March 24th, 2017, 08:17 AM  #21  
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0  Quote:
For the next, i have to show that "Betadistribution provides a conjugate prior for a Bernouilli trial". Is it correct my demonstration: P(xθ) = θ^(x)* (1 − θ) ^(1−x) P(θx) = P(xθ)P(θ) /P(x) P(θx) = θ ^(x)* (1 − θ)^( 1−x )*θ ^(α−1)* (1 − θ)^( β−1)/ ( P(x)B(α, β)) with B(α, β)= beta function. P(θx) ∝ θ ^(α+x−1)* (1 − θ)^( β+(1−x)−1) = Beta(â, ê) Thanks a lot again  
March 26th, 2017, 01:29 PM  #22 
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0 
For the next exercise, i have to show that "Betadistribution provides a conjugate prior for a Bernouilli trial". Is it correct my demonstration: P(xθ) = θ^(x)* (1 − θ) ^(1−x) P(θx) = P(xθ)P(θ) /P(x) P(θx) = θ ^(x)* (1 − θ)^( 1−x )*θ ^(α−1)* (1 − θ)^( β−1)/ ( P(x)B(α, β)) with B(α, β)= beta function. P(θx) ∝ θ ^(α+x−1)* (1 − θ)^( β+(1−x)−1) = Beta(â, ê) Thanks a lot again 
March 26th, 2017, 02:28 PM  #23 
Senior Member Joined: Sep 2015 From: USA Posts: 2,264 Thanks: 1198  
March 26th, 2017, 03:46 PM  #24 
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0  The probability of success π follows a beta distribution : π∼Beta(a,b). x=(x1,…,xn) in which the xi are IID. xi/π∼Bernoulli(π). The prior distribution for π has the form: fπ(π) ∝ π^(a−1)*(1−π)^(b−1) ,0<π<1. fx(x∣π)=π^(y)*(1−π)^(n−y) where y= sum (i=0 to n) {xi} (the likelihood) f(π/x)=fx(x/π)*fπ(π)/f(x) ∝ fx(x/π)fπ(π) since the denominator is independant of π. f(π∣x)∝ π^(a+y−1)*(1−π)^(b+n−y−1),0<π<1 witch is proportional to a beta distribution. This is of course proportional to a beta distribution with posterior hyperparameters â=a+y,b*=b+n−y,y counts the number of successes in n trials. To conclude,if the prior distribution had hyperparameters a and b, and I observe y successes in n trials, the posterior distribution is again beta, but with new hyperparameters. Thanks for your help. 
March 27th, 2017, 09:05 AM  #25 
Newbie Joined: Mar 2017 From: lyon Posts: 18 Thanks: 0 
The probability of success π follows a beta distribution : π∼Beta(a,b). x=(x1,…,xn) in which the xi are IID. xi/π∼Bernoulli(π). The prior distribution for π has the form: fπ(π) ∝ π^(a−1)*(1−π)^(b−1) ,0<π<1. fx(x∣π)=π^(y)*(1−π)^(n−y) where y= sum (i=0 to n) {xi} (the likelihood) f(π/x)=fx(x/π)*fπ(π)/f(x) ∝ fx(x/π)fπ(π) since the denominator is independant of π. f(π∣x)∝ π^(a+y−1)*(1−π)^(b+n−y−1),0<π<1 witch is proportional to a beta distribution. This is of course proportional to a beta distribution with posterior hyperparameters â=a+y,b*=b+n−y,y counts the number of successes in n trials. To conclude,if the prior distribution had hyperparameters a and b, and I observe y successes in n trials, the posterior distribution is again beta, but with new hyperparameters. Thanks for your help. canfly is online now Report Post 

Tags 
baysian, econometrics 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Financial econometrics  exam question  Ku5htr1m  Economics  0  March 11th, 2016 12:36 PM 
Applied Statistics and Econometrics  sarwar  Probability and Statistics  0  September 19th, 2014 10:27 AM 
Linear algebra & Econometrics  BartL  Math Books  0  October 26th, 2010 12:17 AM 
Econometrics  princessanna57  Economics  1  September 11th, 2009 07:29 PM 