March 24th, 2017, 07:17 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, 12:29 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, 01:28 PM  #23 
Senior Member Joined: Sep 2015 From: USA Posts: 1,941 Thanks: 1008  
March 26th, 2017, 02:46 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, 08:05 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 

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