There is going to be a subjective risk factor of how much a person will be willing to gamble with. That will vary from person to person and doesn't say anything about the objective gain/loss.

You are only willing to make that traded because it is only 10 bucks. That is not much to lose in the first place. It is a subjective criteria and it does not say anything about the actual probability.

What would you do if you opened envelope A and saw 1000 bucks?

This is not a paradox and in fact the solution is straight forward and simple. This is an exercise in observational bias. Separating the objective and subjective without all those mitigating safeguards we are so use to.

People come up with an idea of how they think it works, then they model their belief and when their model matches their beliefs they decide that confirms their beliefs, it turns into a type of confirmation bias. Obviously if you model your subjective beliefs then your model will confirm your subjective beliefs, that is no paradox. You have to separate the objective process from the subjective process.

The secret is to realize that $Y$ is a trap.

You never need to know the distribution or even guess at it.

We have $X=f(x_i)$ for some unknown population. Where $x_i$ is some positive real number on a countable domain.

Then you have two more variables, the envelopes $A$ and $B$, which the contents of have a relationship.

Which can be displayed with this makeshift table.

1 and 2 represent which case you are in.

It is important to understand the significance of this relationship,
as $P(A_i=x_i) = P(B_i=2x_i)$
and $P(A_i=2x_i) = P(B_i=x_i)$
which means $P(Y_a)=P(Y_b)$

Knowledge of the distribution is not needed, nor is a prior, as the value in envelope $B$, has the same exact chance of occurring as the value in envelope $A$.

Consider an example were we know the contents of both envlopes

$A=10 =x_i$ and $B=20 = 2x_i$

You get $A$ and you see the 10, what you didn't know, however, that there was only a 2% of getting this 10 bucks, which means, due to the relationship there was only a 2% chance that $B$ ended up with the 20 it did. This equality will be true in every single case for every possible distribution.

Now you might be tempted to think well what about the chance of 5 bucks? I have ten the other one may have 5. It won't matter, the equality would hold, it would just be 2% chance for 10 and likewise 2% chance for 5. Whatever the case may be and whatever you think might be in the other envelope that equality will always be true: That is one thing you know and can count on.

This means $P(a_i) = P(b_i)$ and since probability must sum to one and you only have two choices then it is a 50/50 split. Some people have been ignoring this relationship and treating the envelopes as merely two different amounts, but that is not congruent with the OP and it is an error when considering the probability.

Whatever the chance of the value you see when opening your envelope it has exactly the same chance as the contents of the other envelope.

Repeating over and over that this is "not a statistics problem" is not a valid response to anything I am doing. That is nothing but blind categorical dismissal.

I will not consider "this is not a statistics problem" as valid or thoughtful criticism and I will continue to evaluate this by the methods I consider most appropriate.

The sample distribution of a statistic is the distribution of that statistic, considered as a random variable, when derived from a random sample of size n. Since we have no such sampling, let alone a statistic, there is no sample distribution, or use for one. Period. This just isn't a statistics problem. — JeffJo

How does this relate to what I am doing? As far as I can tell it doesn't.

I am not talking about a model of the probability, I am talking about modeling the problem. We build models for two reasons, to make a predictions or to understand relationships. By working though the model we come to a better understanding.

Andrewk had a completely valid point about things being ill-defined and one that has never been fully addressed. I don't agree with the methods so far that have been used to try and map this out.

This is not well-defined. — andrewk

Let me give you an example of what I mean.

The OP has two random variables that have yet to be defined, $X$ and $Y$.

Since $X$ is undefined, for the moment I will stick with my previous definition. Although I am not fully satisfied it.

For $Y$ maybe something like this:

$y_i = \beta_0 +I_1\beta_1A_i+I_2\beta_2B_i+\varepsilon_{ij}$

This would just be an example of what I mean, I am still hammering it out. Something to think about during the dull points in my day.



I will work at defining everything as long as my interest hold. Mapping everything out is good practice and it helps one to understand everything much better.

Also even without a sample distribution a theoretical model can still be set up. That is where my interests are atm. However, sigma is still a problem when going the Bayesian route.

The normal distribution is widely used to make probabilistic claims about unknown populations, which have discrete values and unknown limits. If you want to know how this magical process works, pick up an introductory statistics book.

*Edit, sometimes I use my phone, and it just turns out a mess.

Doesn't matter, I am not using it as proxy.

There is a problem with the third chance mechanism, the one where we get the chance to switch, and if it should be included in the model, as it is subjective, the criteria used will vary from person to person. I could treat it as an arbitrary do/don't switch logical vector, but that would not really reflect the subjectivity inherited in that final step. It may be better to cut the model off at the point when the envelope is handed to the player.

There are actually two random variables in the OP, which may need two models.

It may need two logical vectors.

