Hello Joseph,

First, since we aim to fit every phrase in a single line, we tend to shorten definitions like this one in interest of simplicity, while maintaining correctness.

Second, I think I actually disagree with your definition.

1. Set theory is more basic than probability theory, that much should be clear. Thus, set operation definitions should not be dependent on the definitions of probability theory of event and sample space. So the vocabulary is, dare I say, anachronic.

2. In set theory, your confusion is actually correct. Due to the trivial existence of the theoretical universal set, the complement of a given set can be used despite not referring to any exterior set. It's just another variable, that may or may not contain every possible element or not. The generality of the statement doesn't have to be lost. So in that sense, the definition we gave is valid, and yours is too restrictive. However, as we discussed in the study group, computers are rarely so theoretical. There is no purely mathematical set in the Wolfram language, only lists. In the same way, there is no purely theoretical complement implemented, only a computationally approximation of that concept.

I would like to suggest another clarification to section 3.

Thanks Marc, thanks a lot!

Thanks Marc.. I do believe there may be some impossible (in the sense of not meaningful) cases. I still try to figure out what makes sense to ask and what doesn't .But ok, here I go with a clarification of my question. I still believe this is applicable in real life, but I may be wrong. I tried to depict the situation with an "entry level skills" notebook, so my question gets easier to be assessed..

The question never states that the game is over as soon as a red ball is taken. Without that information, I interpreted the question as there being three outcomes at the end after all balls are taken, one set for "you" and another for the "friend":

{{r,r,b},{b,b,b}}
{{r,b,b},{r,b,b}}
{{b,b,b},{r,r,b}}

This gives a ⅔ probability of having a red ball in "your" set at the end.

Is that line of thinking correct given my assumptions?

For Section 2 Exercise 3. how do all the paths to (4,2) add up to 7 segments? It looks like 6 to me.

Attachments:

Exercse2_3.nb

Hello Joseph,

Indeed, on problems 1 and 3, there seems to be text from another part of the course. This is an error, as mentionned by Juan Ortiz Navarro, posted 2 days ago Regarding excercises-02.nb The solution to problem 1 (...)

So yes the answer is 25!/15!.

The solution for exercise 1 from section 2 seems disconnected. Q: Twenty-five runners compete in the 200m event. How many top 10 arrangements are possible? A: The arrangements are ordered, but you were only asked to consider 10 elements out of 25. Thus, this corresponds to a permutation: y1 = Integrate[(-s + 0), {s, 0, x}]

What does this statement have to do with the problem?

Shouldn't the answer be 25!/15! ?

On Excercise 4 of excercises-06.nb, "A smoker is twice as likely to have an ectopic pregnancy as a non-smoking pregnant woman." is denoted as P(E|S)=2P(E). I thought it to be denoted as P(E|S)=2P(E|S'), and P(S|E)=P(E|S)P(S)/(P(E|S)P(S)+P(E|S')P(S'))=0.461538

How can P(E|S)=2P(E|S') be said in words?

Hello all,

For a bit of context for the exponential family of distributions, here are some ressources:

A well made short video series introducing the subject, by Mutual Information.

An introduction article to get familiar with this, from Berkeley EECS.

A short textbook of Statistical Theory focused on the exponential family, from the University of Oxford.

A paper on the link with machine learning, from Princeton University.

The exponential family is usually covered in any course of Statistical Theory. Go satisfy your curiosity!

Thanks, Marc. I'll keep that advice in mind!

Hello Alex,

As noted in the post by John Burke, there is a mistake and the answer is 40, with Binomial[4,3]*Binomial[5,2].

William:
I agree with you but I see it as which parts of the tree will give at least one red, taking into account that as soon as I have three balls the game is over. So in the first round I have a 2/6 probability of getting a red one. If I do not succeed on that, then on the second round I have 4/62/5 probability of getting a red one. And on the third round, I have 4/63/51/2 of getting a red one and the game is over since the other player took the other three balls:
2/6 + 4/62/5 + 4/63/52/4=4/5
or the complement of not getting red balls, which will be all black balls: 1-4/63/51/2.
Does that make sense?

EDIT: I seem to have answered that half asleep, I can't even read my answer. Please disregard it.

Oh that's interesting, as the assumption for the multinomial is that all possible Ads are "exhausted" at the last observation. My intuition was that, while surfing the internet, I used a finite amount of time, and during that undetermined time Window I saw only 13 Ads, like if I keep browsing I may perfectly see more ads. Let say if I would double the browing time, I may end seeing 26 ads or 30 ads in total, .. So I guess I get your point if all possible ad instances are contained in the limited length sample (of 13 total ads). Otherwise, if not limiting length is stated, I guess it may be fair to assume that As Bs or Cs are never exhausted, and browsing more...means watching more ads..(like the "YouTube Premium thing...that seems to never end.. ;) ....) Thanks Marc for the clarification

Hello J,

That's a pretty fun question actually. Let's go through it.

