Factorization Theorem for Sufficient Statistics
ฝัง
- เผยแพร่เมื่อ 11 พ.ย. 2024
- Step by step, I work through the definition of sufficient statistic (intuitive vs formula, showing them equal) then interpreting the claim made in the factorization theorem, and then line by line, I give a proof of the discrete case in both directions.
Thx, Doc. We in the streets turning up to this one. Using exponential families right now for most sufficient statistic work, but I am hoping to get this theorem down well.
Also, you looking sharp, homie. I like the haircut and fit. Fresh.
This is the most exact sufficiency video ever.
Thank you very much! When I struggle to learn an idea, I try to spare others the same grief.
Watching this 30 mins before my final exam. Thank u 😭😭😭
I'm studying to be an actuary, a big thanks to you for making these!
Your research behind the video, is crystally clear, making very easy to understand.
Kindly we want more statistical videos. 😄
This video was incredibly helpful! Thank you for posting this.
Hey, great stuff. Please keep up the good work. Much love!
thank you mr cruikshank
Great explanation! Thank you!
This video helps a lot, thanks
Glad to hear it!
Thank you !
brilliant
I love you.
Man you are confusing as hell
Sorry the video didn't work for you. It IS a confusing topic. Where did you get lost? The first couple of minutes explain the definition of sufficient, and I cover that separately in at least one other video.
What do you think would help? Some examples?
@@robertcruikshank4501 I will try to explain where I got confused at 1:09 you explain P(theta|x)=p(theta) is definiton of independence. I understand that.
After that you say 'that means' and write some equation after that. How does that mean the next equation( P(theta and X= p(theta).p(X))?
@@robertcruikshank4501 hay man u just dissapeared everything fine with wife and family?
@@Dupamine Sorry I missed this until just now. There are three equivalent definitions of independence. Remember the definition of conditional probability P(A|B) = P(A and B)/P(B), so saying P(A|B) = P(A) is the same thing as saying P(A) = P(A and B)/P(B) or P(A) P(B) = P(A and B).
@@robertcruikshank4501 thanks man