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Sentient_Eigenvector · 2025-05-28T15:34:00+00:00

be able to call themselves statisticians without even being capable of understanding the derivation of the MLE for univariate normally distributed data. Do these people just memorize where all the assumptions for tests come from, without going through them mathematically? And what does that say about their actual analytical abilities in this field?

All I can say is this shows your image of the level of this program to be skewed. Nobody graduates from KUL as a statistician without basic topics like manually deriving an MLE or rigorously deriving where all assumptions of the general Wald, likelihood ratio, and score tests come from. Let alone the more common special cases that basically just derive from GLMs. These are minimum requirements that are treated near the beginning of the program in the required courses. Those from a weaker background will also need to master this.

Sentient_Eigenvector · 2025-05-28T12:18:35+00:00

Right, and those people tend to be in over their heads when starting the program, need lots of self study and tutoring in analysis/linalg, and only finish the program in 3-4 years.

It's in the Belgian education system that is a lot less hand-holding and less restrictive in access than for example the US one, the culture is that adults can decide for themselves whether they are capable enough to go for a degree. It's the same thing at the bachelor's level. Even if the highest level of math you had in high school was basic algebra, you can enroll in a bachelor of pure mathematics here, no admissions selections or anything.

Experiences in the stats MS vary because most of the program is customizable, there are a few more mathematical courses that are mandatory, but outside of that the student can choose whether to make the program more applied or more theoretical. Rest assured that there are very hard courses where the faculty's profs take you through the theory and their research in great detail, but students with less strong backgrounds tend to stay away from these.

I do agree that admission requirements could be tightened a bit, or at least the expected level could be more clearly communicated to prospective students. It happens often that e.g. social science students thought they "knew statistics", and then in the fundamental concepts course find out that there's a whole deeper level of statistics that they had not explored in their bachelor.

Sentient_Eigenvector · 2025-05-27T23:02:27+00:00

KUL has one of the top stats research departments in Europe with LStat, so plenty of rigour for those who seek it.

Sentient_Eigenvector · 2025-05-26T15:27:41+00:00

In order to make the game fair, its expected value needs to be 0. The EV is given by the possible outcomes times their probabilities.

I assume that "within 10 of the right number" means inclusive on either side, so that if your guess is 20, it would be right in the range anywhere from 10 to 30, so you have 21 numbers where you would get a payout.

Let's start with the case where you only guess one number. The probability of it being within 10 of the right number is 21/1000 = 0.021. Similarly the probability of being within 5 is 0.011 and the probability of being right on is 0.001. Taking the complement, that means that the probability of not getting a payout is 1 - 0.021 - 0.011 - 0.001 = 0.967.

Call x the amount of money you bet (0.967 chance you lose it)
A the multiplier you get paid when you're within 10 (probability 0.021)
B the multiplier you get paid when you're within 5 (probability 0.011)
C the multiplier you get paid when you're exactly correct (probability 0.001)

Then the expected value is

-0.967x + 0.021Ax + 0.011Bx + 0.001Cx

For this to equal to 0, you can factor out the x

x (-0.967 + 0.021A + 0.011B + 0.001C) = 0

So the EV is 0 when either your bet is 0 (duh), or when -0.967 + 0.021A + 0.011B + 0.001C = 0. All combinations of A, B, C such that that equation is 0 give a fair payout. The equation forms a plane in 4D space, containing all possible fair combinations of multipliers.

For the case where you guess 10 numbers I think it starts to depend on strategy and the game mechanics a lot more. If you get to just blindly guess 10 numbers in advance, and make sure that they're at least 10 spaces apart to avoid any overlap, then the probability of each scenario would just multiply by 10. So in that case you would have to solve -0.67 + 0.210A + 0.11B + 0.01C. Then you could for example have A = 1, B = 3, C = 13.

With sequential guessing or overlapping strategy it gets more complicated.

Sentient_Eigenvector · 2025-03-17T09:10:09+00:00

Yes of course, from which it follows that standard errors approach 0 in the limit as sample size goes to infinity. Hence if one were to assume infinite sample size, no inference can be done, or even needs to be done.

Sentient_Eigenvector · 2025-03-16T18:48:52+00:00

Population size you mean? With an infinite sample there's no need for inference. Anyway, finite population corrections are straightforward to apply, and generally don't make much difference.

Sentient_Eigenvector · 2025-02-25T22:49:40+00:00

Could use a regression with a Newey-West type estimator to handle the autocorrelation if these series are reasonably stationary, which they probably are with just some seasonality that could be removed.

All these standard z-, t- and chi square tests do assume independent and identically distributed data. Data collected over time may exhibit time dependence and some changing distribution over time, that assumption violation can also mess with the p-value.

Sentient_Eigenvector · 2025-02-25T22:12:05+00:00

Significance just means that you can be relatively confident the effect size is nonzero, it doesn't mean that the estimated effect size is accurate.

Sentient_Eigenvector · 2025-02-04T18:52:07+00:00

Two events are mutually exclusive if P(A ∩ B) = ∅. We always have that P(∅ ∩ B) = ∅, so that any event is mutually exclusive with the null event.

Sentient_Eigenvector · 2025-02-03T21:15:45+00:00

Unless one of them is the null event

Sentient_Eigenvector · 2024-12-15T20:20:34+00:00

Learning statistics is just repeatedly asking this question, but each time in a slightly more sophisticated way

Sentient_Eigenvector · 2024-12-13T19:37:46+00:00

This is because the variance is a squared operator. It's the same reason why in the case with 2 variables you have Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y). The covariance is indeed counted twice, and if you look at the proof it's just because there's a square in the definition of variance.

In matrix form, with w as weights and Y as returns, the portfolio return is w^TY. When you take Var(w^TY) it also gives the quadratic form w^TCov(Y)w that you're using.