My simulation can give a visualization of this.The resulting distribution from the process of filling the envelopes changes the possible outcomes from an independent event to possible outcomes that have a relationship.

Using my simulation from before for an example,

Here is a plot of the values after they have been put into envelopes :

https://ibb.co/nGsEOz

Now for comparison:

Here is the selection before the envelopes are filled.

https://ibb.co/m0uuOz

The second chance mechanism transforms the distribution into one where now we have a relationship to account for.

I actually have been thinking about how to model this problem and I believe I will need to include a logical true/false vector.

The problem here is that at the first chance mechanism $x_{i,1}$ is an independent observation. However, when you get to the second chance mechanism, when the envelopes are filled, this could be thought of as another function which transforms the distribution into one where the values have a relationship. It is not as simple as trying to determine where on the distribution I am, as I also have to account for that relationship.

I am not advocating for this approach but I think this is your best shot at estimating the median of the distribution from a single sample.

People have been assuming a prior that fits their expectations the best; however, if you are gonna model off such assumptions you need to at least try to mitigate observational bias, which is why I think you should use a normal distribution. The point is to lessen the impact of false assumptions, and due to robustness of the normal distribution, I feel this makes it a better choice. Also because of the empirical rule one can better estimate the relative position of $Y$ on the distribution and the normal distribution has a common occurrence in nature. For all these reasons I feel it is a good choice for a prior.

If you assume a normal distribution (already transformed by chance mechanism number 2) as the prior then when you see Y you could say by the empirical rule:

$P(\sigma+\mu \geq Y \geq \mu) = 34\%$
$P(\sigma-\mu \leq Y \leq \mu) = 34\%$
$P(2\sigma+\mu \geq Y \gt \sigma+\mu) = 13.5\%$
$P(2\sigma-\mu \leq Y \lt \sigma-\mu) = 13.5\%$
$P(3\sigma+\mu \geq Y \gt 2\sigma+\mu) = 2.4\%$
$P(3\sigma-\mu \leq Y \lt 2\sigma-\mu) = 2.4\%$

Here is a picture.

Now you just need an estimator for $\sigma$.

Here is a link for the emprical rule, which I think makes a normal distribution a better choice for a prior. https://www.investopedia.com/terms/e/empirical-rule.asp

I have to think about it more.

I was thinking you could use Y to justify your priors. It is the only information you have. Maybe include justification on the emprical rule which says 68% of all observations will be within one SD of the mean; 95% within 2 SD and 99% within 3 SD. If I am remembering that right.

My continue interest here is not in debating the different approaches, but in the modeling of X. If I was going to build a theoretical Baysian model I would use a normal distribution as a prior since when dealing with an unknown population of this sort it would be the most robust. This would include a prior mu and SD.

https://support.sas.com/documentation/cdl/en/statug/63033/HTML/default/viewer.htm#statug_introbayes_sect009.htm

Highest Posterior Density (HPD) Interval

The HDI is 95% of the area under the curve. It is a cut off range.

It'll be fine once we use a MCMC and get the HDI. You forgot the part of my argument where we need a sample distribution.

A normal distribution does not have to have a mid of 0, nor do they need negative values.

There are two mindsets here, people who hinge their expections on what is known and people who hinge their expections on what they feel they can safely assume about the unknown. However, these additional assumptions are not cost free and I am not sure everyone here gets that. When you start making assumptions, about unknowns, which are not backed by a probability model you inflate your uncertainty in a way that can't be quantified.

When dealing with unknown populations even priors needs to be tempered with a sample distribution before you have a posterior distribution that is reliable enough to make probabilistic inferences about the unknown population.

Without a sample distribution, then your assumptions of the unknown distribution carry additional uncertainty that cannot be quantified. These additional assumptions come at a cost.

A normal prior would actually make more sense, as empirical investigations have shown it robust against possible skewness.

No, it is not.

A random variable is defined by a real world function.

Each observed x is an independent event.

It does not even matter, as the second chance mechanism transforms the distribution.

You use samples to make probabilistic inferences about an unknown population. I roll a die 10 times and 9 of those times get a 6. Now it is not impossible to get 6 nine times but it is highly improbable, therefore I decide the die is loaded. That is how it works; you use the samples to make probabilistic claims of an unknown population. Samples we are lacking.

Now not only do we have an unknown population but we have an unknown function and that means unknown exploratory variables. Maybe Farmer Bob counted how many eggs his chickens laid the night before and that is how x was decided.

But in the ideal case — Pierre-Normand

Claiming this case is "ideal" is an entirely subjective standard pumped full of observational bias.

The thing to recognized here is that we have a nested structure. There are three chance mechanisms, with each housed in the last.

