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March 24th, 2017, 07:17 AM   #21
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Originally Posted by romsek View Post
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
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March 26th, 2017, 12:29 PM   #22
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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
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March 26th, 2017, 01:28 PM   #23
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For the next exercise, n
make a new thread for this please
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March 26th, 2017, 02:46 PM   #24
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make a new thread for this please
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.
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March 27th, 2017, 08:05 AM   #25
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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.
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