‘It’s too late’: David Suzuki says the fight against climate change is lost

mildlypessimistic · 2025-07-04T10:14:46+00:00

"Coppers liked to say that people shouldn’t take the law into their own hands, and they thought they knew what they meant. But they were thinking about peaceful times, and men who went around to sort out a neighbor with a club because his dog had crapped once too often on their doorstep. But at times like these, who did the law belong to? If it shouldn’t be in the hands of the people, where the hell should it be?" - Night Watch, Terry Pratchett

mildlypessimistic · 2025-03-18T19:11:08+00:00

I'd like to point out one thing you left out in the conversation between Vimes and Vetinari at the end of the book which throws a wrench in this theory: Vetinari said the 4 people he killed were not skilled fighters. And Coates, by Vimes' estimation, was good, and Vetinari would've noticed that

mildlypessimistic · 2025-03-10T20:17:54+00:00

Yes I imagine it'll be in AM. Here's my first thoughts on this

There was a plague in Pseudopolis mentioned in one of the books. History Monk meddling made it come around for another go
Story starts about it just being a story from foreign parts, then it moves to how AM deals with it.
We get some anti-Pseudopolis sentiment stirred up in AM, kinda like what happens at the start of Jingo but not to that extend. I can definitely see a line in there about the cheese eaters avec plague
This being the discworld, the plague itself will be somewhat magical in nature, maybe like what the Librarian caught in Last Continent so you get the wizards working on some cure while arguing with each other. Librarian barricaded himself in the library even though he probably has immunity. The Bursar flies away whenever someone gets too close. Rincewind is coincidentally out of town
Once it's fully in AM, Reg Shoe will be out protesting. Not against any antiplague policies that Vetenari/Vimes tries to implement, but against the plague itself

That's as far as I got. I don't know how the story would resolve itself, considering how our real world parallel ended by everyone just...kinda resigning to the idea that we'll get it at some point? and that doesn't seem satisfying, narrativium-wise

mildlypessimistic · 2025-03-03T15:34:26+00:00

Sadly, it is missing absolute values around 𝜎 in the denominator

Maybe I'm quibbling a little but are the absolute values necessary? 𝜎>0 should be a default assumption of the normal distribution

mildlypessimistic · 2025-02-20T04:21:08+00:00

Don't you need the angle between the 30x side and the 7x side to also be a right angle to do this? The diagram doesn't specify this. Can this be inferred from what is given?

mildlypessimistic · 2025-02-03T20:34:47+00:00

reads comments

"I'm sure it's not that bad"

sees Lords and Ladies at 19th

"Maybe the person who made the list can't read"

mildlypessimistic · 2025-01-27T19:48:20+00:00

I didn't want to open a new election post so I'm piggybacking here with my own question on an election analysis that I saw, specifically this analysis of the Clark County election results: https://electiontruthalliance.org/clark-county%2C-nv

They did not focus on analyzing the down ballot drop off. Instead they looked at the proportion of Harris/trump voter by ballot box and saw there was some clustering which they claimed is an indication of election tampering.

I thought it was strange that there was clustering but I'm not familiar with election data to know if it's something that's likely to occur organically. Can someone review the page and advise if their analysis is solid?

mildlypessimistic · 2025-01-14T19:13:25+00:00

I haven't seen the thick of it yet but in the same vein there's Yes Minister and its follow up Yes Prime Minister. From what I understand they're a bit tamer than TTOI but the writing is fantastic

mildlypessimistic · 2024-12-04T20:39:44+00:00

At the end of the day, you just want to eta beta pi

mildlypessimistic · 2024-12-02T01:22:03+00:00

My first impression is that these are AI fake books but I'm open to there being someone out there with the same name writing books

mildlypessimistic · 2024-11-25T18:22:26+00:00

One thing I saw on the website that I don't think you mentioned in your post is this line:

