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I was thinking about the meaning of location-scale family. My understanding is that for every $X$ member of a location scale family with parameters $a$ location and $b$ scale, then the distribution of $Z =(X-a)/b$ does not depend of any parameters and it's the same for every $X$ belonging to that family.

So my question is could you provide an example where two random from the same distribution family are standardized but that does not results in a Random Variable with the same distribution?

Say $X$ and $Y$ come from the same distribution family (where with family I mean for example both Normal or both Gamma and so on ..). Define:

$Z_1 = \dfrac{X-\mu}{\sigma}$

$Z_2 = \dfrac{Y-\mu}{\sigma}$

we know that both $Z_1$ and $Z_2$ have the same expectation and variance, $\mu_Z =0, \sigma^2_Z =1$.

But can they have different higher moments?

My attempt to answer this question is that if the distribution of $X$ and $Y$ depends on more than 2 parameters than it could be. And I am thinking about the generalized $t-student$ that has 3 parameters.

But if the number of parameters is $\le2$ and $X$ and $Y$ come from the same distribution family with the same expectation and variance, then does it mean that $Z_1$ and $Z_2$ has the same distribution (higher moments)?

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    $\begingroup$ Yes, they can. But, you would need at least 3 parameters in a generalized distribution. $\endgroup$ – Carl Dec 23 '17 at 15:24
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    $\begingroup$ @Carl One parameter will suffice. $\endgroup$ – whuber Dec 23 '17 at 17:46
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    $\begingroup$ @Carl It's unclear what you mean by "same distribution." Literally, that would refer to a unique distribution, with one law and therefore a unique expectation, unique variance, and unique moments (to the extent they are defined). If you mean "same distribution family," then your remark is meaningless, because the family is whatever you define it to be. $\endgroup$ – whuber Dec 23 '17 at 19:29
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    $\begingroup$ @HardCore Since it seems you feel your question has been answered, please see What should I do when someone answers my question? $\endgroup$ – Glen_b Dec 24 '17 at 23:17
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    $\begingroup$ @Carl I did upvote your answer too. The OP's usage seems to support the notion of $Z=(X-a)/b$ as having the same standard distribution for all choices of $X$ in the family. Let's see which answer the OP accepts (if the OP ever reads Glen_b's comment and acts on it). $\endgroup$ – Dilip Sarwate Dec 25 '17 at 16:17
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There is apparently some confusion as to what a family of distributions is and how to count free parameters versus free plus fixed (assigned) parameters. Those questions are an aside that is unrelated to the intent of the OP, and of this answer. I do not use the word family herein because it is confusing. For example, a family according to one source is the result of varying the shape parameter. @whuber states that A "parameterization" of a family is a continuous map from a subset of ℝ$^n$, with its usual topology, into the space of distributions, whose image is that family. I will use the word form which covers both the intended usage of the word family and parameter identification and counting. For example the formula $x^2-2x+4$ has the form of a quadratic formula, i.e., $a_2x^2+a_1x+a_0$ and if $a_1=0$ the formula is still of quadratic form. However, when $a_2=0$ the formula is linear and the form is no longer complete enough to contain a quadratic shape term. Those who wish to use the word family in a proper statistical context are encouraged to contribute to that separate question.

Let us answer the question "Can they have different higher moments?". There are many such examples. We note in passing that the question appears to be about symmetric PDFs, which are the ones that tend to have location and scale in the simple bi-parameter case. The logic: Suppose there are two density functions with different shapes having two identical (location, scale) parameters. Then there is either a shape parameter that adjusts shape, or, the density functions have no common shape parameter and are thus density functions of no common form.

Here, is an example of how the shape parameter figures into it. The generalized error density function and here, is an answer that appears to have a freely selectable kurtosis.

enter image description here

By Skbkekas - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6057753

The PDF (A.K.A. "probability" density function, note that the word "probability" is superfluous) is $$\dfrac{\beta}{2\alpha\Gamma\Big(\dfrac{1}{\beta}\Big)} \; e^{-\Big(\dfrac{|x-\mu|}{\alpha}\Big)^\beta}$$

