the use of a particular Cauchy principle value...

I need to find a derivation of the Cauchy distribution and find out on what grounds they justify the existence of the integral in the first place !!! or if there is a way to work around the issue.

 It's best if you leave the 1/X distribution in terms of the e mentioned above so that later you can see how this e effects the calculation of the mean of Y/X: this solves your problem of analyticity and you can use this to compare how many standard deviations you need to get a mean of a particular value, but looking at how the epsilon affects the final calculation of E[Y/X] where Y is Gaussian and 1/X is the transformation of the inverse of your censored distribution.
I did come up with something interesting by a different route -- when attempting 1/x.
The censoring window is roughly equivalent to truncating the tails of the Cauchy at *very* large values.

The non-existence of the integral of the mean, for example, is based on a-symmetrical areas for the left and right hand portions of the integral; but notice -- truncating the zero denominator point symmetrically -- has the effect of forcing the limit of the left and right hand values to have a symmetric end point.

Therefore your idea (nearly) reduces to mine as the window shrinks to zero.

 You can simulate this extremely easily by using a method to simulate from a censored distribution (an MCMC approach will do this) and then simply simulating from the Gaussian giving a simulation for Y/X.
Yes. That works fine...

 The assumption of censoring is one that can quantified in the context of more general assumptions in the domain (i.e. engineering) by considering the nature of what is being calculated (i.e. scales of things, what these things are) in relation to the epsilon used in the censoring process. I think that the above suggestion will help you not only derive a distribution and ultimately a mean using censorship around 0 for the denominator RV, but also to actually quantify the characteristics and how the epsilon changes the value of not only the mean, but also the other moments as well.
It introduces the same question we originally had -- but in a different context.

The symmetry of the censoring window is equivalent to my idea -- but the flaw in my idea is that the limit of the integral used to compute the mean can have be chosen to be asymmetrical as one approaches infinity. Just so, your censoring window is symmetrical -- but we could choose an asymmetrical window...

I see no way, based on engineering considerations, how to justify this window's symmetry or a-symmetry. In the ideal case, it wouldn't exist -- and is just a fix-up to work around a problem not defined by engineering at all...

 Quote by Stephen Tashi I don't understand your resistance to stating your results in terms of conditional gaussian distributions. Suppose we are using Cauchy principal value integration and computing the duck mean of a density $f(x)$ duck_mean $= lim_{h \rightarrow 0+} ( \int_{-\infty}^{a-h} x f(x) dx + \int_{a+h}^{\infty} x f(x) dx )$
So, this is the density of the Cauchy, I suppose. But, if f(x) is the density of a Cauchy -- then there's a mistake in the idea; The failure of the integration isn't near zero -- it's because the definite integral goes to infinity at infinity. One can take the limit as the two half integrals approach infinity with the approach toward the limit occurring at different scalar rates; This difference in rates gives a result which is proportional to ln( ratio ).

see my comments to chiro regarding the windowing issue in general, and not being able to solve integrals with a cut-out window.

 It's no suprise in mathematics when a limit of things of one type is not of the same type. ( For example, the limit of a sequence of continuous functions need not be a continuous function. The limit of a sequence of rational numbers need not be a rational number etc. ) So it's OK when the limit of means isn't a mean. You can always say the duck mean is infinitely close to an actual mean if that would make the NIST happy.
The duck mean is equivalent to taking the Cauchy distribution and truncating the tails at some grotesquely large values -- but symmetrically. When the tails become suddenly "zero", it is no longer possible that the limit of half of the Cauchy distribution produces a value different on the left side from the right side.

 Ask your engineering professor about whether the "default" interpretation of integration in a mathematics text includes Cauchy prinicpal value integration. Ask him if the online documents and internet strangers who say the Cauchy distribution has no mean are wrong. .
Really, it's not my habit to walk into a buz-saw, even if someone else asks me to. I'm sure he'd agree with you on these particular questions.

 Quote by Stephen Tashi In your particular case, you believe that the concept of "mean value of a distribution" has a reality beyond the formal definition, so you allow yourself to reason about this reality and reach conclusions based on your private vision of it. I think almost everybody does this to some degree.
There are two issues here, whether a distribution has a mean -- and whether the distribution itself accurately reflects the process used to arrive at that result. In the former, I was unaware I had made a mistake -- and if I could change the opening post, I would, to reflect that. But it is no longer my option to do that...

In the second question, I think the issue (likely) resides around whether we define the distribution to be the limit of the ratio set of finite binomials, where we take the mean of the finite ratio -- and then let the number of elements in the binomials approach infinity -- or whether we take the limit before computing the mean, and end up computing the mean of the continuous distribution.

Sure.

 Physicists and engineers often take the Platonic view of mathematics and I suspsect the reason that physics and engineering are able to cruise along with the Platonists on board is that most concepts they deal with don't depend on a legalistic and precise application of logic.
Or, one has a deadline -- and can't work out the real answer. A curve fit to the most common case or example available is often the best that can be done.