You encounter A 7 times, B 3 times and C 3 times, 13 in total. How count this have happened? You have your limited ressource of 7 As, 3 Bs and 3 Cs, so all the orders are going to be 13!, not 3^13, because this would imply you may not have encountered the given number of ads.

But, can you distinguish the difference between A and A? No, the elements are not distinguishable. So you need to divide by 7! for all the orders of A, 3! for B, and 3! for C. You get 13!/(7!3!3!) which is the Multinomial[7,3,3].

Thanks. I meant to write 25!/15! indeed.

Hi J,

This is a bit complicated, so let's discuss this one thing at a time. This textbook expresses a sample space where each data point is in itself a sequence of multiple numbers. In that sense, this is a multivariate random variable. Within a single sample, or data point, there are multiple times, which are the periods of time between breakdowns. Let's say there are n such periods in each sample. Thus, the domain of any single outcome is R(>0)^n, that is, the positive reals in n dimensions (periods of time are always positive).

In this context, the sample space is the set of all possible sequences of periods between breakdowns, possibly R(>0)^n itself.

So overall, I believe you are right with your sample space, but also that you are wrongly interpreting their explaination of the sample space. Hopefully that explanation helped.

Regarding excercises-04.nb

On problem 4, I think the solution is not complete. I believe, the answer displayed is for P(R' intersection B'), which one needs to add 1-P(R) and 1-P(B).
Or P(R' union B')=1-P(R intersection B)=1-.1=.9.

What mistake am I doing?

While trying to confirm concepts in the course with 3rd party reference, I found the attached. It states that the "Sample Space" for the times to failure for a certain machine is a sequence (T1...Tn). In my opinion, sample space must contain all possible times to failure, as opposed to certain specific sample. So I'd say the "space" would be all the real numbers, or perhaps all the real numbers below the maximum allowable age for the equipment being analyzed. Am I right on that the reference is confusing a "Sample space" with a specific sample?

I'd appreciate your comments. Jorge

a

Hi Marc, In Lesson 7, Slide 10 and 11, you showed two different examples of how to load and aggregate. I have difficulty in understanding how to aggregate :

height[h_?(60 <= # <= 76 &)] := aggdata[[2, h - 59]]/Length[data] and roll[n_Integer?(2 <= # <= 12 &)] := dice[[n - 1, 2]]/36

Could you elaborate a bit, or use simpler and understandable codes?

Thanks, Lewis

Indeed, that is correct.

There are 6*6 combinations, 6 with doubles, 6 starting with the number 6 and 6 ending with the number 6. The three events have the occurrence (6,6) in common, so one occurrence is counted three times, thus:

(6 + 6 + 6 - 2)/(6*6)

is indeed 4/9, thus odds of 4 to 5.

Thank you for informing us, it will be corrected.

Marc,

You asked for mistakes in the documents. So, please have a look at “Lesson 2, Slide 9”: Your equations are wrong, which means LHS is not equal to RHS. Right?

Thanks Marc for your prompt consideration. I have another for you! In Exercise 1 of exercises-05.nb, the event you describe out of the 36 possible equally likely outcomes is E={{1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6}, {1, 6}, {6, 1}, {2, 6}, {6, 2}, {3, 6}, {6, 3}, {4, 6}, {6, 4}, {5, 6}, {6, 5}}. Hence, P(E) =16/36=4/9. Thus, the odds are 4 to 5. Would you please look into this also. Thanks.

Indeed! You caught my second mistake!

P(A'[Intersection]B')=P((A[Union]B)')=1-P(A[Union]B)=1-0.4=0.6

Thus, the probability that neither A nor B occurs is 0.6.

Thank you for noticing, it will be corrected.

Does anybody have the link to the course materials that they could post on the community thread?

I want to get started on the exercises but I missed the last meeting and the recording does not show the chat pane with the links.

Thanks

Hello all,

We had a more difficult question today, given as:

Given a vector with m element (real numbers) I want to compute the PDF for the sum of the m elements considering that each element will have positive or negative sign with equal probability.

First, consider that each element must have its own probability distribution. Since it is real, let me for example assume any element is uniformly distribution between -10 and 10.

