Gaussian mixture model - does an improper uniform prior give a proper posterior?

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We draw $n$ i.i.d. points $x_1 , x_2 , ..., x_n$ from the following Gaussian mixture:

$$p(x|mu_1,mu_2) = frac12 textN (x|mu_1,1) + frac12 textN (x|mu_2,1).$$

The prior is the improper prior $p(mu_1, mu_2) propto 1$. Does the posterior have a finite integral (i.e., is it a proper distribution)?

prior posterior gaussian-mixture conjugate-prior

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up vote
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We draw $n$ i.i.d. points $x_1 , x_2 , ..., x_n$ from the following Gaussian mixture:

$$p(x|mu_1,mu_2) = frac12 textN (x|mu_1,1) + frac12 textN (x|mu_2,1).$$

The prior is the improper prior $p(mu_1, mu_2) propto 1$. Does the posterior have a finite integral (i.e., is it a proper distribution)?

prior posterior gaussian-mixture conjugate-prior

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edited 2 hours ago

Ben

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asked 3 hours ago

aastha agrrawal

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aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote

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up vote
2
down vote

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We draw $n$ i.i.d. points $x_1 , x_2 , ..., x_n$ from the following Gaussian mixture:

$$p(x|mu_1,mu_2) = frac12 textN (x|mu_1,1) + frac12 textN (x|mu_2,1).$$

The prior is the improper prior $p(mu_1, mu_2) propto 1$. Does the posterior have a finite integral (i.e., is it a proper distribution)?

prior posterior gaussian-mixture conjugate-prior

share|cite|improve this question

edited 2 hours ago

Ben

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asked 3 hours ago

aastha agrrawal

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aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

We draw $n$ i.i.d. points $x_1 , x_2 , ..., x_n$ from the following Gaussian mixture:

$$p(x|mu_1,mu_2) = frac12 textN (x|mu_1,1) + frac12 textN (x|mu_2,1).$$

The prior is the improper prior $p(mu_1, mu_2) propto 1$. Does the posterior have a finite integral (i.e., is it a proper distribution)?

prior posterior gaussian-mixture conjugate-prior

prior posterior gaussian-mixture conjugate-prior

share|cite|improve this question

edited 2 hours ago

Ben

17.5k22286

asked 3 hours ago

aastha agrrawal

111

New contributor

aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 hours ago

Ben

17.5k22286

asked 3 hours ago

aastha agrrawal

111

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aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

share|cite|improve this question

edited 2 hours ago

Ben

17.5k22286

edited 2 hours ago

Ben

17.5k22286

edited 2 hours ago

Ben

17.5k22286

17.5k22286

asked 3 hours ago

aastha agrrawal

111

New contributor

aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

aastha agrrawal

111

asked 3 hours ago

aastha agrrawal

111

111

New contributor

aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

aastha agrrawal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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From your specified model you have likelihood function:

$$L_mathbfx(mu_1,mu_2) = (8 pi)^-n/2 prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg].$$

To facilitate the analysis, let $mathscrB equiv 1,2 ^n$ be the set of all $n$-length binary vectors of values of the two means. For any vector $mathbfb in mathscrB$ let $mathcalS(mathbfb) equiv i=1,...,n $ be the set of index values that use the first mean. With the improper uniform prior $pi (mu_1, mu_2) propto 1$ you get the posterior:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto L_mathbfx(mu_1,mu_2) \[6pt]
&propto prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg] \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i=1^n (x_i-mu_b_i)^2 Big) \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big) exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big). \[6pt]
endaligned endequation$$

Now, letting $barx_1^(k) (mathbfb) equiv sum_i in mathcalS(mathbfb) x_i^k$ and $barx_2^(k) (mathbfb) equiv sum_i notin mathcalS(mathbfb) x_i^k$ we have:

$$beginequation beginaligned
exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big)
&= exp Big( -frac12 Big( mu_1^2 + 2 mu_1 sum_i in mathcalS(mathbfb) x_i + sum_i in mathcalS(mathbfb) x_i^2 Big) Big) \[6pt]
&= exp Big( -frac12 ( mu_1^2 + 2 mu_1 barx_1^(1) (mathbfb) + barx_1^(2) (mathbfb) ) Big) \[10pt]
&= exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big)
exp Big( -frac12 (mu_1 - barx_1^(1) (mathbfb))^2 Big) \[10pt]
&propto exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1), \[10pt]
exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big) &propto exp Big( -frac12 ( barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2 ) Big) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[10pt]
endaligned endequation$$

So, letting $H(mathbfb) equiv barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 + barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2$ we have:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[6pt]
endaligned endequation$$

This kernel has a finite integral, and yields the proper posterior density:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&= fracsum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big). \[6pt]
endaligned endequation$$