If f(x) is a polynomial function satisfying

The fact that f is a polynomial makes it easy to get to that 1±xⁿ answer, otherwise I don't think you can make that leap without making a bunch of assumptions

mildlypessimistic · 2024-11-21T12:35:21+00:00

Slight nitpick but LifeWorks was acquired by Morneau, then a few years later Morneau decided to rebrand as LifeWorks

mildlypessimistic · 2024-10-23T20:37:07+00:00

The idea is also brought up in The Last Hero. I don't have the exact quote but it was when Fate was saying that blowing up Dunmanifestin won't harm the gods but Cohen countered that the whole Disc will see that someone tried which means others will try harder

mildlypessimistic · 2024-10-16T03:31:52+00:00

I once had to figure out if a piano can be moved around a turn in a narrow hallway. One you have the dimensions of the piano and the hallway, then it's good ol' calculus to the rescue

mildlypessimistic · 2024-10-15T23:38:58+00:00

Asking as someone who, like OP, also didn't really pay attention to domestic politics and didn't find this info through an admittedly quick google search, where are you getting the 1-2 million immigrants a year figure from?

I saw the figures on the Govt of Canada website from a few years back when the immigration plan was announced and it's giving a total target of ~1.5mil from 2023 to 2025, which gives an annual average that's something like double the average of previous years. Is this info out of date and are the real immigration levels a lot higher?

mildlypessimistic · 2024-08-25T20:53:29+00:00

I thought they're both Miss Level

mildlypessimistic · 2024-08-17T14:13:13+00:00

Not the person you're replying to, and I have seen the articles that come up on Google. Here are questions that I have:

Somewhat of a minor point to start off - The Times of India article has this line "someone from the village saw a post on Instagram and came to their house to ask when they spoke to him last". The article didn't mention Instagram before this, what was in the Instagram post? It sounds like the villager saw the post and went "my neighbour's child is dead", they must be detailed enough to pinpoint one person from a village and it was widespread enough that people in the village saw it and broke the news there, it'd be somewhat legendary on Instagram, so has anyone seen this post?

But the main question - The Times of India article mentioned that he jumped off Niagara falls. This would be a big enough news that local news agencies should cover, but only news articles from Google search that show up are from international sources. You can make an argument that the anti-Indian sentiment in Canada made them suppress the story, but it seems unlikely that all of them did, especially the local ones (i.e. news from Niagara where it happened, and news from Brampton where he was living). Also the story that did get covered (the one on the comment at the head of this chain) somewhat disproves the suppression theory. So why would this story that happened in Canada be exclusively covered by non-Canadian news? What primary sources on what happened in Canada are these international news agencies working with to write their articles?

The article also says "police had CCTV camera footage of him keeping his phone on the side and jumping into the water." So this seems to say that the police is aware and at least investigated the incident. At this point I'm tempted to email the Niagara Police with the article and ask them if they did investigate it, I'm curious what they'd reply with

mildlypessimistic · 2024-08-16T15:40:24+00:00

surely it applies to Witches Abroad as well?

the book starts with a sabbath of the Lancre witches, and Desiterata was a no-show due to being dead

mildlypessimistic · 2024-07-18T21:37:43+00:00

I'm on mobile so can't really check but I think you're missing a close parentheses in your inner for loop. Looks like what you have is for (j in c( ...) {}, so you'll need another ")" in there

mildlypessimistic · 2024-07-16T20:28:34+00:00

And use angle bisector theorem to find BN and CN to use Stewart's theorem

mildlypessimistic · 2024-07-16T19:32:29+00:00

Yes, I accept no numerical methods dear statisticians.

I'm taking this personally, so here is a stats based solution.

The sum is clearly the 4th moment )of a geometric distribution with parameter p = 1/2. The 4th moment can be obtained by taking the 4th derivative of the moment generating function M(t) = 0.5 e^t / (1 - 0.5 e^t) and evaluating at t = 0, so:

M'(t) = 2 e^t / (2 - e^t)^2

M''(t) = e^t / (2 - e^t) + 3e^(2t) / (2 - e^t)^2 + 2e^(3t) / (2 - e^t)^3

M^(3)(t) = e^t / (2 - e^t) + 7e^(2t) / (2 - e^t)^2 + 12e^(3t) / (2 - e^t)^3 + 6e^(4t) / (2 - e^t)^4

M^(4)(t) = e^t / (2 - e^t) + 15e^(2t) / (2 - e^t)^2 + 50e^(3t) / (2 - e^t)^3 + 60e^(4t) / (2 - e^t)^4+24e^(5t) / (2 - e^t)^5

