How do you marginalize in graphical model elimination? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Method of Moments on a Uniform distribution (a,b)Multivariate normal density function of function of random variableMath Intuition and Natural Motivation Behind t-Student DistributionMultivariate Conditional Survival FunctionM. Ross problem 12 chapter 5 - Exponential distributionPDF/CDF of max-min type random variableDistribution of a random real with i.i.d. Bernoulli(p) binary digits?How to derive the marginal distribution based on a join distribution of X and Y?Distribution of $min(X_1+X_2+X_3,X_2+X_3+X_4,X_3+X_4+X_5,X_4+X_5+X_6)$how to find Marginal probability function for Piecewise joint probability density
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How do you marginalize in graphical model elimination?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Method of Moments on a Uniform distribution (a,b)Multivariate normal density function of function of random variableMath Intuition and Natural Motivation Behind t-Student DistributionMultivariate Conditional Survival FunctionM. Ross problem 12 chapter 5 - Exponential distributionPDF/CDF of max-min type random variableDistribution of a random real with i.i.d. Bernoulli(p) binary digits?How to derive the marginal distribution based on a join distribution of X and Y?Distribution of $min(X_1+X_2+X_3,X_2+X_3+X_4,X_3+X_4+X_5,X_4+X_5+X_6)$how to find Marginal probability function for Piecewise joint probability density
$begingroup$
I'm reading Michael I. Jordan's book on probabilistic graphical models, and I don't understand the elimination algorithm presented in chapter 3. To narrow the question down, consider page 6. In equation (3.10), we see that
$$m_5(x_2,x_3) = sum_x_5p(x_5|x_3)p(barx_6|x_2,x_5)$$
where the $x_i$ are random variables and $barx_i$ indicates a fixed/realized value of $x_i$.
Given that all $x_i$ are discrete random variables (as is the case in chapter 3), both $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ are represented by two-dimensional matrices. And since $m_5$ is a function of as-yet unrealized variables $x_2$ and $x_3$, it is also a two-dimensional matrix.
How then do we perform the multiplication and the summation above?
probability-distributions conditional-probability marginal-distribution
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
$endgroup$
add a comment |
$begingroup$
I'm reading Michael I. Jordan's book on probabilistic graphical models, and I don't understand the elimination algorithm presented in chapter 3. To narrow the question down, consider page 6. In equation (3.10), we see that
$$m_5(x_2,x_3) = sum_x_5p(x_5|x_3)p(barx_6|x_2,x_5)$$
where the $x_i$ are random variables and $barx_i$ indicates a fixed/realized value of $x_i$.
Given that all $x_i$ are discrete random variables (as is the case in chapter 3), both $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ are represented by two-dimensional matrices. And since $m_5$ is a function of as-yet unrealized variables $x_2$ and $x_3$, it is also a two-dimensional matrix.
How then do we perform the multiplication and the summation above?
probability-distributions conditional-probability marginal-distribution
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
$endgroup$
add a comment |
$begingroup$
I'm reading Michael I. Jordan's book on probabilistic graphical models, and I don't understand the elimination algorithm presented in chapter 3. To narrow the question down, consider page 6. In equation (3.10), we see that
$$m_5(x_2,x_3) = sum_x_5p(x_5|x_3)p(barx_6|x_2,x_5)$$
where the $x_i$ are random variables and $barx_i$ indicates a fixed/realized value of $x_i$.
Given that all $x_i$ are discrete random variables (as is the case in chapter 3), both $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ are represented by two-dimensional matrices. And since $m_5$ is a function of as-yet unrealized variables $x_2$ and $x_3$, it is also a two-dimensional matrix.
How then do we perform the multiplication and the summation above?
probability-distributions conditional-probability marginal-distribution
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
$endgroup$
I'm reading Michael I. Jordan's book on probabilistic graphical models, and I don't understand the elimination algorithm presented in chapter 3. To narrow the question down, consider page 6. In equation (3.10), we see that
$$m_5(x_2,x_3) = sum_x_5p(x_5|x_3)p(barx_6|x_2,x_5)$$
where the $x_i$ are random variables and $barx_i$ indicates a fixed/realized value of $x_i$.