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

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active

oldest

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up vote
2
down vote

From your specified model you have likelihood function:

$$L_mathbfx(mu_1,mu_2) = (8 pi)^-n/2 prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg].$$

To facilitate the analysis, let $mathscrB equiv 1,2 ^n$ be the set of all $n$-length binary vectors of values of the two means. For any vector $mathbfb in mathscrB$ let $mathcalS(mathbfb) equiv i=1,...,n $ be the set of index values that use the first mean. With the improper uniform prior $pi (mu_1, mu_2) propto 1$ you get the posterior:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto L_mathbfx(mu_1,mu_2) \[6pt]
&propto prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg] \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i=1^n (x_i-mu_b_i)^2 Big) \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big) exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big). \[6pt]
endaligned endequation$$

Now, letting $barx_1^(k) (mathbfb) equiv sum_i in mathcalS(mathbfb) x_i^k$ and $barx_2^(k) (mathbfb) equiv sum_i notin mathcalS(mathbfb) x_i^k$ we have:

$$beginequation beginaligned
exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big)
&= exp Big( -frac12 Big( mu_1^2 + 2 mu_1 sum_i in mathcalS(mathbfb) x_i + sum_i in mathcalS(mathbfb) x_i^2 Big) Big) \[6pt]
&= exp Big( -frac12 ( mu_1^2 + 2 mu_1 barx_1^(1) (mathbfb) + barx_1^(2) (mathbfb) ) Big) \[10pt]
&= exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big)
exp Big( -frac12 (mu_1 - barx_1^(1) (mathbfb))^2 Big) \[10pt]
&propto exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1), \[10pt]
exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big) &propto exp Big( -frac12 ( barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2 ) Big) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[10pt]
endaligned endequation$$

So, letting $H(mathbfb) equiv barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 + barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2$ we have:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[6pt]
endaligned endequation$$

This kernel has a finite integral, and yields the proper posterior density:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&= fracsum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big). \[6pt]
endaligned endequation$$

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answered 2 hours ago

Ben

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up vote
2
down vote

From your specified model you have likelihood function:

$$L_mathbfx(mu_1,mu_2) = (8 pi)^-n/2 prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg].$$

To facilitate the analysis, let $mathscrB equiv 1,2 ^n$ be the set of all $n$-length binary vectors of values of the two means. For any vector $mathbfb in mathscrB$ let $mathcalS(mathbfb) equiv i=1,...,n $ be the set of index values that use the first mean. With the improper uniform prior $pi (mu_1, mu_2) propto 1$ you get the posterior:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto L_mathbfx(mu_1,mu_2) \[6pt]
&propto prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg] \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i=1^n (x_i-mu_b_i)^2 Big) \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big) exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big). \[6pt]
endaligned endequation$$

Now, letting $barx_1^(k) (mathbfb) equiv sum_i in mathcalS(mathbfb) x_i^k$ and $barx_2^(k) (mathbfb) equiv sum_i notin mathcalS(mathbfb) x_i^k$ we have:

$$beginequation beginaligned
exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big)
&= exp Big( -frac12 Big( mu_1^2 + 2 mu_1 sum_i in mathcalS(mathbfb) x_i + sum_i in mathcalS(mathbfb) x_i^2 Big) Big) \[6pt]
&= exp Big( -frac12 ( mu_1^2 + 2 mu_1 barx_1^(1) (mathbfb) + barx_1^(2) (mathbfb) ) Big) \[10pt]
&= exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big)
exp Big( -frac12 (mu_1 - barx_1^(1) (mathbfb))^2 Big) \[10pt]
&propto exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1), \[10pt]
exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big) &propto exp Big( -frac12 ( barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2 ) Big) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[10pt]
endaligned endequation$$

So, letting $H(mathbfb) equiv barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 + barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2$ we have:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[6pt]
endaligned endequation$$

This kernel has a finite integral, and yields the proper posterior density:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&= fracsum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big). \[6pt]
endaligned endequation$$

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answered 2 hours ago

Ben

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up vote
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down vote

up vote
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From your specified model you have likelihood function:

$$L_mathbfx(mu_1,mu_2) = (8 pi)^-n/2 prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg].$$

To facilitate the analysis, let $mathscrB equiv 1,2 ^n$ be the set of all $n$-length binary vectors of values of the two means. For any vector $mathbfb in mathscrB$ let $mathcalS(mathbfb) equiv i=1,...,n $ be the set of index values that use the first mean. With the improper uniform prior $pi (mu_1, mu_2) propto 1$ you get the posterior:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto L_mathbfx(mu_1,mu_2) \[6pt]
&propto prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg] \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i=1^n (x_i-mu_b_i)^2 Big) \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big) exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big). \[6pt]
endaligned endequation$$

