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March 24th, 2017, 08:17 AM   #21
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From: lyon

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 Originally Posted by romsek The problem is correct as written. I'm just a retired engineer who actually used some of this stuff from time to time.
Good, i would like to pursuit on phd but i need a strong skills in statistics and proba. If you have some advice is welcomed for me to improve my skills and to get your level at this end.

For the next, i have to show that "Betadistribution provides a conjugate prior for a Bernouilli trial".
Is it correct my demonstration:

P(x|θ) = θ^(x)* (1 − θ) ^(1−x)

P(θ|x) = P(x|θ)P(θ) /P(x)

P(θ|x) = 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(θ|x) ∝ 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(x|θ) = θ^(x)* (1 − θ) ^(1−x) P(θ|x) = P(x|θ)P(θ) /P(x) P(θ|x) = 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(θ|x) ∝ θ ^(α+x−1)* (1 − θ)^( β+(1−x)−1) = Beta(â, ê) Thanks a lot again
March 26th, 2017, 02:28 PM   #23
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Quote:
 Originally Posted by canfly For the next exercise, n

March 26th, 2017, 03:46 PM   #24
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From: lyon

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Quote:
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.

 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

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