The mean and location is $\mu$, the scale is $\alpha$, and $\beta$ is the shape. Note that it is easier to present symmetric PDFs, because those PDFs often have location and scale as the simplest two parameter cases whereas asymmetric PDFs, like the gamma PDF, tend to have shape and scale as their simplest case parameters. Continuing with the error density function, the variance is $\dfrac{\alpha^2\Gamma\Big(\dfrac{3}{\beta}\Big)}{\Gamma\Big(\dfrac{1}{\beta}\Big)}$, the skewness is $0$, and the kurtosis is $\dfrac{\Gamma\Big(\dfrac{5}{\beta}\Big)\Gamma\Big(\dfrac{1}{\beta}\Big)}{\Gamma\Big(\dfrac{3}{\beta}\Big)^2}-3$. Thus, if we set the variance to be 1, then we assign the value of $\alpha$ from $\alpha ^2=\dfrac{\Gamma \left(\dfrac{1}{\beta }\right)}{\Gamma \left(\dfrac{3}{\beta }\right)}$ while varying $\beta>0$, so that the kurtosis is selectable in the range from $-0.601114$ to $\infty$.

That is, if we want to vary higher order moments, and if we want to maintain a mean of zero and a variance of 1, we need to vary the shape. This implies three parameters, which in general are 1) the mean or otherwise the appropriate measure of location, 2) the scale to adjust the variance or other measure of variability, and 3) the shape. IT TAKES at least THREE PARAMETERS TO DO IT.

Note that if we make the substitutions $\beta=2$, $\alpha=\sqrt{2}\sigma$ in the PDF above, we obtain $$\frac{e^{-\frac{(x-\mu )^2}{2 \sigma ^2}}}{\sqrt{2 \pi } \sigma }\;,$$

which is a normal distribution's density function. Thus, the generalized error density function is a generalization of the normal distribution's density function. There are many ways to generalize a normal distribution's density function. Another example, but with the normal distribution's density function only as a limiting value, and not with mid-range substitution values like the generalized error density function, is the Student's$-t$ 's density function. Using the Student's$-t$ density function, we would have a rather more restricted selection of kurtosis, and $\textit{df}\geq2$ is the shape parameter because the second moment does not exist for $\textit{df}<2$. Moreover, df is not actually limited to positive integer values, it is in general real $\geq1$. The Student's$-t$ only becomes normal in the limit as $\textit{df}\rightarrow\infty$, which is why I did not choose it as an example. It is neither a good example nor is it a counter example, and in this I disagree with @Xi'an and @whuber.

Let me explain this further. One can choose two of many arbitrary density functions of two parameters to have, as an example, a mean of zero and a variance of one. However, they will not all be of the same form. The question however, relates to density functions of the SAME form, not different forms. The claim has been made that which density functions have the same form is an arbitrary assignment as this is a matter of definition, and in that my opinion differs. I do not agree that this is arbitrary because one can either make a substitution to convert one density function to be another, or one cannot. In the first case, the density functions are similar, and if by substitution we can show that the density functions are not equivalent, then those density functions are of different form.

Thus, using the example of the Student's$-t$ PDF, the choices are to either consider it to be a generalization of a normal PDF, in which case a normal PDF has a permissible form for a Student's$-t$'s PDF, or not, in which case the Student's$-t$ 's PDF is of a different form from the normal PDF and thus is irrelevant to the question posed.

We can argue this many ways. My opinion is that a normal PDF is a sub-selected form of a Student's$-t$ 's PDF, but that a normal PDF is not a sub-selection of a gamma PDF even though a limiting value of a gamma PDF can be shown to be a normal PDF, and, my reason for this is that in the normal/Student'$-t$ case, the support is the same, but in the normal/gamma case the support is infinite versus semi-infinite, which is the required incompatibility.

