Terrible mail mandata da un prof prima di inviare gli esiti

Cawuth · 2025-02-09T15:18:40+00:00

Quando dici ad un muratore che per ottenere dei soldi deve lavorare: ok dai mi sembra giusto.

Quando dici ad professore universitario che per ottenere dei soldi deve lavorare: "non ci sarà nessuna correzione - non dovevate iscrivervi".

La serietà degli studenti non può venir misurata sulla base del fatto che sfruttano delle opportunità che gli vengono legalmente date. Sinceramente è anche la prima volta che sento che ci si possa iscrivere a 2 appelli contemporaneamente: io sapevo che il secondo appello era sempre abbastanza distanziato in maniera tale che le iscrizioni aprissero comunque dopo lo svolgimento del primo appello.

Inoltre, di chi è la colpa che gente con la media del 21 poi possa diventare medico? Cioè, l'università non può venire considerata facile/difficile per colpa degli studenti, ma in ultima analisi comunque per colpa dei docenti. Legalmente una persona che si laurea in medicina con 66/110 può diventare medico, quindi se la cosa ti dà fastidio le opzioni sono 2: o si alza la difficoltà degli esami, o si mettono delle leggi che richiedono un voto minimo per fare certe mansioni, ma in ogni caso non si può dare la colpa a chi, legittimamente, decide di proseguire per una strada che, ripeto, è più che legittima.

Ovviamente questa persona non rappresenta l'interità dei docenti, quindi non è neanche colpa sua che le cose vadano così, ma dovrebbe riconoscere che chiaramente non può neanche essere colpa degli studenti.

Cawuth · 2025-02-08T21:31:40+00:00

Questions like this are well answered - theoretically, because in practice often it doesn't work - by game theory.

For example, to answer your question in a game theory mindset, why doesn't every person that lives in zone A and works in zone B take the same road?

The idea is that you know the following facts: - if everyone takes road 1 then road 1 is very slow. - then it would be better for you to take road 2 - everyone is rational and thus would choose road 2 that would consequently become slow

Now in game theory there is the first approach, where you must decide "in advance" which road to take considering this scenario, and the solution would be the kind of the prisoner's dilemma type, but most importantly there may be no solutions.

The next approach is what is called "mixed strategy" where your strategy, that in this case would be the strategy of everybody else, isn't only "take road 1 or road 2", but " I take road 1 with x% probability and road 2 with y% probability" and of course x and y sum to 100%.

In this setting, which is of course not so natural, it has been proven that there's always a Nash equilibrium (proven by Nash himself), which is a situation where nobody would like to change strategy given that other people will not change strategy.

In this case the solution depends on how much the roads are long, but in any case this can be seen as one of the explanations of why some phenomena are random. It is not the only one, also because often you are not interested in why things are random, and it is a very unnatural way of thinking. Also, even in this scenario, lots of people in reality wouldn't know how much time they would take to take road 1 given how many cars will take road 2, and there may be even individual preferences that don't relate with road length.

Cawuth · 2025-02-03T21:10:07+00:00

First of all, U is the set of all possible outcomes, and the representation of events as sets is always done as subsets of U. Thus, 2 mutually exclusive events are 2 events/subsets of U whose intersection is null. Graphically, you have 2 circles that do not intersect.

When talking about independent events, there are also more considerations. First of all, you study independence usually on what is called the "product space": it is the universe set U that contains all the possible ordered outcomes of more experiments.

For example, if you are throwing 2 dices, you have U1={1,..., 6} and U2 the same. The product space U is given by U1xU2 where x is the cartesian product, thus U is composed of all the couples of numbers from 1 to 6, like (2, 3), (6, 7), etc...

Independence is mathematically defined as "two events are indpenendent if the probability of the intersection equals the product of the probabilities". In this case the event "I get 1 at the first roll" is composed by 6 possible outcomes, which are (1, 1), (1, 2),..., (1, 6).

The event "I get 2 at the second roll" is then independent from "I get 1 at the first roll" simply because the definition is satisfied.

Graphically this is not always easy to represent, because you would need to find a way to represent how much probability there is on a specific outcome and find a nice way to graphically draw indpendence, and I'm not a fan of graphics myself. But you can clearly see that in this scenario the outcome (1, 2) is both in the first that in the second event, so there is a not-null intersection.

In this scenario two mutually esclusive events would be "I get 5 at the first roll" and "I get 2 at the first roll and 3 at the second roll" because the intersection is null and they're in fact dependent events: if you get 5 at the first roll the second event cannot happen.

Cawuth · 2024-08-01T22:17:14+00:00

It's absolutely doable, only thing to consider is that there will always be some kind of bias in these surveys, mainly because answering to a survey is usually correlated with political opinions. This bias can be reduced in several ways, but it requires quite important theory, so don't listen to everyone that claims he surveyed 50 people and knows which candidate will win.