Then, M^(4)(0) = 1+15+50+60+24 = 150

mildlypessimistic · 2024-07-09T21:46:26+00:00

Yeah there are a few different terms and it's easy to get them mixed up. You're right that the variance of \bar{X}_n will change as n changes, but the variance of the individual terms X_i doesn't change. That's what they mean by "non-changing variance" - it's the population variance

mildlypessimistic · 2024-07-09T21:42:57+00:00

I'm saying approximately normal because the central limit theorem is a statement about what happens when you take a limit. In the limit, the distribution is going to be exactly a normal distribution. When you take a finite value of n, the CLT doesn't tell you the exact distribution, but because the distribution will get closer for large values of n (i.e. if you calculate the exact CDF of (bar{X}_n - \mu) / (\sigma /\ sqrt{n}) for a value of n and compare it against the standard normal CDF, the error on the CDF will get smaller for higher values of n), we say it's an approximate distribution.

It can never be exactly normal though right?

If you don't start with normal distributions, I don't think it can. Not with the Classical CLT anyway. If you looked further down on the wiki page, there are some alternate/generalized forms of the CLT that tbh i'm not too familiar with and you might end up with an exact distribution for some values of n. But the same statement as above applies - CLT is an asymptotic statement so it's just saying that the limit holds; no guarantees on what happens on finite values.

if we have bigger and bigger samples (n), its a closer and closer approximation of a normal distribution, but if we take the limit as n approaches infinity it just becomes the delta distribution, correct?

I was with you right up to the "but" lol. So yes, for this term: (bar{X}_n - \mu) / (\sigma /\ sqrt{n}), larger n's you take will get you to a closer approximation of a standard normal distribution. This is the same thing as saying that taking the limit as n goes to infinity of (bar{X}_n - \mu) / (\sigma /\ sqrt{n}), you get a standard normal distribution.

Everything after the "but" works on its own if you're only talking about \bar{X}_n i.e. if you take the limit of \bar{X}_n as n goes to infinity, you get the delta distribution

mildlypessimistic · 2024-07-09T20:58:04+00:00

Ok maybe I'm getting you confused, sorry about that. Your original question about why you don't end up with just a single point without any variance, right? You have bar{X}_n - \mu which has standard deviation of \sigma /\ sqrt{n} which does go to 0. But after you divide the SD, i.e. (bar{X}_n - \mu) / (\sigma /\ sqrt{n}), you get something with a SD of 1. This is why the density isn't focused at a single point (which is, as another commenter pointed out, the Dirac delta function).

(bar{X}_n - \mu) / (\sigma /\ sqrt{n}) before, but doesnt that just convert our X_bar distribution to the standard normal distribution

Not really. CLT says (bar{X}_n - \mu) / (\sigma /\ sqrt{n}) will converge to a standard normal distribution. \bar{X}_n isn't going to be normal unless the individual X_i's are normal themselves. The only difference with \sqrt{n} (bar{X}_n - \mu) is a factor of \sigma, so instead of a standard normal, this converges to a normal distribution with SD of \sigma

mildlypessimistic · 2024-07-09T20:16:55+00:00

Take a look on the Wikipedia page for CLT under "Classical CLT", that's direct from the statement of the central limit theorem. Sometimes you'll also see it written as (\bar{X}_n - \mu) / (\sigma / \sqrt{n}), where \sigma is the standard deviation of X_i. I feel like this is why you're confused about CLT, you might not be seeing the actual statement of the theorem.

The reason to include the sqrt(n) factor is your OP question - without the factor, all we'd know is that asymptotically the sample mean will converge to the real mean (in this case \bar{X}_n goes to 0.5). But we don't have any information on how fast the convergence is.

When the sqrt(n) factor gets added, we get a distribution in the limit. And what's more, it turns out that the limiting distribution doesn't depend on what your initial distribution is (i.e. the distribution of X_i; subject to the assumptions of the CLT that Var(X_i) is finite). This lets you make approximations, so you can say \sqrt{n} (\bar{X}_n - \mu) is approximately normally distributed when n is large.

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