UniformDistribution[{-10, 10}]

If all elements are distributed in the same way, the PDF of the sum will be a result of m elements added. So we get:

PDF[TransformedDistribution[
  m*x, x \[Distributed] UniformDistribution[{-10, 10}]], x]

However, if you assume a normal distribution centered as 0 for your elements, then you could write:

PDF[TransformedDistribution[
  m*x, x \[Distributed] NormalDistribution[0, 1]], x]

Finally, your elements may be distributed in different ways, which could look like something like this, assuming let's say m is 4:

PDF[TransformedDistribution[
  a + b + c + d, {a \[Distributed] NormalDistribution[0, 1], 
   b \[Distributed] NormalDistribution[0, 4], 
   c \[Distributed] NormalDistribution[0, 19], 
   d \[Distributed] UniformDistribution[{-10, 10}]}], x]

Another way to intepret your question is to each element is given a binary choice between -1 or 1, but the values are constant. In that case, you may want to apply many distributions to a single sum, which results in this, assume for example this vector:

v = {1.77, 5.65, 10.14, 195.14}
PDF[TransformedDistribution[
  v[[1]]*(-1)^a + v[[2]]*(-1)^b + v[[3]]*(-1)^c + 
   v[[4]]*(-1)^d, {a, b, c, d} \[Distributed] 
   Table[BernoulliDistribution[0.5], Length[v]]], x]

Hopefully that answers the question. Transformed distributions will be seen in Lesson 20.

Hi Bob,

The textbook of the proof is Probability - An Introduction by Geoffrey Grimmett and Dominic Welsh. It is concise but a bit short on examples. A great book for examples would be Introduction to Probability by Charles Grinstead and Laurie Snell. Less mathematical, more examples, more discussion. It's also freely available.

Hopefully that helps.

Hi,

It may be best for you to prioritize other engagements, as this course is made to be taken at any time.

The course will soon be fully released with all materials and lessons freely accessible. Moreover, the daily study group sessions are all recorded and also freely accessible. There is no planned second study group for this course for now, but it may happen if there is interest for it. Feel free to also continue asking questions in this post even after the study group has ended.

Will this course be repeated? I am having struggles with it right now, and would like to work on Computational X-plorations or something else. Thanks.

Would you please provide details of the proof for the inequality on Slide 12 of Lesson 10? Thanks very much.

Thanks, Marc, for clarifying. I'll get used to your precise definitions, including that of modeling. Again, thanks 10^6 for a so-far wonderful course.

Hello Joseph,

Basically yes. It indicates you're making jumps of 1, and the fact you're jumping over values indicates you are taking discrete steps; thus, you are using a discrete variable.

Notice you could also theoretically also jump of 2, or 0.5, and this would still be a discrete distribution. The point is that you are jumping values instead of using a continuous range.

In lesson 8, "Discrete Random Variables" we use the statements

children[n_] := 1/((n + 1)^3*Zeta[3]);
distChildren = 
 ProbabilityDistribution[children[n], {n, 0, Infinity, 1}]

Does the "1" in the ProbabilityDistribution function indicate that n is a discrete variable?

Some weird behavior when trying to use a PDF

Attachments:

Using a PDF.nb

Hi everybody!! I'm having some issues while trying to Plot a PDF. I see some unexpected changes at the staring of the PDF plot. Am I doing something wrong in the PDF definition or invoking the Plot WLFunction? Thanks in advance!

MODERATOR NOTE: notebook Some wierd behavoir when trying to use a PDF was moved to the attachment below and also can be viewed at https://www.wolframcloud.com/obj/29289cee-c540-4d5e-9507-dbfcc2001739 Wolfram cloud notebook.

Attachments:

Some wierd behav...nb

Hello Mitchell,

No, these are not equivalent. Those are for very different contexts.

Conditioned is only used in the context of Probability to symbolize the classical P(A|B) or probability of A given B, basically replacing the word given.

Condition is used in pattern recognition, always jointly with a pattern, giving an iterative test to accomplish on a list usually, as a much more core mechanic of the language.

Marc

Thanks for your response. I agree that the set of all possible outcomes should not include repeats. But is the probability of an even sum 1/2? .

I'm enjoying these study group sessions!

Joe

Hello J,

A valid question for sure. Let me try to be as explicit as possible.

First, why must a probability be a constant measure? Well, a quick Google search gives us that a measure is "A reference standard or sample used for the quantitative comparison of properties." In other words, for something to be a measure, it needs to be reliable, a reference. An uncertain number that has some convergence is not evidently reliable, thus not a good measure. For us to define an entire branch of mathematics, we need more reliable things.

Second, why is the assumption of convergence of any frequency problematic? Simply, because it is too much of a strong statement. If I start by assuming a big and complicated statement, then any possible exception to that statement may render worthless all the theorems I’ve built on that statement. It's really a question of what is considered conventionally true.

Look at it this way. We want the most basic statements for axioms. You can prove the convergence of frequency through Kolmogorov's axioms. However, you cannot prove Kolmogorov's axioms from the assumption of the convergence of frequency. This implies Kolmogorov's axioms are more "basic".

