Hi, really great video series, much appreciated! One question - when you divided the expression inside the exponential by alpha, why did (alpha - 1)log x and log(gamma(alpha)) also not get divided by alpha?
hey dan, i think its bc she divided the whole thing by 1/alpha after dividing the terms by alpha. this means that the term is now equivalent to what it was at the beginning (bc the alphas cancel out)
I have a question ! my professor expresses α (φ) as τ ^ 2 ..... could you ever get τ ^ 2 equal to the -1/α that you come out as a result? due to the fact that α is positive?
Your professor is right. Although she gets the answer, that quantity specifies most of the exponential family their dispersions, hence we all know dispersions are greater than zero (i.e. positive). She just needed to put the negative value at the theta's position and find the reciprocal of the beta in the log function to make it -b(theta) (or not find the reciprocal but put double negatives which is positive). And the answer will come out beautifully.
To begin with, it's not a member of the natural exponential family, so this simplified approach will not work. Begin by finding T(x) - the minimal sufficient statistic for the Weibull; you will need to assume that the shape parameter is known [when it is not, the Weibull is not a member of the exponential family]. Then rearrange f(x) to be f(x) = h(x)exp{n(theta)T(x)-A(theta)} if you allow T(x)=x^k [k is the shape parameter] and allow h(x) = x^(k-1) everything else follows quite easily. The trickier part of this process is showing that x^k is a sufficient statistic for the Weibull distribution where k is known (so you may want to watch the series on finding minimal sufficient statistics]. Hope this helps.
I have tried it but ended up proving that summation of x^k is the minimal sufficient statistic not x^k. Again if T(x) = summation x^k than the pdf f(x) can't be written in the form you said. Am I going in a wrong way ?
This was so well explained it made me wonder why I was ever confused in the first place. Thank you!
thank you. im still wondering where you got phi at 4:14. it felt arbitrary.
Hi, really great video series, much appreciated!
One question - when you divided the expression inside the exponential by alpha, why did (alpha - 1)log x and log(gamma(alpha)) also not get divided by alpha?
hey dan, i think its bc she divided the whole thing by 1/alpha after dividing the terms by alpha. this means that the term is now equivalent to what it was at the beginning (bc the alphas cancel out)
I was not able to understand how u got the value of phi at 4:14
Thank you for the very bright demonstration!
How did you know that phi=1 and a(phi)=-1/alpha? Couldn't phi have been just alpha and a(phi) -1/alpha?
very goood👏
thankyou very much mchinkuuu
really very nice .
and well explained..
thank you..
Thank-you so much!
I have a question ! my professor expresses α (φ) as τ ^ 2 ..... could you ever get τ ^ 2 equal to the -1/α that you come out as a result? due to the fact that α is positive?
Your professor is right. Although she gets the answer, that quantity specifies most of the exponential family their dispersions, hence we all know dispersions are greater than zero (i.e. positive). She just needed to put the negative value at the theta's position and find the reciprocal of the beta in the log function to make it -b(theta) (or not find the reciprocal but put double negatives which is positive). And the answer will come out beautifully.
Actually so helpful
how does phi=1/alpha?
can you show how does weibull distribution belong of exponential family of distribution ?
To begin with, it's not a member of the natural exponential family, so this simplified approach will not work. Begin by finding T(x) - the minimal sufficient statistic for the Weibull; you will need to assume that the shape parameter is known [when it is not, the Weibull is not a member of the exponential family].
Then rearrange f(x) to be f(x) = h(x)exp{n(theta)T(x)-A(theta)} if you allow T(x)=x^k [k is the shape parameter] and allow h(x) = x^(k-1) everything else follows quite easily.
The trickier part of this process is showing that x^k is a sufficient statistic for the Weibull distribution where k is known (so you may want to watch the series on finding minimal sufficient statistics].
Hope this helps.
Thank you so much........
I have tried it but ended up proving that summation of x^k is the minimal sufficient statistic not x^k. Again if T(x) = summation x^k than the pdf f(x) can't be written in the form you said. Am I going in a wrong way ?
ty
Thanx a looot
tkm