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    $\begingroup$ (-1) As has been stated in other comments, the issue is "what does a distribution family mean?". I can easily define a new "family" of distributions that are simply rescaled t-distributions to have mean = 0, sd = 1, with a single parameter: df. Then the 1st and 2nd moments are equal for all df, but for different values of df, they have different higher moments. $\endgroup$ – Cliff AB Dec 25 '17 at 4:26
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    $\begingroup$ Hard Core, that comment is difficult to fathom, given that your title itself contains the word "family"! Moreover, if you deny that a family is meaningful, then the question makes no sense. Please clarify by editing your question to reflect your intentions. $\endgroup$ – whuber Dec 26 '17 at 19:29
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    $\begingroup$ -1 because you start by saying "The answer is NO." and then proceed to give an example that effectively answers Yes (another example is given in kjetilbhalvorsen's answer that you favourably mention). This does not make sense to me. I think the math here is clear to all of us, so my downvote is only for the lack of consistency in presentation. $\endgroup$ – amoeba Dec 26 '17 at 19:51
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    $\begingroup$ Carl, there is a stark inconsistency between the question and Hard Core's comments. The question is explicit: to "provide an example where two random [variables] from the same distribution family are standardized but that does not result in ... Random Variable[s] with the same distribution." Obviously some meaning of "family" is intended. The usual meaning is clear, despite there being various technical variants around, and the (easily demonstrated) correct answer is "yes, there are many such examples." $\endgroup$ – whuber Dec 26 '17 at 19:52
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    $\begingroup$ Thank you. Clearly you have a good conception of what you're writing about, but unfortunately your post propagates quite a bit of confusion about what the meanings of "distribution," "shape," "form," and "parameter" might be. As one example of the subtleties, consider a family of distributions created by any distribution law $F$ that has nonzero third central moment. The family is indexed by two real numbers $(\mu,\sigma\ne 0)$ and consists of all laws $x\to F(\sigma x+\mu)$. It is a location-scale family, but the shapes of these laws differ depending on the sign of $\sigma$. $\endgroup$ – whuber Dec 26 '17 at 20:05
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If you want an example which is an "officially named parameterized distribution family, you can look into the generalized gamma distribution, https://en.wikipedia.org/wiki/Generalized_gamma_distribution. This distribution family has three parameters, so you can fix mean and variance and still have freedom to vary higher moments. From the wiki page, the algebra do not look inviting, I would rather to do it numerically. For statistical applications, search this site for gamlss, which is an extension of gam (generalized additive models, in itself a generalization of glm's) which have parameters for "location, scale and shape".

Another example is the $t$-distributions, extended to be a location-scale family. Then the third parameter will be the degrees of freedom, which will wary the shape for a fixed location and scale.

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    $\begingroup$ Although the generalized error distribution may have been a better choice. $\endgroup$ – Carl Dec 23 '17 at 21:55
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    $\begingroup$ Thank you very much for your answer!! I choose Carl's one because it was more detailed but this was fine too .. thank you very much !!! $\endgroup$ – Hard Core Dec 26 '17 at 0:37
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There is an infinite number of distributions with mean zero and variance one, hence take $\epsilon_1$ distributed from one of these distributions, say the $\mathcal{N}(0,1)$, and $\epsilon_2$ from another of these distributions, say the Student's $t$ with 54 degrees of freedom rescaled by $\sqrt\frac{1}{3}$ so that its variance is one, then $$X=\mu+\sigma\epsilon_1\qquad\text{and}\qquad Y=\mu+\sigma\epsilon_2$$ enjoy the properties you mention. The "number" of parameters is irrelevant to the property.

Obviously, if you set further rules to the definition of this family, like stating for instance that there exists a fixed density $f$ such that the density of $X$ is $$\frac{1}{\sigma^d} f(\{x-\mu\}/\sigma)$$ you may end up with a single possible distribution.

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  • $\begingroup$ thank you for the answer but I think that this is not what I asked $\endgroup$ – Hard Core Dec 23 '17 at 17:46
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    $\begingroup$ I think it does because if the family of distributions is defined by the reunion of both the distributions of the $X$'s and the $Y$'s, then you have a contradiction to the property. A "family" of distributions is quite a vague notion. $\endgroup$ – Xi'an Dec 23 '17 at 17:48
  • $\begingroup$ yes in fact is quite vague but if you read my question I wrote that in this context with family I mean for example both Normal or both Gamma and so on .. You made an example with one normal and one t student $\endgroup$ – Hard Core Dec 23 '17 at 18:13
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    $\begingroup$ Hard Core, you seem to confuse the name of a family with its concept. This answer is a fine one and nicely illustrates the concept. Your question doesn't ask that the solution be a location-scale family. If you need it to be one, you can always take this answer--or any other answer--and prolong it to a location-scale family by allowing arbitrary translations and rescalings. Xi'an's point about the number of parameters still holds. $\endgroup$ – whuber Dec 23 '17 at 19:32
  • $\begingroup$ @whuber I think it is confused as an answer. Student's-t by itself would be a better answer, rather than use the extreme answer of $df=3,\infty$ and not specify it. Indeed, it is $df$ which is the third parameter. $\endgroup$ – Carl Dec 23 '17 at 21:59
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I think you are asking whether two random variables coming from the same location-scale family can have the same mean and variance, but at least one different higher moment. The answer is no.