Let $x_{ij}, i = 1, 2, 3,...,n, j=1,2,3$ ( j is which chance mechanisms we are at) denote the observed value of the random variable $X$. Then $P(X=x_{i,1})$, which is the sum of probabilities for the sample points that are assigned to $x_{i,1}$.

Sorry if my nested notation is a bit off, it has been awhile and I don't feel like looking it up, but basically we need another index to tell which chance mechanisms we are at, in our case I am using j.

Since $X$ is a random variable then it is a response variable of some unknown real-world function in which the unknown domain is the sample space. The domain would actually spreads between two unknowns. I know zero has often been the assumed min, but we actually have no idea what that is; as far as we know 100 bucks could be the lowest value.


Fist chance mechanism.

By whatever function of $X$ an $x_{ij}$ will be selected, we'll call this value $x_{1,1}$.

Second chance mechanism

We have two envelopes, lets call one $A_i$ and the second $B_i$. We need to define $A_i$ and $B_i$ at this point. So using a simple if then statement let's define them as: If $A_i = \{x_{i,2}\}$ then $B_i = \{2x_{i,2}\}$ or if $A_i = \{2x_{i,2}\}$ then $B_i = \{x_{i,2}\}$.

At this level of the nested structure $x_{1,2}$ is put into one envelope while $2x_{1,2}$ is put into the other envelope.

The possible assignments are:

$A_1 = \{x_{1,2}\}$
$B_1 = \{2x_{1,2}\}$
$A_2 = \{2x_{1,2}\}$
$B_2 = \{x_{1,2}\}$

So the two possible cases are:

$E_1 = \{A_1, B_1\}$
$E_2 = \{A_2, B_2\}$

Then $P(E_i) = 1/2$, with $P(A_1) = 1/4 = P(x_{1,2}) = P(B_2)$ and $P(A_2) = 1/4 = P(2x_{1,2}) = P(B_1)$, meaning $P(x_{1,2}) = 1/2$ and $P(2x_{1,2}) =1/2$.

Third chance mechanism

The third chance mechanism is the subjective one, as it up to us to decide by what means what we want to do and I think this is the real trick to this so-called "paradox". It says, "What should you do?", and that point is really where the conflicts arises.

I still maintain that our gain/loss is: $P(-x_{1,3}) = 1/2, P(x_{1,3}) = 1/2$. However, I set this up, because I think it is important people understand where and how probabilities are being applied.

The probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer from this the characteristics of an unknown population.

Mathematical Statistics with Applications, Wackerly, Mendenhall, Scheaffer

This is an important separation to keep in mind, as the conceptual difference between these approaches to probability is not the number of observations, I to fell into that misleading line of thought myself, but the real difference is in the direction in which they move. One moves from a known population to the sample, while the other moves from the sample to the unknown population. To take a "probabilist" approach to the unknown population is to go in ill equipped to do that job. I think a few of the people here have little experience in dealing with unknown populations and in that naiveness they have moved in the wrong direction.

Calling me a troll is a personal attack.

Also. . . ,

Since I have already displayed in this thread that x can come from any number of continuous distributions, this means we have no clue how and from where x was selected, and if we don't know then we should apply the the principle of indifference, right? Of course there are infinite possible distributions, and one over infinity is 0.

Something else I would like to point out, is that assuming that the probability of any given x must come from a discrete distribution is not necessarily true. In fact I used a selection method where the actual chance mechanism was applied to a continuous distribution in this very thread.



I used a continuous uniform distribution to randomly selected an x then formatted it into real dollar values. Now who is to say that such a chance mechanism was not used to fill the envelopes?

If a probabilistic approach is to reason from a known population then the known population is x or 2x. Where x came from and how it was chosen is something we don't know.

you haven't refuted them. — JeffJo

That's because I accept that you use a different vernacular.

What is the difference between experimental data and observational data?

There are some interesting aspects to note if you diagram the 1.25X argument in comparison to the other two. Which clearly shows why it is a faulty argument.

When considering the unselected envelope under the {x,2x} sample space objectively you have one true value and one false.

Likewise, when considering the unselected envelope under the {x/2.2x} sample space objective, you have one true value and one false.

However, it get's different when you consider the sample space {{x/2,x},{x,2x}}. Here you have objectively one true set and one false set. If I end up in the false set then I have two statements that are false. If I end up in the true set then I have one statement that is true, which only has a 1/4 weight in my consideration when it should have a 1/2 weight. So I reduce my chance of getting the true value, meaning I inflate my possibility of error.

Here I sketched it out by hand to show it.

https://ibb.co/dB7BUT

I have also been thinking about the [0,M] argument.

One of my books makes this distinction.

The probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer from this the characteristics of an unknown population.

Mathematical Statistics with Applications, Wackerly, Mendenhall, Scheaffer

I would argue that the known population are the amounts in the envelope x and 2x and the unknown population is the distribution that x was selected from.