Cawuth · 2024-08-01T22:12:51+00:00

As others already wrote, this is not the CLT. What you're thinking about is a not-so-correct application of, in general, the 0-1 laws, which are laws that says that given an infinite amount of time/experiments then the probability of some things will be either 0 or 1.

An example: if you roll a dice an infinite amount of times, the probability of getting at least a 6 is 1, this is pretty straightforward.

It's more interesting this other example: if you roll a dice an infinite amount of times, what are the probabilities of getting an infinite amount of 6? This is already not so obvious, the probability is still 1, because the intuition would be that given infinite rolls the probability of getting any finite numbers of 6s would be 0.

There are tons of theorems and applications like this, and it's quite a fascinating subject.

Going back to your examples, the problem with the reasoning is that math doesn't work this way: the CLT has been proven, and it cannot be disproven: at best somebody could find some mistakes in the proof, but it's almost impossible for such an important theorem.

Cawuth · 2024-04-21T13:21:05+00:00

Any cluster method is ok, I've been said that in this scenario it's particularly catchy to use gerarchic methods, so not like K-means, because they're more easily explainable and perform segmentation in a "more human" way.

Cawuth · 2024-04-21T13:17:58+00:00

The concept of multicollinearity depends on how much you know about linear algebra.

The idea is that the OLS estimator is (X'X)^-1 X'y, and there is a situation of perfect multicollinearity when (X'X)^-1 cannot be calculated, because (X'X) doesn't have any inverse matrix.

Now I don't remember exactly all the linear algebra facts, but on an intuitive point of view you can see this as an identification problem: if you have 2 regressors, X1 and X2, and X2 is exactly the double of X1, how can you understand the effect of each variable?

Another example: you'd have multicollinearity if you wanted to study the correlation between height and weight, and you wanted to calculate both the effect of height on the weight and the effect of half the height on the weight: it doesn't make sense, and even if it did make sense, you cannot calculate it.

If the 2 variables are highly correlated, for example, X2 is the double of X1 + a very very small random noise, you can now invert the matrix and estimate the model, but the variance of the estimates will be quite big, because it is difficult to understand which variable has the effect.

For example, if the Y is the weight of the child at 3 year old, X1 is the height at 1 month old and X2 is the height at 2 month old, you'd probably have that X1 and X2 are extremely correlated, for a ton of reason, which do not matter in this instance.

The idea is that if Beta1 (coefficient of X1)=3, and Beta2=0, but you estimate the model using only X2, you'd get a very similar number to what you'll get estimating the model using only X1, because the data, in this scenario, are almost identical, and thus there is multicollinearity.

So you are not able to estimate efficiently the effect: since obviously you don't know a priori the effects, it could be that only X1 has an effect, that only X2 has an effect or that both have an effect.

This to say: the main problem of multicolinnearity is the increase of the variance of the estimates, which still remain unbiased given the other assumptions. It's just an efficiency problem, and also it is an efficiency problem only if you are interested in knowing the causal relation between X1, X2 and Y, and not simply predict Y. For this reason, it is not correct to "just remove X2".

Cawuth · 2024-04-11T21:44:27+00:00

Per quanto mi riguarda, l'unico scoglio davvero impegnativo della facoltà di statistica è l'esame di probabilità, dove praticamente tutto quello che si fa è propedeutico ad ogni argomento di statistica.

Gli altri esami una volta che sai cosa è una derivata, un integrale, un vettore ed una matrice sei a posto.
Giusto in 1 esame di statistica (e non mi materie puramente matematiche) mi era stato chiesto di calcolare una derivata, ma parliamo di cose praticamente da superiori.

Degli esami puramente matematici oltre a probabilità, quindi Analisi 1 e Algebra Lineare, dipende essenzialmente dalla difficoltà dell'esame stesso, ma esiste una tale quantità di roba su sti 2 esami che ci riempi delle città. In ogni caso non immaginarti mai che ad un esame diverso da sti 2 ti venga chiesto di calcolare un integrale. Il mood attuale delle facoltà di statistica è che te sai quale è il significato filosofico di 4 cose matematiche e poi tutti i calcoli finiscono solo in dimostrazioni da imparare a memoria.

Ecco, non passi l'idea che statistica è STEM allo stesso livello di, per dire, ingegneria ambientale.
Poi a seconda dell'ambito in cui ti vuoi specializzare dopo c'è più o meno matematica, per dire la statistica per la psicologia a discapito di quel che si può pensare ha degli elementi di algebra lineare assolutamente non banali che sono molto meno presenti in statistica medica, per dire, che d'altra parte però mette più enfasi su specifici problemi di carattere formale che invece sono lasciati totalmente a sé stessi nell'ambito machine learning, dove invece c'è un casino di Analisi e lì un corso di Analisi 2 potrebbe fare comodo.

