Master Yaser and Dr. Pinheiro: S1-Convexity

Master Yaser Maleki                               Master Science (Mathematics) Tarbiat Modares University Tehrān  E-mail yasermaleki71@gmail.com Web: Research Gate

Dr. Marcia Pinheiro Lecturer at IICSE University Certified Translator and Interpreter Portuguese & English NAATI  40296          Member: PROz, RGMIA, Ancient Philosophy PhD in Philosophy and Mathematics Master in Philosophy Certified TESOL/TEFL professional Licentiate in Mathematics PO Box 12396 A’Beckett St Melbourne, VIC, AU, 8006 Tel 0416915138 E-mail drmarciapinheiro@gmail.com Web Research Gate

Master Yaser, have you heard of Hudzik, Maligranda, and their S₁-Convexity before? Have you heard of convex functions? 

Yes, I know convex functions but I haven't heard anything about S₁-Convexity. 

Usually people like the concept of convex function because it is a lot graphical, is it not? In the Universe of the Real Numbers, a convex function is a function with a graph that is built in such a way that regardless of which couple of points we pick on its line (it will be a line for our eyes, right?), the curve representing the function will always be either over or under that line. 

Yes it is graphical and this property helps us understand convex functions better than other types of function. Also people who do not have a mathematical knowledge like convex function more than other in fact they understand their eyes. But if we want to talk about Mathematics, in particular Pure Mathematics, we know that it is not necessary for a function to have a graphical property, but yes it is interesting.

As we can see in the graph above, whatever that is part of the curve (blue) between one intersection of the straight line (red) with the curve and another is under the straight line. It does not matter where we draw this straight line that contains two points of our original curve; it is always going to be the same: It is all under the line, at most over it, that is, never above it. We can then say that, in analytical terms, that, for any two elements of the domain of the function we pick, so say x and y, it is always true that the image of ax + by is always less than or equal to a times the image of x plus b times the image of y if a+b=1. 

In our first internationally known paper on S-convexity, which got published by the WSEAS group, due to their conference in Cancun, which we did not actually attend, we wrote: 

That was in 2004, Master Yaser. You can already notice some difference between this picture, from the WSEAS paper, and what I wrote before. Can you?

No, I can't see any differences between them: Everything you say is shown in this definition. I just zoom on the (f:X--->R)². 2 is for footnote or it is part of the definition? 

I think you are talking about a footnote, Master Yaser. You are right, it is all very subtle, but I have been working on this for a while because I think it all matters quite a lot. Please observe the coefficients: In one definition, you see a and b, therefore two constants. In another definition, you see lambda only. You are also immediately told, in the second definition, the one presented at the WSEAS, that it is lambda and 1-lambda, and therefore the sum of the coefficients leads to 1, so that you don't need to write that down, like not only we have reduced the constants to one (we had two constants, now we have one), in terms of coefficients, but we also deleted the extra piece of information: That the coefficients together give us 1. That saves us and makes the definition look more elegant, which can only be part of the objectives of Science when we talk about refining mathematical definitions: more objectivity, simplest presentation as possible, more immediate application, etc. 

The first difference between convex and S-convex functions is their domain. The second difference is the way of choosing a and b, and the third is the conditions on a and b. For instance, a+b=1. The difference between S₁-convex and S₂-convex is in the condition on a+b, perhaps that a+b=1 for one of them.  I can imagine some things, as you can see. Are they true? In the convex function, we see that if we choose two elements in the domain, x and y, then the image of the graph for all points between x and y is under the line (red line) across f(x) and f(y), but I imagine that this S-convex concept does not lead to a line!! I think it is a curve and its  Curvature depends on S. 

                                                                    

You are very right, Master Yaser. It is precisely that, the main points are precisely those. I saw things in the same way you see them in that 2001, since it is all very much obvious. One more detail catches our eyes: Because S is between 0 and 1, and both a and b would have to be between 0 and 1, and you will notice that I actually produced a proof for that in my paper with WSEAS, the first one on the topic that got published by a major vehicle, and the proof follows this paragraph, a to the s would have to be greater than a. That was the key for my understanding of the shape of S-convexity: It is actually a lift on the limiting line for Convexity, and that is why both Hudzik and Maligranda thought that they had a proper extension of the concept. See the proof regarding the coefficients first:

Before we talk about the rest, I think I would like to know if you agree with what I stated before after seeing the proof we presented at the WSEAS (a and b would both be between 0 and 1, and, therefore, we can rewrite the definition of S-convexity given by Hudzik and Maligranda in the way you see in the last picture of the paper presented at the WSEAS in that 2004). Do you agree with that, please, Master Yaser?

