3 ms·

Sure, but you can associate events with indicator rvs: X = 1[Heads], Y = 1[Tails]

mturmon 2y ago
Definitely. Taking this to its logical end leads to a beautiful concise notation that I learned from David Pollard: - You literally identify sets with their indicators: they are the same. - You identify the “P” operator as expectation (integration) with respect to the underlying measure, and next… - You note that integration is linear, so you use linear operator notation everywhere you’d use P. So if “A” is a set, you just write P A = 0.5 This is equivalent to: P A = P 1[ω ∈ A] = ∫ 1[ω ∈ A] dP(ω) in Lebesgue notation. There’s an example on page 2 of http://www.stat.yale.edu/~pollard/Courses/241.fall2014/notes2014/Probability.pdf http://www.stat.yale.edu/~pollard/Courses/241.fall2014/notes... although he’s not using measure theory there. It can be really clean and terse especially when doing bounds for random variables.
- dfeng 2y ago
  Wow, I did not expect to see David's notation here on HN. The only problem with the notation is that it becomes so second nature that you forget it's not standard!
knappa 2y ago
A lot is lost here by using notation that looks like it is rigorous math, but is actually pretty vague. For example, are X and Y indicators for the same flip? If so, they are mutually exclusive, X=Y is contradictory, and hence P(X=Y)=0. If they are samples from different flips (and your coin is the usual idealized one) then X and Y are independent random variables and P(X=Y)=0.25. It's just like if X~N(0,1), Y~N(0,1) and you want to know the distribution of X-Y. You need to know what the PDF of (X,Y) looks like. Well, you don't know. X and Y could be correlated or they might not be. e.g. if could be that (X,Y)~N( (0,0), [(1,0),(0,1)] ) or maybe (X,Y)~N( (0,0), [(1,1/2),(1/2,1)] ). The distribution of X-Y cares how correlated X and Y are.
- setopt 2y ago
  > If they are samples from different flips (and your coin is the usual idealized one) then X and Y are independent random variables and P(X=Y)=0.25. Isn’t it 0.5?

Definitely. Taking this to its logical end leads to a beautiful concise notation that I learned from David Pollard: - You literally identify sets with their indicators: they are the same. - You identify the “P” operator as expectation (integration) with respect to the underlying measure, and next… - You note that integration is linear, so you use linear operator notation everywhere you’d use P. So if “A” is a set, you just write P A = 0.5 This is equivalent to: P A = P 1[ω ∈ A] = ∫ 1[ω ∈ A] dP(ω) in Lebesgue notation. There’s an example on page 2 of http://www.stat.yale.edu/~pollard/Courses/241.fall2014/notes2014/Probability.pdf http://www.stat.yale.edu/~pollard/Courses/241.fall2014/notes... although he’s not using measure theory there. It can be really clean and terse especially when doing bounds for random variables.

dfeng 2y ago
Wow, I did not expect to see David's notation here on HN. The only problem with the notation is that it becomes so second nature that you forget it's not standard!

knappa 2y ago

A lot is lost here by using notation that looks like it is rigorous math, but is actually pretty vague. For example, are X and Y indicators for the same flip? If so, they are mutually exclusive, X=Y is contradictory, and hence P(X=Y)=0. If they are samples from different flips (and your coin is the usual idealized one) then X and Y are independent random variables and P(X=Y)=0.25. It's just like if X~N(0,1), Y~N(0,1) and you want to know the distribution of X-Y. You need to know what the PDF of (X,Y) looks like. Well, you don't know. X and Y could be correlated or they might not be. e.g. if could be that (X,Y)~N( (0,0), [(1,0),(0,1)] ) or maybe (X,Y)~N( (0,0), [(1,1/2),(1/2,1)] ). The distribution of X-Y cares how correlated X and Y are.

setopt 2y ago
> If they are samples from different flips (and your coin is the usual idealized one) then X and Y are independent random variables and P(X=Y)=0.25. Isn’t it 0.5?