Proof: Let $X_1$ and $X_2$ be two such random variables. Since $X_1$ and $X_2$ are in the same location-scale family, there exist a random variable $X$ and real numbers $a_1>0, a_2>0, b_1, b_2$ such that $X_1 \stackrel{d}{=} a_1 X + b_1$ and $X_2 \stackrel{d}{=} a_2 X + b_2$. Since $X_1$ and $X_2$ have the same mean and variance, we have:

  1. $E[X_1] = E[X_2] \implies a_1 E[X] + b_1 = a_2 E[X] + b_2$.
  2. $\operatorname{Var}[X_1] = \operatorname{Var}[X_2] \implies a_1^2 \operatorname{Var}[X] = a_2^2 \operatorname{Var}[X]$.

If $\operatorname{Var}[X] = 0$, then $X_1=E[X_1]=X_2=E[X_2]$ with probability $1$, and hence the higher moments of $X_1$ and $X_2$ are all equal. So we may assume that $\operatorname{Var}[X] \neq 0$. Using this, (2) implies that $|a_1|=|a_2|$. Since $a_1>0$ and $a_2>0$, we have in fact that $a_1=a_2$. In turn, (1) above now implies that $b_1=b_2$. We therefore have that: $$ E[X_1^k] = E[(a_1X+b_1)^k] = E[(a_2X+b_2)^k] = E[X_2^k], $$ for any $k$, i.e., all moments of $X_1$ and $X_2$ are all equal.

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    $\begingroup$ (+1) I cannot find fault with this answer. Apparently someone does, and they also find fault with mine. I do not understand this unexplained behaviour. $\endgroup$ – Carl Dec 24 '17 at 18:28
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    $\begingroup$ @Carl This answer is incorrect--that's why it's being downvoted. Xi'an has already provided a counterexample. $\endgroup$ – whuber Dec 24 '17 at 20:22
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    $\begingroup$ @whuber Please see my comments under Xi'an's answer. I do not agree with him but did not downvote because both he and you have a right to your opinion, even if I consider it to be incorrect. $\endgroup$ – Carl Dec 25 '17 at 1:31
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    $\begingroup$ @Carl After re-reading this answer, I need to retract my original assessment: this answer is correct (and +1 for that), and it is correct because it clearly explains how it is interpreting the original question. (Specifically, there is a common yet narrow concept of a "location-scale family" as consisting of just a single standard distribution along with all its translates and positive rescalings.) I believe the original question was intended to ask something a little different; the basis of that belief is the reference to more than two parameters in the post. $\endgroup$ – whuber Dec 25 '17 at 23:30
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    $\begingroup$ I am sorry if I have not been very clear and I thank you for the time you have spent for looking into this but that is not what I asked. $\endgroup$ – Hard Core Dec 26 '17 at 0:33
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Since the question can be interpreted in multipe ways I will split this answer into two parts.

  • A: distribution families.
  • B: location-scale distribution families.

The problem with case A can be easily answered/demonstrated by many families with a shape parameter.

The problem with case B is more difficult since one and a half parameters seem to be sufficient to specify location and scale (location in $\mathbb{R}$ and scale in $\mathbb{R_{>0}}$), and the problem becomes whether two parameters can be used to encode (multiple) shapes in addition as well. This is not so trivial. We can easily come up with specific two parameter location scale families and demonstrate that you do not have different shapes, but it does not proof that this is a fixed rule for any two parameter location scale family.

A: Can two different distributions from the same 2 parameter distribution family have the same mean and variance?

The answer is yes and it can already be shown using one of the explicitly mentioned examples: the normalized Gamma distribution

Family of normalized gamma distributions

Let $Z = \frac{X-\mu}{\sigma}$ with $X$ a Gamma distributed variable. The (cumulative) distribution of $Z$ is as below:

$$F_Z(z;k) = \begin{cases} 0 & \quad \text{if} & z < -\sqrt{k}\\ \frac{1}{\Gamma(k)} \gamma(k, {z\sqrt{k}+k}) & \quad \text{if} & z \geq -\sqrt{k} \end{cases} $$

where $\gamma$ is the incomplete gamma function.