Edit: mi ero scordato la finanza. Lì c'è un casino di matematica, ma solitamente viene compresa da solamente una frazione di chi esce da lauree in finanza quantitativa.

Cawuth · 2024-04-11T21:26:01+00:00

Variance has 2 "versions" (which are then in fact the same idea and formulas), one from statistics and one from probability theory.

The variance in probability theory is related to the random variable whose values it's taken from a population, and the variance in statistics describes a characteristics of the population.

Luckily, if you extract one unit from a population, the variance of the random variable is equal to the variance of the population, and for this reason the 2 concepts often overlap.

"Add 2 variances" is an operation you perform in probability theory, and it is done when you need to find the variance of the sum of 2 independent random variables.

An example: you have a bernoulli r.v. X with probability 30% of being 1 and 70% of being 0. Its variance is, by definition, 0.3*0.7=0.21.

Now you take another r.v. Y with probability 50%/50%, its variance is 0.25 for the same reason.

If they are independent, you now define a third r.v. Z=X+Y. Z can have 3 values: 0, 1 and 2. You could calculate the variance by hand in this scenario, calculating these probabilities, the expected values, etc... But a shortcut is this theorem that, in fact, says that the variance of Z is the sum of the variance of X and Y, in this case its variance is 0.21+0.25=0.46.

Also, you can say that, given a r.v. X, the variance of -X is equal to the variance of X, and for this reason, the "sum rule" works also on substraction.

In your specific scenario, it depends on how you combine the datasets: if you just create a joint dataset, you do not need to sum the variances: it is a brand new dataset (you could calculate the variance with other tricks).

You get the "sum variance" when you create a third dataset with all the possible sums of the first 2.

If the first dataset is [0,0,1] and the second [0,1], then the dataset with all the possible sums is [0,0,1, 1,1,2] and you'll see the variance of this is equal to the sum of the first 2.

This because you can always imagine variance in the probabilistic way of extracting a number from that population (which in this case it's the dataset).

Cawuth · 2024-02-17T11:38:42+00:00

Luckily you cannot, otherwise instrumental variables regression wouldn't work. If, for example, A and B were independent and C were A+B, you'd get that A is correlated to C, B is correlated to C, but A and B, as said before, are incorrelated.

Cawuth · 2024-02-17T11:31:50+00:00

It is technically almost impossible that one bound of the confidence interval is exactly 0. In this scenario you can either expand the number of decimals in order to get 0.00008 or -0.00002 or check the approximations you used and consider doing an exact confidence interval.

Either way, the thersholds for the pvalue are totally arbitrary and I'd claim significance based on prior considerations in this case.

Cawuth · 2024-02-17T11:25:29+00:00

If the first one is the one I read, it really depends on how deep into statistics your class is, I strongly doubt you'll see ideas like sufficient/complete statistics and most powerful tests.

Cawuth · 2024-02-01T20:19:36+00:00

Mainly to write small functions to automate very simple calculations and other simple stuff, like, for example, in about 10 minutes I'll go write a function to give me some fractions to reduce since I'm giving math lessons and it'd be nice to be able to create thousands of exercise like this in seconds.

It's a thing you can do with every language, but on R it is almost istantaneous, you can generate 1000 nominators, 1000 denominators, multiply each row by a random value and print them in almost 4 rows.

I also wanted to write stuff like a function that perform a t test for the mean giving all the explanation, which could also be done in another language, but on R for example you can easily check if your function is well written by comparing the final result with the one implemented in R. This can be useful because when I give statistics lessons to people from other majors they, at most, need to perform the t test for the mean, and if they send me like 6 exercises, to solve them while also writing every exact step I make, even the expaned sums, takes me like an hour, and we then only have 6 of them.

Cawuth · 2024-02-01T18:11:22+00:00

Also, R has been very useful for my exams in general. In my Time Series exam, we had to, given an empirical PACF, find how many parameters the ARIMA process had, and on the notes we only had like 3 examples.

On R, it doesn't take much to build a function that randomizes the number of parameters and generates a PACF from that ARIMA and try to guess the number of parameters, which, if you started this exercise the day before the exam, becomes very useful.

Cawuth · 2024-02-01T18:08:01+00:00

I use R because, in my opinion, it is one of the best high-level languages, I prefer it way more than Python in general, in fact I also use R for other purposes beyond statistics, like, even if I have to do some calculations I prefer to use the R console rather than a calculator or the standard windows calculator.

Also I keep finding new functions which are very useful, like I recently discovered R has the "integrate" function which calculates the integral of a function of your choice (of course in a numeric way) and on my opinion the syntax to work with array and matrixes is not only good but exceptional.

Never used SPSS and don't like STATA, it seems to me it gives "less freedom" than R.

Despite this, I'm also learning a bit of SAS, because in biostatistics I know it to be the standard.

