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I got a B in probability because I didn't write a proof of the central limit theorem on the allowed cheat sheet for the final exam. So of course it's the first

I got a B in probability because I didn't write a proof of the central limit theorem on the allowed cheat sheet for the final exam. So of course it's the first thing I looked for on this one. It's not there.

creddit 7y ago
Aren't most proofs of the CLT pretty extensive?
- AstralStorm 7y ago
  There are a few short ones going from area integral comparisons.
  - crdrost 7y ago
    The easiest is to look at characteristic functions and cumulants; for a random variable T with PDF p(t) we say T ~ p and define φ_T[f] = ∫ dt p(t) exp(-2πift) = ⟨ exp(-2πifT) ⟩ If two variables X ~ r and Y ~ s are independent then you can prove [from ⟨f(X) g(Y)⟩ = ⟨f(X)⟩ ⟨g(Y)⟩ or X+Y ~ q where q(z) = ∫ dx r(x) s(z — x)] that their sum has a characteristic function φ_{X+Y} = φ_X + φ_Y And therefore the “sample mean” M of n IID variables is itself a random variable with characteristic function φ_M[f] = ( φ[f/n] )^n. So we find that log φ_M[f] = n log φ[f/n] ≈ 0 + i a f – b f²/n + O(f³/n²). These terms [a, b] from expanding the log of the characteristic function constitute the cumulant expansion and for large n the other terms shrink to zero, so that the characteristic function is to first order in 1/n a Gaussian. The characteristic function was a Fourier transform of a PDF, so an inverse Fourier transform gets it back: p(t) = ∫ df φ_T[f] exp(2πift) But the Fourier transform of a Gaussian is just a Gaussian.
    - ngcc_hk 7y ago
      Surprise you can use those character like “∫ φ π”. Thought you need to say pig etc.
lucb1e 7y ago
A B is like 80% or so, judging from A being the best iirc? I'm not familiar with alphanumeric grading systems.
- barrkel 7y ago
  Things can get a bit customized at either end of the spectrum, but generally each letter covers 15%, with A..D covering the span 100% to 40%. So a B would be 70 to 85.
  - nutjob2 7y ago
    Shouldn't that sort of grading be on a bell curve? That would seem to be the fairest.
    - kaitai 7y ago
      Many smaller (<60 students) math classes, in my experience, have bimodal distributions of scores, so a bell curve by definition simply doesn't make sense. In addition, there's been a move to standards-based or mastery-based grading, that is, grading according to what you know of the material rather than how you compare to your neighbor. This allows comparisons over time and consistency with regard to subsequent classes -- if you have a C as a prerequisite for the next class, then a C should indicate the same mastery of material rather than the same relative position in the class.
    - dboreham 7y ago
      Based on my memory of what the teacher said in my statistics class in school, exam results in Scotland are normalized as z-scores vs that year's population. At least they were a zillion years ago.
    - daveFNbuck 7y ago
      Let's say that most math majors take this course in the first semester of their sophomore year. Then the bell curve grading gives you a much lower grade for identical work if you take the class in the first semester vs second semester. You also get a lower grade for going to a better school. You even get a lower grade for helping your classmates. This makes the grades unfair and not useful for judging mastery of the subject. It only makes sense if the goal of a course is to beat the other students. For most courses, the goal should be learning.
Mithriil 7y ago
Question asking for the central theorem proof in a probability class happened to me too. Not answering this one gave me an A, though.

creddit 7y ago

Aren't most proofs of the CLT pretty extensive?

AstralStorm 7y ago
There are a few short ones going from area integral comparisons.
- crdrost 7y ago
  The easiest is to look at characteristic functions and cumulants; for a random variable T with PDF p(t) we say T ~ p and define φ_T[f] = ∫ dt p(t) exp(-2πift) = ⟨ exp(-2πifT) ⟩ If two variables X ~ r and Y ~ s are independent then you can prove [from ⟨f(X) g(Y)⟩ = ⟨f(X)⟩ ⟨g(Y)⟩ or X+Y ~ q where q(z) = ∫ dx r(x) s(z — x)] that their sum has a characteristic function φ_{X+Y} = φ_X + φ_Y And therefore the “sample mean” M of n IID variables is itself a random variable with characteristic function φ_M[f] = ( φ[f/n] )^n. So we find that log φ_M[f] = n log φ[f/n] ≈ 0 + i a f – b f²/n + O(f³/n²). These terms [a, b] from expanding the log of the characteristic function constitute the cumulant expansion and for large n the other terms shrink to zero, so that the characteristic function is to first order in 1/n a Gaussian. The characteristic function was a Fourier transform of a PDF, so an inverse Fourier transform gets it back: p(t) = ∫ df φ_T[f] exp(2πift) But the Fourier transform of a Gaussian is just a Gaussian.
  - ngcc_hk 7y ago
    Surprise you can use those character like “∫ φ π”. Thought you need to say pig etc.

lucb1e 7y ago

A B is like 80% or so, judging from A being the best iirc? I'm not familiar with alphanumeric grading systems.

barrkel 7y ago
Things can get a bit customized at either end of the spectrum, but generally each letter covers 15%, with A..D covering the span 100% to 40%. So a B would be 70 to 85.
- nutjob2 7y ago
  Shouldn't that sort of grading be on a bell curve? That would seem to be the fairest.
  - kaitai 7y ago
    Many smaller (<60 students) math classes, in my experience, have bimodal distributions of scores, so a bell curve by definition simply doesn't make sense. In addition, there's been a move to standards-based or mastery-based grading, that is, grading according to what you know of the material rather than how you compare to your neighbor. This allows comparisons over time and consistency with regard to subsequent classes -- if you have a C as a prerequisite for the next class, then a C should indicate the same mastery of material rather than the same relative position in the class.
  - dboreham 7y ago
    Based on my memory of what the teacher said in my statistics class in school, exam results in Scotland are normalized as z-scores vs that year's population. At least they were a zillion years ago.
  - daveFNbuck 7y ago
    Let's say that most math majors take this course in the first semester of their sophomore year. Then the bell curve grading gives you a much lower grade for identical work if you take the class in the first semester vs second semester. You also get a lower grade for going to a better school. You even get a lower grade for helping your classmates. This makes the grades unfair and not useful for judging mastery of the subject. It only makes sense if the goal of a course is to beat the other students. For most courses, the goal should be learning.

Mithriil 7y ago

Question asking for the central theorem proof in a probability class happened to me too. Not answering this one gave me an A, though.