20 - Beta conjugate prior to Binomial and Bernoulli likelihoods

This video sketches a short proof of the fact that a Beta distribution is conjugate to both Binomial and Bernoulli likelihoods.
If you are interested in seeing more of the material, arranged into a playlist, please visit: • Bayesian statistics: a... Unfortunately, Ox Educ is no more. Don't fret however as a whole load of new youtube videos are coming soon, along with a book that accompanies the series: www.amazon.co.uk/gp/product/1... Alternatively, for more information on this video series and Bayesian inference in general, visit: ben-lambert.com/bayesian-lect... For more information on econometrics and Bayesian statistics, see: ben-lambert.com/

[@mohammedismail308]

Today is Apr 12, 2019. I am following this playlist for about 5 hours so far. Thank you for bounding the knowledge and ordering the concepts. Your explanation is filling a lot of gaps.

[@jamesclark3459]

THANK YOU SO MUCH FOR AD-FREE AND EXCELLENT LECTURES!

[@anthonyarmour1812]

Sir, this is exactly the explanation I was looking for. Thank you very much.

[@caseyli5580]

Super clear in 5 minutes. Thank you!

[@gracelaryea2682]

Very well explained. Thanks for posting!

[@Tuupoification]

Thanks! The concepts and the derivation were clearly explained.

[@parametersofstatistics2145]

Can u please share a link where u explained posterior of binomial with prior poison ..please sir i need it desperartely

[@AMAN-dt9ry]

Thanks a lot

[@NehadHirmiz]

Excellent lecture :) Thank you very much :)

[@maximegrossman2146]

brilliant

[@user-et4om7zr4s]

简洁明了 赞👍

[@zainabesa4339]

Please, How can solve the following equation
Gamma(x)= 1+2x^2
Thank you

[@HimanshuKumar-vg3io]

Use Gamma(x+1) = x*Gamma(x)
You will get a cubic equation.
Substitute the three values so obtained in the primary equation, one of them will be the answer!

[@TheUgoren]

For pythonistas, see python implementation of various conjugate priors : github.com/urigoren/conjugate_prior

[@bnglr]

i think it needs a proof to before saying the denominator is B(a', b')

[@yonghongbai5023]

Yes. Do you know how to prove it? Or could you share a proof from other guy?