So here it is clearly the case that different $Z_1$ and $Z_2$ (distributions from the family of normalized gamma distributions) can have same mean and variance (namely $\mu=0$ and $\sigma=1$) but differ based on the parameter $k$ (often denoted 'shape' parameter). This is closely linked to the fact that the family of gamma distributions is not a location-scale family.

B: Can two different distributions from the same 2 parameter location-scale distribution family have the same mean and variance?

I believe that the answer is no if we consider only smooth families (smooth: a small change in the parameters will result in a small change of the distribution/function/curve). But that answer is not so trivial and when we would use more general (non-smooth) families then we can say yes, although these families only exist in theory and have no practical relevance.

Generating a location-scale family from a single distribution by translation and scaling

From any particular single distribution we can generate a location-scale family by translation and scaling. If $f(x)$ is the probability density function of the single distribution, then the probability density function for a member of the family will be

$$f(x;\mu,\sigma) = \frac{1}{\sigma}f(\frac{x-\mu}{\sigma})$$

For a location-scale family that can be generated in such way we have:

  • for any two members $f(x;\mu_1,\sigma_1)$ and $f(x;\mu_2,\sigma_2)$ if their means and variances are equal, then $f(x;\mu_1,\sigma_1) = f(x;\mu_2,\sigma_2)$

Can for all two parameter location-scale families their member distributions be generated from a single member distribution by translation and scaling?

So translation and scaling can convert a single distribution into a location-scale family. The question is whether the reverse is true and whether every two parameter location-scale family (where the parameters $\theta_1$ and $\theta_2$ do not necessarily need to coincide with the location $\mu$ and scale $\sigma$) can be described by a translation and scaling of a single member from that family.

For particular two parameter location-scale families like the family of normal distributions it is not too difficult to show that they can be generated according to the process above (scaling and translating of single example member).

One may wonder whether it is possible for every two parameter location-scale family to be generated out of a single member by translation and scaling. Or a conflicting statement: "Can a two parameter location-scale family contain two different member distributions with the same mean and variance?", for which it would be necessary that the family is a union of multiple subfamilies that are each generated by translation and scaling.

Case 1: Family of generalized Students' t-distributions, parameterized by two variables

A contrived example occurs when we make some mapping from $R^2$ into $R^3$ (cardinality-of-mathbbr-and-mathbbr2) which allows the freedom to use two parameters $\theta_1$ and $\theta_2$ to describe a union of multiple subfamilies that are generated by translation and scaling.

Let's use the (three parameter) generalized Student's t-distribution:

$f(x;\nu,\mu,\sigma) = \frac{\Gamma \left( \frac{\nu + 1}{2} \right) }{\Gamma \left( \frac{\nu}{2} \right) \sqrt{\pi\nu}\sigma} \left(1 + \frac{1}{\nu} \left( \frac{x-\mu}{\sigma} \right)^2 \right)^{-\frac{\nu+1}{2}}$

with the three parameters changed as following $$\begin{array}{rcl} \mu &=& \tan (\theta_1)\\ \sigma &=& \theta_2\\ \nu &=& \lfloor 0.5+\theta_1/\pi \rfloor \end{array}$$

then we have

$f(x;\theta_1,\theta_2) = \frac{\Gamma \left( \frac{\lfloor 0.5+\theta_1/\pi \rfloor + 1}{2} \right) }{\Gamma \left( \frac{\lfloor 0.5+\theta_1/\pi \rfloor}{2} \right) \sqrt{\pi\lfloor 0.5+\theta_1/\pi \rfloor}\theta_2} \left(1 + \frac{1}{\lfloor 0.5+\theta_1/\pi \rfloor} \left( \frac{x-\tan(\theta_1)}{\theta_2} \right)^2 \right)^{-\frac{\lfloor 0.5+\theta_1/\pi \rfloor+1}{2}}$

which may be considered a two parameter location-scale family (albeit not very useful) that can not be generated by translation and scaling of only a single member.

Case 2: Location-scale families generated by negative scaling of a single distribution with nonzero skew

A less contrived example, than using this tan-function, is given by Whuber under the comments of Carl's answer. We can have a family $x \mapsto f(x/b + a)$ where flipping the sign of $b$ keeps the mean and variance unchanged but possibly changing the uneven higher moments. So this gives a bit more easily a two parameter location-scale family where members with the same mean and variance can have different higher order moments. This example from Whuber can be split into two subfamilies each of which can be generated out of a single member by translation and scaling.