 On the other hand, mathematics gets into a mess if it tries to develop results based on Platonic arguments.
I'd say physics is a mess anyhow, and Engineering is a tamed mess...

 I don't want to discourage you from Platonic reasoning. I just want to make the point that whatever conclusions you reach by that reasoning have to reconciled with the formal mathematical definitions and presented in those terms in order for them to be accepted as mathematics.
I appreciate that.

 The sample mean of the Cauchy distribution is a statistic that does have a distribution. For a sample size of 1, it is obviously just the Cauchy distribution. It's an interesting question what the distribution is for larger size samples.
Yes, and sorely difficult. The CDF allows me to predict a "worst" case, but not a typical mean...

 The Central Limit Theorem (that the mean of an independent samples of size n is approximately normally distributed for large n) doesn't apply to the Cauchy distribution since that theorem requires that the distribution being sampled have a (finite) variance.
I think I'll try to graph a sampling distribution, just to see what it looks like. But, even intuitively -- I'm pretty sure it isn't Gaussian...

 The non-existence of the mean of Cauchy distribution involves ... the entire distribution ... when doing the integration.
Yes, the entire *continuous* distribution.

 I'm sure you've studied integrals invovling infinite limits and various theorems about when they exist or not. Some function die-off quick enough so that the integral from 0 to infinity exists, other's die-off but not quickly enough. That's the type of thing involved in the integral for the mean of the Cauchy. In the integral for the mean of the ratio of Gaussians, the problem is that the integrand is unbounded.
The Cauchy integral goes infinite at infinity... yes. Although, as a NOTE, I didn't compute the mean from the distribution itself -- but directly from the ratio of two independent Gaussians using a change of variables to polar coordinates. In that particular form, the infinity shows up at a finite location; I could discuss some alternative approaches, that by clever re-arrangement can eliminate the infinities ... but that still is just hiding the fact that they are there without the re-arrangement.

 (A density f(x) defined on a the real line can be used to define another density g(x) that leaves out intervals of the real line. On the part that is not left out, define the modified density to be g(x) = f(x)(/ 1 - P) where P is the probability of the left-out part. On the part that is left out, you define g(x) = 0.)
I'll have to think about it -- but as the part that it is desired to leave out is at "infinity" -- I'm not sure how to do this idea of yours.

Stephen Tashi!

Now, I'm making progress!

You said...
 The Central Limit Theorem (that the mean of an independent samples of size n is approximately normally distributed for large n) doesn't apply to the Cauchy distribution since that theorem requires that the distribution being sampled have a (finite) variance.
When I graph the equation, I discover something that ought to have been obvious ... but didn't occur to me.

The graph is for the distribution of 100,000 means of 200 point samples from a true Cauchy.
The resulting shape -- is another Cauchy.
I tried many more points, many less points, many more and less repetitions -- the shape is exactly the same... !

So what you are saying is correct; but the reason is simply that a Cauchy added to a Cauchy is .... a Cauchy; and that has implications....

Analogically:
When something is Gaussian, say -- with a mean of mu, and a sigma of s; Then If we add two independent samples of the distribution together -- The result's distribution is again a Gaussian:

Gaussian + Gaussian = Gaussian.

So:
N(mu,s) + N(mu,s) = N( 2mu, 1.414213...s )

When doing an average, we simply add (as above) and then divide by 2: The result is NOW:
N(mu, 0.7071...s).

The significant detail is that the sigma has gotten *smaller* after the average. Hence: The new result is "closer" on average to mu.

As long as the measure of the width of the distribution (which does not have to be sigma, but any "x" scaling one can invent) shrinks with each averaging; the result converges toward an "average".

It really doesn't matter what the shape is -- Gaussian, Cauchy, etc.

In the OP, I originally said that I thought the Cauchy had a mean in the "limit"; but I think I edited that out... It's important to recognize, however, that a ratio of Gaussians is *not* a Cauchy in the strict sense.
We've been mixing ideas carelessly...

Only when a=0 and b=0 is it truly a Cauchy.

In all other cases, I'm pretty certain, the "scale" of the distribution is not preserved on addition.
So, convergence may not happen with the Cauchy -- but could with the others. (even if very slowly...)

I have a comment to make to Chiro, but I know enough that I probably ought to re-start the thread with more consistent and accurate labeling; At this point, I think certain things being confounded early on in the thread are preventing a wider participation; and clarifying, and summing up the useful things said in a concentrated new OP would be best.
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 If you want to check theoretically if it is preserved and you have an exact analytic representation then you should firstly look at the nature of the MGF. If you can combine the MGF's to get something that has the same form then that's it. Also with regards to the sigma, when you are estimating the mean you want the sigma to get smaller and this is what you should get since the average of the sample is an unbiased estimator for the mean. I know in this example you have two distinct random variables so you are not actually doing it in the above context, but the average of distributions will always make the variance smaller than the sum by using standard variance operator laws. I'd be interested though if you derived an analytic form of a distribution by using the censoring technique we discussed earlier in this thread, because that would really nail the behaviour of what is going on when you start to allow values close to 0. Did you look into this if not analytically, via simulation of sorts?