Factorization Theorem for Sufficient Statistics
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- Опубликовано: 21 авг 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.
I'm studying to be an actuary, a big thanks to you for making these!
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 😭😭😭
This video was incredibly helpful! Thank you for posting this.
Your research behind the video, is crystally clear, making very easy to understand.
Kindly we want more statistical videos. 😄
Hey, great stuff. Please keep up the good work. Much love!
Great explanation! Thank you!
thank you mr cruikshank
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