Date: 2020-01-08 08:17 pm (UTC)
From: [personal profile] sassa_nf
Compare whether means of two distributions are the same - something like this: http://www.stat.yale.edu/Courses/1997-98/101/meancomp.htm There's a bunch of approaches, but this will do for the sake of the discussion.


"What exactly does it mean that T(t) is "non-linear"?"

Maybe I am confused. Let me try and untangle what I was thinking.

So, we have a weather model, M(t). This model is a function that is not linear - that is, M(x+y)=M(x)+M(y) ∧ M(k*x)=k*M(x) does not hold. In other words, there is no f(x)=a+b*x such that M(x)=f(x)+random noise. Put yet another way, we've spent so much effort modelling weather systems because they are far more complex than just a straight line.

Now we want to show that temperature has "a trend". The proposal is to estimate a and b such that the trend T(t)=a+b*t. Then we have some test procedure (which I understand).

My question is, aren't the two propositions at odds with each other? If we prove M(x) is the best fit, why do we bother with "a trend", T(t)? If we prove T(t) is the best fit, why do we bother with the model, M(x)?


Now: of course we can find a and b using least mean squares; this will estimate the slope that describes the temperature growth with the least error. But isn't this supposed to not be a good fit, ever, so it is meant to have a non-random noise added to it ("the least error" is not guaranteed to be distributed normally with mean 0)? We've got to reject "a trend" every time we accept the model to be non-linear!
This account has disabled anonymous posting.
If you don't have an account you can create one now.
HTML doesn't work in the subject.
More info about formatting

Profile

juan_gandhi: (Default)
Juan-Carlos Gandhi

May 2025

S M T W T F S
    1 2 3
456 7 8 9 10
11 121314151617
181920 21 222324
25262728293031

Most Popular Tags

Style Credit

Expand Cut Tags

No cut tags
Page generated May. 24th, 2025 09:26 am
Powered by Dreamwidth Studios