I understand the new definition of   S₂-convex  and agree with this, but I don't understand S₁-convexity. In fact I don't understand why you use (1-lambda ^ s)^ (1/s)!!!!

Yes, exactly. Perhaps the first intuition is that it is important to keep the coefficients unaltered because we want to keep the percentages we take in the mix unaltered in terms of base. You will notice however that easy counter-examples exist to prove to us that S₁-Convexity is not a proper extension of Convexity. Perhaps the main question to be asked was always what does extension mean? When we extend something in Mathematics, that means that we have included what we had before in what we have after the extension and we have added a little bit, as a minimum thing, is it not? If we lose something that was part of the something we claim to now be extending, then we must not be extending: We must be creating another class instead. 

See the counter-example to the claim that S₁ extended convexity, which is right below this line. I want to know if you agree with all that is said here. Perhaps you could give your take on extension as well. 

  

I agree with your take on extension in Mathematics. We can also notice that if we find a new class of something, then we also extend the Mathematics involved. However maybe we don't extend an old definition. Now something to make my mind busy: Is S-convex function a subclass of convex function? In other words, is every S-convex function a convex function? 

Master Yaser, the reason for your confusion is probably the fact that you agreed that we have a genuine counter-example to the claim that S₁-convexity extends convexity. I wonder if you have validated my every step in the proof above. Please confirm. 

Yes. Every part of the above extract proves that it is all true and I can't add any comments to that. 

Great, Master Yaser! I think I was eager to get more people saying yes to my results in a meaningful manner, people who are not journal editors. Thanks for that. That means we both agreed that S₁-convexity cannot extend Convexity because, for instance, the group of functions we have just mentioned is part of the class convex real functions but is not part of the class S₁-convex functions. As said before, if a class extends another, we should have at least the group we claim to be extending inside of it. As for convexity, please observe that whenever s=1 you would be recovering this notion both in the definition of S₁- and in the definition of S₂-convexity.

                            References

Pinheiro, M. R. (2004). Exploring the Concept of S-convexity – Proceedings of the 6th WSEAS Int. Conf. on Mathematics and Computers in Physics (MCP '04).

Pinheiro, M. R. (2015). Second Note on the Definition of S1-Convexity. Advances in Pure Mathematics, 5, 127–130.

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Master Angela, Dr. Pinheiro, and the Monty Hall Puzzle: Part 2, Discussing Dr. Pinheiro's Solution

Mrs. Angela Kotsiras Secondary School Mathematics Teacher Founder and editor of mathsteachhersonly.com Master of Education (Mathematics) University of Melbourne Melbourne, VIC, Australia E-mail mrskotsiras@gmail.com

Dr. Marcia Pinheiro Lecturer at IICSE University Certified Translator and Interpreter Portuguese & English NAATI  40296          Member: PROz, RGMIA, Ancient Philosophy PhD in Philosophy and Mathematics Master in Philosophy Certified TESOL/TEFL professional Licentiate in Mathematics PO Box 12396 A’Beckett St Melbourne, VIC, AU, 8006 Tel 0416915138 E-mail drmarciapinheiro@gmail.com Web Research Gate

It seems like we need to reach common ground first and then work from there.

Firstly do you agree that we need to prove that

Pr( winning by switching given the host opens a particular door)= 2/3 ?

No, Angela, I don’t. I will repeat here what I have written on the blog post recently:

In short, if we go Mathematics with the Month Hall Problem, we get ½ of chance of winning, since if the door we chose first is the door they chose, only sticking to our first choice will return win, but we have stick or swap (2 possible choices, only one wins). If the door we chose first is not the one they chose, we only win if we swap, what is one in two again (2 possible choices, only one wins). In this way, 2 wins/4 possible choices or 1/2. If you see things differently, you will have to explain to me how you get to your conclusions. Sorry.

This is how I originally thought too but now I wish to find some common ground and then see all possible outcomes on a tree diagram or a table with the corresponding probabilities.  I agree with the 12 outcomes listed from your first post that were all considered equally likely. The probability of these outcomes may be where I now see things differently.

Firstly I need to know if you agree with the following statements.

1.      The car is randomly allocated to a particular door.

2.      The contestant randomly chooses a door.

3.      The host knows which door will reveal the car.

4.      The door revealed by the host is always a donkey.

5.      The door revealed by the host depends on the door chosen by the contestant.

6.      If the contestant chooses the door that will reveal the car then the host will need to randomly choose to open one of the other two remaining doors.

Problem: What is the probability of winning by switching? 

Yes, Angela. I do agree with your premises, all of them, from 1 to 6, I reckon. I cannot immediately see what could be wrong with any of them, definitely not.