Now, letting $barx_1^(k) (mathbfb) equiv sum_i in mathcalS(mathbfb) x_i^k$ and $barx_2^(k) (mathbfb) equiv sum_i notin mathcalS(mathbfb) x_i^k$ we have:

$$beginequation beginaligned
exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big)
&= exp Big( -frac12 Big( mu_1^2 + 2 mu_1 sum_i in mathcalS(mathbfb) x_i + sum_i in mathcalS(mathbfb) x_i^2 Big) Big) \[6pt]
&= exp Big( -frac12 ( mu_1^2 + 2 mu_1 barx_1^(1) (mathbfb) + barx_1^(2) (mathbfb) ) Big) \[10pt]
&= exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big)
exp Big( -frac12 (mu_1 - barx_1^(1) (mathbfb))^2 Big) \[10pt]
&propto exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1), \[10pt]
exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big) &propto exp Big( -frac12 ( barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2 ) Big) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[10pt]
endaligned endequation$$

So, letting $H(mathbfb) equiv barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 + barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2$ we have:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[6pt]
endaligned endequation$$

This kernel has a finite integral, and yields the proper posterior density:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&= fracsum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big). \[6pt]
endaligned endequation$$

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answered 2 hours ago

Ben

17.5k22286

From your specified model you have likelihood function:

$$L_mathbfx(mu_1,mu_2) = (8 pi)^-n/2 prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg].$$

To facilitate the analysis, let $mathscrB equiv 1,2 ^n$ be the set of all $n$-length binary vectors of values of the two means. For any vector $mathbfb in mathscrB$ let $mathcalS(mathbfb) equiv i=1,...,n $ be the set of index values that use the first mean. With the improper uniform prior $pi (mu_1, mu_2) propto 1$ you get the posterior:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto L_mathbfx(mu_1,mu_2) \[6pt]
&propto prod_i=1^n Bigg[ exp Big( -frac12 (x_i-mu_1)^2 Big) + exp Big( -frac12 (x_i-mu_2)^2 Big) Bigg] \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i=1^n (x_i-mu_b_i)^2 Big) \[6pt]
&= sum_mathbfb in mathscrB exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big) exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big). \[6pt]
endaligned endequation$$

Now, letting $barx_1^(k) (mathbfb) equiv sum_i in mathcalS(mathbfb) x_i^k$ and $barx_2^(k) (mathbfb) equiv sum_i notin mathcalS(mathbfb) x_i^k$ we have:

$$beginequation beginaligned
exp Big( -frac12 sum_i in mathcalS(mathbfb) (x_i-mu_1)^2 Big)
&= exp Big( -frac12 Big( mu_1^2 + 2 mu_1 sum_i in mathcalS(mathbfb) x_i + sum_i in mathcalS(mathbfb) x_i^2 Big) Big) \[6pt]
&= exp Big( -frac12 ( mu_1^2 + 2 mu_1 barx_1^(1) (mathbfb) + barx_1^(2) (mathbfb) ) Big) \[10pt]
&= exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big)
exp Big( -frac12 (mu_1 - barx_1^(1) (mathbfb))^2 Big) \[10pt]
&propto exp Big( -frac12 ( barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 ) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1), \[10pt]
exp Big( -frac12 sum_i notin mathcalS(mathbfb) (x_i-mu_2)^2 Big) &propto exp Big( -frac12 ( barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2 ) Big) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[10pt]
endaligned endequation$$

So, letting $H(mathbfb) equiv barx_1^(2) (mathbfb) - barx_1^(1) (mathbfb)^2 + barx_2^(2) (mathbfb) - barx_2^(1) (mathbfb)^2$ we have:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&propto sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 | barx_1^(1) (mathbfb), 1) cdot textN(mu_2 | barx_2^(1) (mathbfb), 1). \[6pt]
endaligned endequation$$

This kernel has a finite integral, and yields the proper posterior density:

$$beginequation beginaligned
pi (mu_1, mu_2 | mathbfx)
&= fracsum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big) cdot textN(mu_1 sum_mathbfb in mathscrB exp Big( -frac12 H(mathbfb) Big). \[6pt]
endaligned endequation$$

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answered 2 hours ago

Ben

17.5k22286

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share|cite|improve this answer

answered 2 hours ago

Ben

17.5k22286

answered 2 hours ago

Ben

17.5k22286

answered 2 hours ago

Ben

17.5k22286

17.5k22286

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aastha agrrawal is a new contributor. Be nice, and check out our Code of Conduct.

 

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aastha agrrawal is a new contributor. Be nice, and check out our Code of Conduct.

aastha agrrawal is a new contributor. Be nice, and check out our Code of Conduct.

aastha agrrawal is a new contributor. Be nice, and check out our Code of Conduct.

 

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