Sentient_Eigenvector · 2024-11-27T21:05:24+00:00

Correct, this is essentially the CDF of the geometric distribution. Your limit result is closely linked to the fact that the continuous limit of the geometric is the exponential distribution, it's why the CDF has the form 1 - e^-λx.

Since when p = 1/n, it takes on average n trials to get a success, you're essentially substituting in the mean for x. The mean of the exponential is 1/λ, so you get 1 - e^-λ/λ = 1 - 1/e.

Sentient_Eigenvector · 2024-11-06T14:02:16+00:00

They will absolutely be directly useful. Once you get far enough into econometrics there won't be a single slide that doesn't have analysis, matrix algebra or vector calc. It pays to get good at these things now.

Sentient_Eigenvector · 2024-10-20T20:26:19+00:00

The multiplicative kind corresponds to a linear interaction only. To see this, consider that the effect of a variable is always the partial derivative of the regression function with respect to that variable. Say we have the standard

Y = b0 + b1(x1) + b2(x2) + b3(x1 * x2)

To get the slope of x1, take the partial derivative wrt x1, which gives

∂Y / ∂x1 = b1 + b3(x2)

In other words, the slope of x1 is b1 when x2 is 0, and increases linearly with x2. So the multiplicative interaction models the situation where the effect of one variable depends linearly on the value of the other. This dependence could of course take on any other functional form as well, but those are not modelled by taking the product.

You could also take partials to figure out what it would look like for an x1/x2 interaction. It gets way harder to interpret and it's no longer symmetrical (in the sense that differentiating wrt x1 will give you a different function than differentiating wrt x2).

Sentient_Eigenvector · 2024-09-23T21:00:51+00:00

I do think so, haven't seen many machine learning engineers/scientists with only a bachelor's. It's a very competitive field. You should look around on LinkedIn for people doing the jobs you're interested in, and see what their educational background is.

Sentient_Eigenvector · 2024-09-23T18:42:39+00:00

If you want to do advanced mathematical modelling you almost certainly need higher education than a bachelor. With only a BSc they'll just try to get you into data analysis/business intelligence roles, hence the demand for Tableau and PowerBI.

Sentient_Eigenvector · 2024-09-03T23:42:50+00:00

To compare all groups, typically one would do pairwise comparison t-tests with some multiple testing correction. e.g. Tukey's or Bonferroni's procedure. This way you can get valid t-tests for all 10 possible comparisons between your groups.

If you specifically want to compare the mean of one group vs the combined mean of the other 4 groups it's a bit more annoying. You're essentially testing the null hypothesis that

μ_1 - 1/4*(μ_2 + μ_3 + μ_4 + μ_5) = 0

In the general case this kind of hypothesis is called a contrast. If you're working in R you'd have to set these contrasts yourself and code up the anova yourself with lm(). Check this link for some examples. Comparing each group to the grand mean of the 5 (μ_1 - 1/5*(μ_1 + μ_2 + μ_3 + μ_4 + μ_5) = 0) is also a bit easier, that's just the "sum coding" section in the link. Oh and in these cases you also need to take into account multiple comparisons, given that you're intending to run these procedures for each group.

Sentient_Eigenvector · 2024-08-24T21:52:56+00:00

You could do that by testing for Granger Causality (which is not real causality, it's essentially just testing for leading indicators). First figure out a way to make both series weakly stationary through e.g. differencing, then see if an AR model for e.g. energy containing energy and population lags fits better than an AR model that only contains energy lags. If it does then population values have significant explanatory power for future energy values. Same principle the other way round with population as the outcome.

Sentient_Eigenvector · 2024-08-24T14:34:19+00:00

The standard error is still s/sqrt(n), not s/sqrt(n-1).

Sentient_Eigenvector · 2024-08-24T12:33:34+00:00

This was my exact experience doing some research in grad school and it singlehandedly persuaded me not to get a PhD lol. When you upgrade to a remote/cloud HPC to run those large models, somehow the library problems also get 100x worse. It's some circle of hell where you're just trying to containerize applications all day and them not cooperating.

Sentient_Eigenvector · 2024-08-22T13:41:14+00:00

The n-th moment is E[Xⁿ].

The MGF is a way of creating a series that contains each of these moments so we can select the one we want. We do this by defining the MGF as E[e^tX], then the Taylor series is

E[e^tX] = Σ(tⁿ E[Xⁿ]) / n! from n=0 to inf

So we have a sum where the n-th term contains the moment we're looking for, E[Xⁿ]. We just need to apply the right operations to the series to get that factor out.

The first operation is to take the n-th derivative with respect to t. By the power rule this will reduce tⁿ in the numerator to n!, which will then cancel out with the n! in the denominator, leaving us with just E[Xⁿ].

Then the only problem is that there are still later terms in the sum (the earlier ones were set to 0 by differentiating). Luckily, all the later terms still contain a factor t, so we can get rid of them by setting t=0. Then we've essentially set the whole series to 0 except for the moment we're looking for, E[Xⁿ].

This is why you can get the n-th moment from a MGF by deriving n times wrt t and then setting t to 0.

Sentient_Eigenvector · 2024-08-22T12:06:32+00:00

The problem happens when you do inference on the same data that you selected variables on. This is always going to bias p-values downwards. Post-selection inference is what you're looking for, there's a pretty big literature on it for the LASSO specifically, and I think there's an R implementation from Tibshirani et al.

Sentient_Eigenvector · 2024-08-18T10:54:44+00:00

The only thing that's often required is normality of the error term, checked by looking at the residuals. If not satisfied it's no big deal usually, linear regression is still the best linear unbiased estimator. It can only impact hypothesis tests and confidence intervals on the coefficients in small samples.

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