Given that all $x_i$ are discrete random variables (as is the case in chapter 3), both $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ are represented by two-dimensional matrices. And since $m_5$ is a function of as-yet unrealized variables $x_2$ and $x_3$, it is also a two-dimensional matrix.
How then do we perform the multiplication and the summation above?
probability-distributions conditional-probability marginal-distribution
probability-distributions conditional-probability marginal-distribution
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
asked Mar 25 at 9:17
Philip RaeisghasemPhilip Raeisghasem
1013
1013
add a comment |
add a comment |
1 Answer
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$begingroup$
I'm realizing now that it's just regular matrix multiplication.
For the case above, let $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ be represented by $r!times!s$ and $s!times! t$ matrices, respectively, where $x_5$ can take on $s$ different values. Then, for each element $m_ij$ of $m_5$, we have that
$$m_ij=sum_k=1^s p_ik(x_5|x_3)p_kj(barx_6|x_2,x_5)$$
where $1leq i leq r$ and $1 leq j leq t$.
This is matrix multiplication. Also, intuitively, the dimensions will always work out, since each factor in the product will always be a function of the variable we're summing over.
I think the notation is what was tripping me up. I usually don't see matrix multiplication written explicitly as a sum of products.
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I'm realizing now that it's just regular matrix multiplication.
For the case above, let $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ be represented by $r!times!s$ and $s!times! t$ matrices, respectively, where $x_5$ can take on $s$ different values. Then, for each element $m_ij$ of $m_5$, we have that
$$m_ij=sum_k=1^s p_ik(x_5|x_3)p_kj(barx_6|x_2,x_5)$$
where $1leq i leq r$ and $1 leq j leq t$.
This is matrix multiplication. Also, intuitively, the dimensions will always work out, since each factor in the product will always be a function of the variable we're summing over.
I think the notation is what was tripping me up. I usually don't see matrix multiplication written explicitly as a sum of products.
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
$endgroup$
add a comment |
$begingroup$
I'm realizing now that it's just regular matrix multiplication.
For the case above, let $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ be represented by $r!times!s$ and $s!times! t$ matrices, respectively, where $x_5$ can take on $s$ different values. Then, for each element $m_ij$ of $m_5$, we have that
$$m_ij=sum_k=1^s p_ik(x_5|x_3)p_kj(barx_6|x_2,x_5)$$
where $1leq i leq r$ and $1 leq j leq t$.
This is matrix multiplication. Also, intuitively, the dimensions will always work out, since each factor in the product will always be a function of the variable we're summing over.
I think the notation is what was tripping me up. I usually don't see matrix multiplication written explicitly as a sum of products.
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
$endgroup$
add a comment |
$begingroup$
I'm realizing now that it's just regular matrix multiplication.
For the case above, let $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ be represented by $r!times!s$ and $s!times! t$ matrices, respectively, where $x_5$ can take on $s$ different values. Then, for each element $m_ij$ of $m_5$, we have that
$$m_ij=sum_k=1^s p_ik(x_5|x_3)p_kj(barx_6|x_2,x_5)$$
where $1leq i leq r$ and $1 leq j leq t$.
This is matrix multiplication. Also, intuitively, the dimensions will always work out, since each factor in the product will always be a function of the variable we're summing over.
I think the notation is what was tripping me up. I usually don't see matrix multiplication written explicitly as a sum of products.
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
$endgroup$
I'm realizing now that it's just regular matrix multiplication.
For the case above, let $p(x_5|x_3)$ and $p(barx_6|x_2,x_5)$ be represented by $r!times!s$ and $s!times! t$ matrices, respectively, where $x_5$ can take on $s$ different values. Then, for each element $m_ij$ of $m_5$, we have that
$$m_ij=sum_k=1^s p_ik(x_5|x_3)p_kj(barx_6|x_2,x_5)$$
where $1leq i leq r$ and $1 leq j leq t$.
This is matrix multiplication. Also, intuitively, the dimensions will always work out, since each factor in the product will always be a function of the variable we're summing over.
I think the notation is what was tripping me up. I usually don't see matrix multiplication written explicitly as a sum of products.
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
answered Mar 25 at 16:21
Philip RaeisghasemPhilip Raeisghasem
1013
1013
add a comment |
add a comment |
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