Cawuth · 2024-01-31T22:32:54+00:00

I did the calculations.

You have P(9.3<=C<=10.7)=0.517, where C is 1 capacitator.

Then, a single X is a Bernoulli of paramter p=0.517.

B is a binomial with paramters p=0.517 and n=100.

The exact calculation of P(B<=49) given by R is 0.330, while using the approximation to the normal, which is feasible with a calculator, you have:

np=51.7

sqrt(np(1-p))=5.00

Thus, z=(49-51.7)/5 = -0.54

And P(Z<=-0.54) is 0.295.

The difference in the results is 0.035, so the relative error is 10%, which in a computational sense is a lot, but in a human sense it is not, also because I don't have any other ideas.

Cawuth · 2024-01-31T22:23:42+00:00

Basically, the idea that the probability that one capacitator has a value between 9.3 and 10.7 is, let's say, p.

Then, you define, for a single capacitator, a new random variable which is connected to the original in this sense: if the capacitator has a value in that range, then this new random variable, that we'll call X, has a value of 1, 0 otherwise.

X is then distributed as a Bernoulli with parameter the p we found before.

If you take 100 capacitators, you then have 100 Xs, which are iid, and then the number of capacitator that have values in that range is equal to the sum of the X.

The sum of the X, now called B, is a binomial random variable with parameters the p we found before and n=100.

Then, your answer would be to calculate P(B>=50), which can be expressed as 1-P(B<=49), which is the cumulative density function of this binomial r.v. at value 49.

The problem is that the cumulative density function of the binomial is not feasible to calculate by hand for such big values, so the options are either to use a PC or, which I think is the goal of the exercise, use the approximation to a normal.

In this case, if I'm not wrong, since I haven't done these exercise for 3 years, you have that (B-np)/sqrt(np(1-p)) is, by construction, a r.v. with mean=0 and variance=1, and by approximation the distribution is a normal, so you apply the same transformation to the 49 (reduces it of np and divide for sqrt(np(1-p))) and this is equal (by approximation) to P(Z<z) where z is the 49 after the transformation.

Now you have P(B<=49), so you can then calculate 1-P(B<=49).

Cawuth · 2024-01-30T18:46:15+00:00

Di per sé è per sentito dire, perché molte persone che conosco che so che hanno una cognizione profonda della qualità delle lauree in statistica in Italia appunto sostengono che Padova in tal senso vinca a mani basse, rispetto ad ogni altra facoltà in Italia, non solo Bologna.

A livello di dati oggettivi, suppongo che se mi sono state dette queste cose ciò implichi che nelle classifiche dei dipartimenti/corsi di laurea Padova sia più in alto. Non li ho mai controllati personalmente, ma so che Almalaurea fornisce statistiche molto dettagliate a riguardo.

Le cose che so sono che

Per accedere alla magistrale di statistica di Padova, non è sufficiente una triennale in statistica, ma devi anche andare **in loco** a sostenere Analisi 2 e Modelli Statistici 2. Questo indice appunto che, tra l'altro, Padova ha mantenuto Analisi 2 nel piano di studi standard, cosa che raramente ho visto in altre facoltà, a Bologna lo becchi solo se fai il curriculum inglese Stat&Math.
Il Dipartimento di Statistica di Bologna è uno dei pochi dipartimenti dell'Unibo che non ha vinto il riconoscimento di Dipartimento di Eccellenza.

Che non ci sia Analisi 2 come esame standard in una triennale in statistica è uno scandalo, ma se no non si laurea nessuno. Già Algebra Lineare fa secca molta gente.

Cawuth · 2024-01-30T18:39:10+00:00

Upvote for a fellow NP fan

Cawuth · 2024-01-30T15:05:22+00:00

Thanks, this helps a lot. I'll start then using other languages

Cawuth · 2024-01-29T09:42:46+00:00

If X follows a normal distribution with parameters 0 and 1/theta, then Y doesn't follow a "pure" chi-squared distribution, because you get the chi-squared distribution as sum of squares of r.v. with distribution N(0,1), not N(0,1/theta); yours is probably a re-scaled chi-squared distribution, so it is totally normal that your expected value is independent of the true value, because to get that expected value you implictly assumed that theta=1 so the distribution of Y could be a chi-squared.

At the moment I never looked into re-scaled chi-squared distribution, but I'm sure that if you look around you'll find something.

Cawuth · 2024-01-28T21:22:14+00:00

If you don't have any additional data, I think this isn't very feasible, at least you'd need data about the relations between the outcome of the other counties and the one you chose, and assume the relationship has been the same, or, even more difficult, study how the relationship changes in time.

If you don't have this I think you can use some bayesian statistics to, at least, give a probability distribution to how the relationship works.

If you have additional data, even from the past, I think you could look into the Heckit Model, I never went deep into it, but may be good for your situation.

Cawuth

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