Smooth families

If we try to make a single smooth two parameter distribution family (smooth: a small change in the parameters will result in a small change of the distribution/function/curve) by somehow making a composition of two or more families that are generated by translation and scaling, then we get into problems to have the two parameters cover both the variation of 'mean' and 'variance', as well as the third parameter 'shape'. A formal proof will have to go along the same lines as the answer to the question: Is there a smooth surjective function $f:\mathbb{R}^2 \mapsto \mathbb{R}^3$? (where the answer is no in the case of smooth, ie. infinitely differentiable, functions although there are continuous functions that would do the job such as Peano curves).

Intuition: Imagine there would be some parameters $\theta_1$, $\theta_2$ that describe the distributions in some location-scale distribution family and by which we can change the mean and variance as well as some other moments, then we should be able to express $\theta_1$, $\theta_2$, in terms of the mean $\mu$ and variance $\sigma$

$$\begin{array}{rcl} \theta_1 &= &f_{\theta_1}(\mu,\sigma) \\ \theta_2 &=& f_{\theta_2}(\mu,\sigma)\end{array}$$

but these need to be multiple valued functions and these can not make continuous transitions, the different values from $f_{\theta_1}(\mu,\sigma)$ for a particular $\mu$ and $\sigma$ are not continuous, and will not be able to model a continuous shape parameter.

I am actually not so sure about this final part. We could possibly use a space-filling curve (such as the Peano curve, if only we knew how to express coordinates on the curve to coordinates of the hypercube) to have a single parameter $\theta_1$ completely model multiple features like mean and variance, without giving up the property that a small change of the parameter $\theta_1$ is equivalent to a small change of the function $f(x;\theta_1)$ at every $x$

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    $\begingroup$ I stopped reading after the initial definitions because they are so unclear and contradictory. By "integrate" you of course mean integration over $x$ only. By "$f,$" though, you must mean the CDF and not the PDF, because the division by $b\ne 1$ changes the integral. By not imposing any restrictions on how $f$ can vary with $\theta$ you also adopt a much broader concept of "family" than is usual. Only that allows you to discuss a "map from $R^2$ to $R^3.$" The problem with these "maps" is they cannot be continuous and will have no statistical meaning. $\endgroup$ – whuber Aug 27 '18 at 11:50
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    $\begingroup$ I'm not objecting to simplicity or the language, but to the confusion that is being sown. The problem with your $R^2\to R^3$ map points out why you need to impose additional mathematical structure--a suitable topology--on the family. Allowing the distributions to change in such a (violently) discontinuous manner with $\theta$ is not only impractical and meaningless, it would likely invalidate useful methods and theorems for no good reason. For instance, MLE is almost always performed under the assumption that the distribution varies with $\theta$ in a piecewise differentiable manner. $\endgroup$ – whuber Aug 27 '18 at 13:32
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    $\begingroup$ The second bullet is incorrect: it neither follows from any of the assumptions nor is it part of the definition of a location-scale family. $\endgroup$ – whuber Aug 28 '18 at 14:11
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    $\begingroup$ It is tremendously confusing because now all references to the $\theta_i$ are superfluous. I believe the quantifiers now in your statement might not convey correctly the idea you have. Why not just drop the $\theta_i$ and simply state that the family consists of the set of distributions $x \to F(bx + a)$ for one given $F$ and all $(a,b)\in\mathbb{R}^2$ with $b\gt 0$? There's no need to refer to means and variances, either--that's just a distraction from the essential idea, which does not require $F$ to have any moments at all. $\endgroup$ – whuber Aug 28 '18 at 14:52
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    $\begingroup$ @whuber if you are generating location-scale family from one single example then indeed it would seem like it is much easier to use $\mu$ and $\sigma$. Here I am however imagining that we already have a family of curves parameterized by some alternative $\theta_1$ and $\theta_2$ and I wonder whether it could be possible that such a family contains more curves than just the curves created by scaling one member with $\mu$ and $\sigma$ (as in the transformation with the tangent). I will see if I can change the formulation somehow again (do you disagree with the idea or with the formulation?). $\endgroup$ – Martijn Weterings Aug 30 '18 at 7:51

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