This doesn't dare to be an "answer", but my interpretation, as I had the same "feeling" when reviewed that slide.

As the explicit statement in the slide is {2,4,6,8,10,12}, and those were examples on the topic of "Sample spaces and Events" I saw that list as the "Event Definition" that one would use to calculate the mentioned probability. That list is actually the Event Definition against which the EvenQ function tests the full range of outputs. And I'd humbly agree the probability of an Even Sum is 1/2

.

The example in lesson 3 for the probability that the sum of two dice will be even comes up with a probability of 6/11. This seemed odd and a little bit of extra thought suggests that you should not delete duplicate outcomes as they are part of the sample space of outcomes. When you consider duplicate outcomes in both numerator and denominator, you get the more intuitively satisfying answer of a probability of 1/2.

Attachments:

Lesson 3 example...nb

Hello J,

1. Yes! This is the priority of operations, as noted throughout the lesson. First parentesis, then negation, then intersection, then union. Like the infamous PEDMAS, but for set operations.

2. Yes, the generalization applies. Imagine it as equivalent operations but operation on different objects.

The logical Or ||, the addition +, the set union, all represent the natural first common operation.

The logical And &&, the multiplication *, the set intersection, all represent the natural second common operation in a ring.

This is a common theme in mathematics, which implies a lot of the basic rules you see, often in linear algebra, are appliable to many many contexts!

Hi...two fast questions: 1- In these two last laws (pls. see the image) .. and in general, should it be assumed that intersection always goes (is processed) first, before that Union. 2. Does it make sense to generalize that, what is applicable to + and * when in regular arithmetic {i.e (A+B)=(B+A) } Is valid when replacing + by Union and * by Intersection {i.e ( A u B ) = (B u A) or as A(B+C) is AB +A*C then A n ( B u C) --> (A n B ) u ( B n C) shoud be ok...)

Thanks!

Hi Tianyi,

Most code for visualizations is available just by downloading the notebook and expanding the cell of interest.

For lesson 1, most of the visualisations are taken from the Wolfram Demonstrations Project, feel free to explore it and use it for your own classes.

Here is the code for the Normal Convergence:

Manipulate[
 Show[
  Histogram[
   Table[Mean[RandomReal[{0, 100}, n]], {200}], {35, 65, 1}],
  Plot[200*PDF[NormalDistribution[50, Sqrt[9999/(12 n)]], x], {x, 35, 
    65}],
  ImageSize -> {400, 200}
  ] , {{n, 20, "Sample size"}, 10, 200, 1, Appearance -> "Labeled"}]

I switched Manipulate to Animate to make it an animation.

Hi John,

You spotted the first error! Indeed, this should be Binomial[4,3]*Binomial[5,2]! You are totally right! The answer is 40.

Thank you for noticing!

Sorry I can't give you a physical medal, but this will have to do. The binomial distribution thanks you for noticing binomial mistakes:

Marc,

I downloaded your class notebooks from the site today.

Slide 7 of lesson 2 "Consider that the 4 Swiss and 5 Ethiopian athletes want to form a team of 5 to represent them in competition. How many different teams are possible given that it must include 3 Swiss and 2 Ethiopians?" Shouldn't the answer be Binomial[4,3]*Binomial[5,2]?

John

Emails have just now been sent to everyone who is registered for the Daily Study Group. Recordings can be accessed from the webinar series landing page.

I am wondering if this course will touch on any aspects of probabilistic statements in quantum mechanics such as the probability the electron is here is 50%. Another application would be the probability the radioisotope decays that expresses radioactivity that is radioactive will be based on the half life.
There are some examples of using Mathematica's probability functions in quantum physics at https://resources.wolframcloud.com/PacletRepository/resources/Wolfram/QuantumFramework/tutorial/ExploringFundamentalsOfQuantumTheory.html.

This course will fit well with my university courses of Applied Probability and Statistics STA 345 and Probability and Statistics I STA 445 and Probability and Statistics II 446.

I think applying probability theory to Catan is interesting. For example there are two dice in Catan with six sides and the most common roll is a 7 because the expectation of a normal dice is 3.5 and 3.5+3.5=7. The 7 causes the robber to make you discard half of your resource and commodity cards rounded down if you have more than the discard limit.

I would like to mention that you can calculate the combinatorial enumeration of all events in the sample space with FactorialPower. For example, with the example from the second presentation Imagine that 4 Swiss and 5 Ethiopian athletes compete for the 200m sprint. How many different top-three winner rankings are possible? FactorialPower[9,3] returns 504 which is the same as 9!/(9-3)!.

The study group starts next week, on February 27th.

If you want to take full advantage of this course's material and get a practical and deep understanding of probability, don't forget to click on the REGISTER HERE link to get registered in this course!

I'm looking forward to your participation and feedback!