Great! We have a common ground.

So if I want to construct a tree diagram for this problem

I can begin by firstly saying that the car can be randomly allocated to either door 1, door 2 or door 3 each with a probability of 1/3.

Secondly, we can assume that the contestant randomly chooses a door, each with a probability of 1/3. So if the car was behind door 1 the contestant could choose either Door 1, 2 or 3. They could also choose door 1,2 or 3 if the car was behind door 2 or door 3. Hence the following tree diagram  illustrates this.

Thirdly since the host knows which door will reveal the car and the door revealed by the host will always be a donkey, then if the car was behind door 1 and the contestant chose door 1 then the host will need to randomly choose to open one of the two remaining doors with a donkey, each with a probability of 1/2.

This would also be the case if the car was behind door 2 or door 3 and the contestant chose door 2 or door 3 respectively.

The partially completed tree diagram illustrates this.

For the remaining cases the host has only one option to choose if the contestant has not chosen a door with the car.  So for example if the car was allocated to Door 1 and the contestant chose door 2, then the only door host Monty could open is Door 3. Hence the probability of Monty opening door 3 given car is allocated to Door 1 and the contestant chooses door 2 is 1.

Hence the completed third stage of the tree diagram is illustrated below

Lastly if the contestant chooses to win by switching after Monty opens a door to reveal a donkey then the final stage of the tree diagram can be completed. For example if the car is behind Door 1 and the contestant chooses Door 1 and Monty opens door 2 then the contestant will choose door 3 if he chooses to win by swapping. In this case however, the contestant would have lost.

The tree diagram below shows the final stage and corresponding probabilities of each of the 12 possible outcomes (win or loss).

Hence

Pr(Win by switching) = 1/9 x 6 = 2/3 that is, if a contestant decided to win by switching his original choice, they would have a higher chance of winning (2/3) than staying with their original choice.

Similarly if the contestant decided to not switch and stick to their original choice then the Pr (Win by not switching)= Pr (loss by switching) 1/18x6=1/3.  

Angela, I do think you guessed the reasoning of the fellows you have mentioned in your first text with me, no doubts about it: That must indeed have been how they got their own result. Here I refer to (Pinheiro and Kotsiras, 2016):

By definition, the conditional probability of winning by switching given that the contestant initially picks door 1 and the host opens door 3 is the probability for the event car is behind door 2 and host opens door 3 divided by the probability for host opens door 3. These probabilities can be determined referring to the conditional probability from the decision tree (Chun 1991; Carlton 2005; Grinstead and Snell 2006:137–138). The conditional probability of winning by switching is  1/3 /(1/3+1/6) which is 2/3. (Selvin 1975b).

The problem with their thinking is that, first of all, conditional probability has to be calculated in a slightly different manner, as you can see on the same blog post of the extract.

Second, we know, from very simple writing, what the problem amounts to, and you can see yourself:

In short, if we go Mathematics with the Monty Hall Problem, we get 1/2  of chance of winning, since if the door we chose first is the door they chose, only sticking to our first choice will return win, but we have stick or swap (2 possible choices, only one wins). If the door we chose first is not the one they chose, we only win if we swap, what is one in two again (2 possible choices, only one wins). In this way, 2 wins/4 possible choices or 1/2. 

(Pinheiro, 2016a).

I still tried to do as you did and find some path to get to the 2/3, as you can see on the same site (Pinheiro, 2016a), paragraphs that follow the above extract.

Notwithstanding, I ended up realizing the mistake that I had committed when I wrote (Pinheiro, 2016c) :

Notice that, in our optimisation table, we forgot to eliminate one door when considering sticking and swapping. If we take away one case from each, we will actually have six cases and we win in three, that is, in half, so that now it is all the same, as it should be. That is for the glory of Maths! 

(Pinheiro, 2016b)

I am sorry Marcia I cannot see why you would take away one case. I can see there are a total of 12 outcomes not six if we take all that we know into consideration. However the probabilities of these outcomes are not all equal and the doors opened by the host are dependent on the contestant’s first choice.

As for your diagram, Angela, we can only see things from one perspective when analysing a problem in Combinatorics, so that we either use the perspective of the presenter, and you then have your 1/3 on his first choice and then 1 on his second choice, or we use the perspective of the audience member, when we have 1/3 on the first choice, then 1/2 (swap or stick) for each possibility from the first set. If they chose the right door the first time and they stick, they win, so that we get 1/6 x 3 for this one. If they chose the wrong door the first time and they swap, they win, so that we get 1/6 x 3 for this one. If they chose the wrong door the first time and they stick, they lose, so that we have 1/6 x 3 for this one. If they chose the right door and they swap, they lose, so that we have 1/6 x 3 for this one. In this case, we have 1/2 chance in the end for swap and 1/2 chance in the end for stick in terms of strategy.

From the presenter’s perspective I agree with the 1/3 in his first choice but it is not always 1 on his second choice since if a contestant chooses a door with the car, the presenter will need to randomly choose one of the remaining 2 doors, each with a probability of 1/2. In this case the presenter’s second choice has a probability of ½ not 1.

Angela, we don’t know what probability you refer to here, I suppose. In Combinatorics, we must have an objective, so say the presenter will have the probability of choosing door number 2 in the end. Then you are probably right. You talked about the presenter choosing one door to conceal the car, if I am not mistaken. That would certainly give him 1/3 for number 2. If the audience member chooses number 2 and the car is there, the presenter cannot open number 2, so that it is 1/3 and 0, what will give us 0. If the audience member chooses number 3 or 1 and the car is on number 2, the presenter has a chance of 1/3 x 0 = 0 of choosing number 2. If the audience member swaps, the reasoning is yet another. If we talk about how many choices the presenter has on each step of the process, we then have 1 in 3 for the first round. In the second round, if the door chosen by the member of the audience contains the car, he has 2 in 2 or 1 chance of getting a door without a car when he opens the door, since the chosen door will never be opened.

I can also add this to the tree diagram by leaving out the last stage of the tree and replacing ‘loss’ with ‘staying’ and ‘win’ with ‘switching’. In that case the Pr (switching to win)=2/3 and Pr(staying to win)=1/3 which is again the same.

Marcia I think it is very important that we take into consideration what we know. If we don’t then I agree with you as in the end we really have 2 doors and the chances of choosing the right door will be 50-50.

That is, as Kalid Azad wrote in his post ‘Understanding the Monty Hall Problem’

‘Suppose your friend walks into the game after you’ve picked a door and Monty has revealed a goat — but he doesn’t know the reasoning that Monty used.

He sees two doors and is told to pick one: he has a 50-50 chance! He doesn’t know why one door or the other should be better (but you do). The main confusion is that we think we’re like our buddy — we forget (or don’t realize) the impact of Monty’s filtering.’

‘The fatal flaw of the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after he filters the other doors.’

This has also helped me understand that if there were 1000 doors and a contestant randomly chose a door, they would have a 1/1000 chance of winning the car if they stayed with their original choice. No matter what the host does, this probability will not change. Hence when the host eliminates 998 doors then the remaining door will have a 999/1000 chance of revealing the car as the probabilities have to add to 1. Hence switching will always give the best chance of winning.

Dear Angela, that cannot be right: If the presenter goes and opens 998 doors, you know that 998 doors do not have the car, so that your universe of choice has been reduced from 1000 to 2 and therefore your probability of winning is now much higher. Whilst in the first choice you had 1/1000, as you yourself said, in the second round you had 1/2, so that it is now tremendously higher, not only higher. 

Otherwise we can just ignore what has happened and see there are now 2 doors to choose from. In this case there will be a 50-50 chance of choosing the door with the car. 

So, that is better. That sounds right.

Thank you for your time Marcia. I appreciate your efforts to make me see things differently. I am grateful for the time I spent in understanding this problem and all the underlying information.

My students had a great time discussing this problem with their peers and their families and were able to see a number of ways this problem could be interpreted and solved.

I really don’t think how this problem could have any result that is different from 50%, Angela.

References

Pinheiro, M. R. (2016). Monty Hall: In Short. Retrieved 29 October 2016 from http://mathematicalcircle.blogspot.com.au/2016/10/monty-hall-in-short.html

Pinheiro, M. R. & Kotsiras, A. (2016). Master Angela, Dr. Pinheiro, and the Monty Hall Puzzle: Part 1, Discussing Dr. Pinheiro’s Solution

http://mathematicalcircle.blogspot.com.au/2016/10/master-angela-dr-pinheiro-and-monty.html

Pinheiro, M. R. (2016a). Monty Hall: In Short. http://mathematicalcircle.blogspot.com.au/2016/10/monty-hall-in-short.html

Pinheiro, M. R. (2016b). Now the Detail: Optimisation Finally Equals Mathematics. http://mathematicalcircle.blogspot.com.au/2016/10/now-detail-optimisation-finally-equals.html

Pinheiro, M. R. (2016c). Master Angela and Inspiration: Our Solution? http://mathematicalcircle.blogspot.com.au/2016/10/master-angela-and-inspiration-our.html

Kotsiras, A. (2016). Better Explained blog post.

https://betterexplained.com/articles/understanding-the-monty-hall-problem/